Reteaching Masters. To jump to a location in this book. 1. Click a bookmark on the left. To print a part of the book. 1. Click the Print button.

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1 Reteaching Masters To jump to a location in this book 1. lick a bookmark on the left. To print a part of the book 1. lick the rint button. 2. When the rint window opens, tpe in a range of pages to print. The page numbers are displaed in the bar at the bottom of the document. In the eample below, 1 of 151 means that the current page is page 1 in a file of 151 pages.

2 NME LSS DTE Reteaching 1.1 The uilding locks of Geometr Skill Identifing and naming points, lines, and planes The undefined terms point, line, and plane represent geometric figures. Such figures do not eist in the real world, but ma be represented b real-world objects or b drawings. point has no size. It ma be represented b a dot and named with a single capital letter. line etends forever in both directions, but has no thickness. line is named using either the names of two points on the line or a single lower-case letter. oints that lie on the same line are called collinear. n two points are collinear. plane is like a flat surface that etends forever in all directions but has no thickness. plane is named using either the names of three points that lie in the plane and are not collinear, or b a single script capital letter. oints that lie in the same plane are coplanar. n three points are coplanar. Eample N R Name each figure. a. b. c. J K M a. point b. JK, KJ, or line m c. plane MN or plane Name the indicated figures in the drawing at the right. 1. two lines 2. two planes 3. three noncollinear points 4. four noncoplanar points Skill Identifing and naming segments, ras, and angles segment is a part of a line that consists of two points, called the endpoints, and all the points between them. The segment with endpoints X and Y is denoted XY. ra is a part of a line that has one endpoint and etends without end in one direction. The ra that has endpoint and contains point is denoted. n angle is a figure formed b two ras that do not lie on the same line, but have the same endpoint. The common endpoint is the verte of the angle. The angle in the figure at the right ma be referred to as QR, RQ, 1, or Q. If two angles have the same verte, the must be named using numbers or three letters. Q 1 Q K M R J Geometr Reteaching 1.1 1

3 NME LSS DTE Eample Name each figure. a. b. c. M Z H E J R a. MZ or ZM b. EH c. JR, RJ, or Refer to the figure at the right. Q 5. Name all the segments. 6. Name four ras with endpoint T. S 2 1 T R 7. Give two other names for Name the ras that form the sides of 2. Skill lassifing and identifing intersections of geometric figures The postulates that follow are fundamental geometric ideas. The intersection of two lines is a point. The intersection of two planes is a line. Through an two points, there is eactl one line. Through an three noncollinear points, there is eactl one plane. If two points are in a plane, then the line containing them is in the plane. Eample Name each of the following. a. the intersection of E and line k b. the intersection of planes and DF c. the line containing points and D d. all the planes shown that contain E a. point b. c. D DF (line k) d. plane E and plane EF Refer to the figure at the right. Name each of the following. 9. the intersection of and G 10. a plane containing points D and G 11. the intersection of planes and M F E F D k G H D 12. the line containing points F and E 2 Reteaching 1.1 Geometr

4 NSWERS Reteaching hapter 1 Lesson N and JN 2. Sample: KN and NJ 3. Sample:, N, Q 4. Sample: K, N,, Q 5. Q, S, SR, QR, R, QS, T, TQ, TR, TS 6. T, TQ, TR, TS 7. TQ, QT 8. S and ST or SQ 9. point 10. R FE Lesson ,, D, and D; and ; D and 8. Q and RS; U and TS; QU and RT; QR and UT 9. NK and NM; JK and JM; LK and LM 10. JM el to atrice: 205 feet; atrice to Zaleika: 95 feet Lesson FD FDE Lesson k and n appear to be perpendicular it is perpendicular to the other line. 3. heck students drawings; the perpendicular bisector of D contains on the perpendicular bisector of the segment. 5. heck students drawings; m appears to bisect D bisects the other angle in the linear pair. Lesson heck students drawings. Lesson heck students drawings. Lesson horizontal move right 1 unit K 2. horizontal move 2 units left, vertical move 1 unit down Q' 3. reflection across the -ais X' Y J Q O Z' O K' O J' ' L R' Z L' X Y' R Geometr nswers 161

5 NME LSS DTE Reteaching 1.2 Measuring Length Skill Finding the length of a segment on a number line If and are the coordinates of points X and Y on a number line, then XY, the length of XY, is given b or Notice that the length of XY is also the distance between X and Y. Eample Find the length of each segment on the number line. Q R X Y a. R b. Q c. QR a. The coordinates of and R are 7 and 3. Thus, R 3 ( 7) b. The coordinates of and Q are 7 and 1. Thus, Q 1 ( 7) 6 6. c. The coordinates of Q and R are 1 and 3. Thus, QR 3 ( 1) 4 4. In Eercises 1 6, use the number line below. Find the length of each line segment. L G D M E F D 2. M M 5. L 6. F Skill Using the Segment ongruence ostulate to identif and measure congruent segments The Segment ongruence ostulate states that if two segments have the same length, the are congruent. If two segments are congruent, the have the same length. In illustrations, congruent segments are indicated b equal X numbers of tick marks. In the figure at the right, for eample, XY and XZ are congruent. You write this as, XY XZ. Eample Y Z Use the Segment ongruence ostulate to complete each sentence. E a. If DE MN, then. b. If DE MN, then. a. If DE MN, then DE MN. b. If DE MN, then DE MN. Geometr Reteaching 1.2 3

6 NME LSS DTE Name all congruent segments Q R 9. K D U T S J M N L 10. In Eercise 9, if JK 27, what else can ou determine? Skill Using the Segment ddition ostulate to solve problems The Segment ddition ostulate states that if,,and are collinear with between and, then Eample cadia, rentwood, and edarville lie along a straight stretch of interstate highwa. rentwood lies between cadia and edarville. The distance from cadia to edarville is 200 miles. If edarville is three times as far from rentwood as rentwood is from cadia, find each distance. Let points,, and represent cadia, rentwood, and edarville. Let the distance in miles from cadia to rentwood. Then 3 the distance in miles from rentwood to edarville. oint is between points and, and distance from cadia to rentwood 50 miles distance from rentwood to edarville miles heck: , which is the distance from cadia to edarville. In Eercises 11 14, point is between points and. Find the indicated value. 11. If 22,, and 6, find. 12. If 6 2, 2 1, and 4, find. 13. oint is on DE, between D and E. If D 12 and E 17, find DE. 14. The members of a drill team are lined up along the sideline of a football field. el is at one end and Zaleika is at the other, 300 feet awa. atrice is between el and Zaleika. Her distance from el is 15 feet more than twice her distance from Zaleika. Find the distance from el to atrice and from atrice to Zaleika. 4 Reteaching 1.2 Geometr

7 NSWERS Reteaching hapter 1 Lesson N and JN 2. Sample: KN and NJ 3. Sample:, N, Q 4. Sample: K, N,, Q 5. Q, S, SR, QR, R, QS, T, TQ, TR, TS 6. T, TQ, TR, TS 7. TQ, QT 8. S and ST or SQ 9. point 10. R FE Lesson ,, D, and D; and ; D and 8. Q and RS; U and TS; QU and RT; QR and UT 9. NK and NM; JK and JM; LK and LM 10. JM el to atrice: 205 feet; atrice to Zaleika: 95 feet Lesson FD FDE Lesson k and n appear to be perpendicular it is perpendicular to the other line. 3. heck students drawings; the perpendicular bisector of D contains on the perpendicular bisector of the segment. 5. heck students drawings; m appears to bisect D bisects the other angle in the linear pair. Lesson heck students drawings. Lesson heck students drawings. Lesson horizontal move right 1 unit K 2. horizontal move 2 units left, vertical move 1 unit down Q' 3. reflection across the -ais X' Y J Q O Z' O K' O J' ' L R' Z L' X Y' R Geometr nswers 161

8 NME LSS DTE Reteaching 1.3 Measuring ngles Skill Measuring angles with a protractor protractor is a tpe of geometr ruler used to find the measure of an angle. If the verte of is placed at the center point of a protractor and and intersect the protractor at a and b, respectivel, then the measure of, written m, is a b or b a. Eample Use a protractor to find the measure of XYZ. a b X Y Z 1. ut the center of the protractor at Y, the verte of the angle. 2. osition the protractor so that YZ passes through point Find the intersection point for YX,36. m XYZ Y X Z Use a protractor to find the measure of each angle. You ma trace the figures or use a piece of paper to etend the ras if necessar Skill Using the ngle ddition ostulate The ngle ongruence ostulate: If two angles have the same measure, then the are congruent. If two angles are congruent, then the have the same measure. In the figure at the right, m m, so and are congruent. This ma be written. The tick marks indicate that the angles are congruent. The ngle ddition ostulate: If a point S is in the interior of QR, then m QS m SQR m QR S R Q Geometr Reteaching 1.3 5

9 NME LSS DTE Eample In the figure, m MNQ 104 and m QN 31. Find m MN. the ngle ddition ostulate, m MN m MNQ m QN M N Q Find the missing angle measures. 3. m JKL 41, m LKM 37, m JKM J 4. m JKL 73, m LKM 29, m JKM L K 5. m JKM 83, m JKL 26, m LKM M not to scale Skill Identifing and using special pairs of angles Two angles are complementar if the sum of their measures is 90. Each is called a complement of the other. Two angles are supplementar if the sum of their measures is 180. Each is called a supplement of the other. If the endpoint of a ra is on a line so that two angles are formed, the angles are called a linear pair. In the figure, 1 and 2 form a linear pair. the Linear air 1 2 ropert, the angles in a linear pair are supplementar. ngles ma be classified according to their measures. right angle has measure 90. n acute angle has measure between 0 and 90. n obtuse angle has measure between 90 and 180. Eample Identif an pairs of complementar angles or supplementar angles. 125 W 55 X Y 35 right angle smbol m X m Y , so X and Y are complementar angles. m W m X , so W and X are supplementar angles. Refer to the figure at the right. omplete each statement. 6. FDE and form a linear pair. 7. and are supplementar angles. 8. and are complementar angles G D F E 28 6 Reteaching 1.3 Geometr

10 NSWERS Reteaching hapter 1 Lesson N and JN 2. Sample: KN and NJ 3. Sample:, N, Q 4. Sample: K, N,, Q 5. Q, S, SR, QR, R, QS, T, TQ, TR, TS 6. T, TQ, TR, TS 7. TQ, QT 8. S and ST or SQ 9. point 10. R FE Lesson ,, D, and D; and ; D and 8. Q and RS; U and TS; QU and RT; QR and UT 9. NK and NM; JK and JM; LK and LM 10. JM el to atrice: 205 feet; atrice to Zaleika: 95 feet Lesson FD FDE Lesson k and n appear to be perpendicular it is perpendicular to the other line. 3. heck students drawings; the perpendicular bisector of D contains on the perpendicular bisector of the segment. 5. heck students drawings; m appears to bisect D bisects the other angle in the linear pair. Lesson heck students drawings. Lesson heck students drawings. Lesson horizontal move right 1 unit K 2. horizontal move 2 units left, vertical move 1 unit down Q' 3. reflection across the -ais X' Y J Q O Z' O K' O J' ' L R' Z L' X Y' R Geometr nswers 161

11 NME LSS DTE Reteaching 1.4 Geometr Using aper Folding Throughout this lesson, ou will need folding paper and a marker or pencil that will write on the paper. For Skill, ou will need a compass as well. Skill Using paper folding to construct and stud perpendicular and parallel lines erpendicular lines are lines that intersect to form a right angle. arallel lines are coplanar lines that do not intersect. You can use paper folding to produce a line perpendicular to, or parallel to a given line. Eample Use paper folding to construct two parallel lines and a line perpendicular to one of the lines. 1. Fold the paper once to make a line. Label it. Draw a point on. Fold the paper through so that matches up with itself. Label the new line ou folded m. m 2. Draw a second point Q on. Fold the paper through Q so that matches up with itself. Label the new line ou folded n. Then lines and m are parallel. 3. Draw a point R on m. Fold the paper through R so that m matches up with itself. Label the new line k. Line k is perpendicular to m. Refer to lines m, n, and k in the eample above. R k Q n m Q n m 1. What appears to be true of lines k and n? 2. omplete the following conjecture about parallel and perpendicular lines. If a line is perpendicular to one of two parallel lines, then. Geometr Reteaching 1.4 7

12 NME LSS DTE Skill Using paper folding to construct and stud segment bisectors and angle bisectors n angle bisector is a line or a ra that divides an angle into two congruent angles. Eample Use paper folding to construct each line. a. the bisector of a segment b. the bisector of an angle. a. First fold line. hoose points and, the endpoints of a segment. Then fold the paper so that matches up with. Label the resulting line m. m is the perpendicular bisector of. b. First fold two intersecting lines j and k. Label the point of intersection. Label a point Q on line k and a point R on line j. Then fold the paper through so that line j matches up with line k. Label the new line n. n is a bisector of QR. j Q n m k R Refer to the constructions in the eample above. 3. Use a new piece of paper. Fold line and choose points and D. Using a compass, draw two congruent arcs with centers and D. Label the intersection of the arcs. Repeat part a of the eample to construct the bisector of D. What do ou notice about point? 4. omplete this conjecture: If a point is equall distant from the endpoints of a segment, then the point lies. 5. Use a new piece of paper. Fold two intersecting lines, labeled as shown in the figure. onstruct the bisector of and label it. Fold the paper through so that matches up with itself. Label the new line m. What appears to be true of m and D? D D. 6. omplete this conjecture: If a line that passes through the verte of a linear pair is perpendicular to a bisector of one of the angles, then the line. 8 Reteaching 1.4 Geometr

13 NSWERS Reteaching hapter 1 Lesson N and JN 2. Sample: KN and NJ 3. Sample:, N, Q 4. Sample: K, N,, Q 5. Q, S, SR, QR, R, QS, T, TQ, TR, TS 6. T, TQ, TR, TS 7. TQ, QT 8. S and ST or SQ 9. point 10. R FE Lesson ,, D, and D; and ; D and 8. Q and RS; U and TS; QU and RT; QR and UT 9. NK and NM; JK and JM; LK and LM 10. JM el to atrice: 205 feet; atrice to Zaleika: 95 feet Lesson FD FDE Lesson k and n appear to be perpendicular it is perpendicular to the other line. 3. heck students drawings; the perpendicular bisector of D contains on the perpendicular bisector of the segment. 5. heck students drawings; m appears to bisect D bisects the other angle in the linear pair. Lesson heck students drawings. Lesson heck students drawings. Lesson horizontal move right 1 unit K 2. horizontal move 2 units left, vertical move 1 unit down Q' 3. reflection across the -ais X' Y J Q O Z' O K' O J' ' L R' Z L' X Y' R Geometr nswers 161

14 NME LSS DTE Reteaching 1.5 Special oints in Triangles In this lesson, ou will be required to do constructions. You will need folding paper and an appropriate marker or geometr software. You will also need a compass. Skill Using the perpendicular bisectors of a triangle to construct the circumscribed circle of the triangle triangle is named using the smbol and the vertices. The perpendicular bisectors of a triangle intersect in a point. The point is equidistant from each of the vertices and is the center of the circumscribed circle of the triangle. This circle is outside the triangle and contains the three vertices of the triangle. Given a triangle, ou can locate the center of the circumscribed circle b constructing two of the perpendicular bisectors. You ma then use the third perpendicular bisector as a check. Eample onstruct the circumscribed circle of DEF. 1. hoose two sides of DEF, for eample, DE and DF as shown in the figure. onstruct the perpendicular bisectors of DE and DF. Label the point where the perpendicular bisectors intersect. 2. lace the point of a compass at. hoose an verte of the triangle, sa E, and open the compass to the width E. 3. Draw the circle with center and radius E. onstruct the circumscribed circle of each triangle E D F Geometr Reteaching 1.5 9

15 NME LSS DTE Skill Using the angle bisectors of a triangle to construct the inscribed circle of the triangle The angle bisectors of a triangle intersect in a point. The point is equidistant from the sides of the triangle and is the center of the inscribed circle of the triangle. This circle is inside the triangle and just touches the three sides. Given a triangle, ou can locate the center of the inscribed circle b constructing the bisectors of two of the angles. You ma then use the bisector of the third angle as a check. Eample onstruct the inscribed circle of JKL. 1. hoose two angles of JKL, for eample, K and L as shown in the figure. onstruct the bisector of K and the bisector of L. Label the point where the angle bisectors intersect. 2. Find the distance from to one side of the triangle, KL for instance. Use the distance from to KL as the setting for our compass. 3. Draw the circle with center and radius equal to the distance ou found in Step 2. K J L onstruct the inscribed circle of each triangle Reteaching 1.5 Geometr

16 NSWERS Reteaching hapter 1 Lesson N and JN 2. Sample: KN and NJ 3. Sample:, N, Q 4. Sample: K, N,, Q 5. Q, S, SR, QR, R, QS, T, TQ, TR, TS 6. T, TQ, TR, TS 7. TQ, QT 8. S and ST or SQ 9. point 10. R FE Lesson ,, D, and D; and ; D and 8. Q and RS; U and TS; QU and RT; QR and UT 9. NK and NM; JK and JM; LK and LM 10. JM el to atrice: 205 feet; atrice to Zaleika: 95 feet Lesson FD FDE Lesson k and n appear to be perpendicular it is perpendicular to the other line. 3. heck students drawings; the perpendicular bisector of D contains on the perpendicular bisector of the segment. 5. heck students drawings; m appears to bisect D bisects the other angle in the linear pair. Lesson heck students drawings. Lesson heck students drawings. Lesson horizontal move right 1 unit K 2. horizontal move 2 units left, vertical move 1 unit down Q' 3. reflection across the -ais X' Y J Q O Z' O K' O J' ' L R' Z L' X Y' R Geometr nswers 161

17 NME LSS DTE Reteaching 1.6 Motions in Geometr For this lesson, ou will need patt paper or tracing paper. Skill Drawing a translation rigid transformation of a plane figure does not change the size or shape of the figure. The original figure, or preimage, and the result, or image, are congruent. translation is a rigid transformation in which ever point Q of the figure moves in a straight line with no turns. ll points Q' move the same distance and in the same direction. In the figure, ', QQ', and RR' are all congruent and parallel. R ' Eample Translate the preimage along the given line. R' preimage preimage image 1. Trace the figure and the line onto a piece of patt paper. 2. Start with our drawing directl over the diagram. Slide the patt paper so the image of the line stas directl over the preimage. Trace the figure a second time. 3. Label the first figure ou drew preimage and the second image. Trace the diagram. Then translate the figure along the given line Skill Drawing a rotation rotation is a rigid transformation in which ever point of a figure moves around a given point, called the center of rotation. ll points rotate through the same angle measure. In the figure, is the center of rotation and ' and ' are congruent. ' ' Geometr Reteaching

18 NME LSS DTE Eample Rotate the preimage about the given point. 1. Trace the preimage and the point onto a piece of patt paper. 2. Start with our drawing directl over the diagram. ress the point of our pencil onto the point to hold preimage the patt paper in place. Then turn the patt paper around the point as far as ou like. Trace the figure a second time. 3. Label the first figure ou drew preimage and the second image. image Trace the diagram. Then rotate the figure around the given point Skill Drawing a reflection reflection is a rigid transformation in which ever point of a figure is reflected across a line. The line is the perpendicular bisector of each segment formed b a point and its image. In the figure, is the perpendicular bisector of JJ', KK' and LL'. Eample Reflect XYZ across the given line. 1. Draw XYZ and line. 2. Fold the paper along. unch holes through points X, Y, and Z to locate the image points X', Y', and Z'. 3. onnect the points to draw the image triangle, X'Y'Z'. Trace the diagram. Then reflect the figure across the given line. J X Y Y' X' L K K' J' L' Z Z' Reteaching 1.6 Geometr

19 NSWERS Reteaching hapter 1 Lesson N and JN 2. Sample: KN and NJ 3. Sample:, N, Q 4. Sample: K, N,, Q 5. Q, S, SR, QR, R, QS, T, TQ, TR, TS 6. T, TQ, TR, TS 7. TQ, QT 8. S and ST or SQ 9. point 10. R FE Lesson ,, D, and D; and ; D and 8. Q and RS; U and TS; QU and RT; QR and UT 9. NK and NM; JK and JM; LK and LM 10. JM el to atrice: 205 feet; atrice to Zaleika: 95 feet Lesson FD FDE Lesson k and n appear to be perpendicular it is perpendicular to the other line. 3. heck students drawings; the perpendicular bisector of D contains on the perpendicular bisector of the segment. 5. heck students drawings; m appears to bisect D bisects the other angle in the linear pair. Lesson heck students drawings. Lesson heck students drawings. Lesson horizontal move right 1 unit K 2. horizontal move 2 units left, vertical move 1 unit down Q' 3. reflection across the -ais X' Y J Q O Z' O K' O J' ' L R' Z L' X Y' R Geometr nswers 161

20 NME LSS DTE Reteaching 1.7 Motion in the oordinate lane Skill Using algebraic operations on the coordinates of a point to describe a translation coordinate plane has a horizontal ais, called the -ais, and a vertical ais called the -ais. The origin is the point where the aes intersect. Each point in the plane is identified with an ordered pair. The first number is the -coordinate. The second is the -coordinate. You can appl algebraic operations to the coordinates (, ) of a point to describe a translation. horizontal translation b h units: H((, )) ( h, ) vertical translation b v units: V((, )) (, v) translation b h units horizontall and v units verticall: T((, )) ( h, v) Eample Describe the effect of the transformation T((, )) ( 3, 2). Then draw the triangle with vertices ( 2, 2), ( 1, 1), and (1, 1) on a coordinate plane and use the rule to transform the figure. T((, )) ( 3, 2) ( ( 3), 2) The horizontal change is 3, or 3 units to the left. The vertical change is 2, or up 2 units. ' T(( 2, 2)) ( 2 ( 3), 2 2) ( 5, 4) T(( 1, 1)) ( 1 ( 3), 1 2) ( 4, 1) T((1, 1)) (1 ( 3), 1 2) ( 2, 3) ' ' O Describe the effect of each transformation. Use the given rule to transform the figure. 1. ((, )) ( 1,) 2. M((, )) ( 2, 1) K J O L Q O R Skill Using algebraic operations on the coordinates of a point to describe a reflection across the -ais or the -ais You can appl algebraic operations to the coordinates (, ) of a point to describe a reflection across the -ais or the -ais. reflection across the -ais: F((, )) (, ) reflection across the -ais: G((, )) (, ) Geometr Reteaching

21 NME LSS DTE Eample Describe the effect of the transformation J((, )) (, ). Then draw the triangle with vertices (1, 1), D(3, 1), and E(4, 3) on a coordinate plane and use the rule to transform the figure. J((, )) (, ) is a reflection across the -ais. E J((1, 1)) (1, 1) J((3, 1)) (3, 1) J((4, 3)) (4, 3) O ' D D' E' Describe the effect of each transformation. Use the given rule to transform the figure. 3. ((, )) (, ) 4. K((, )) (, ) X M Y O Z N O Skill Using algebraic operations on the coordinates of a point to describe a 180 rotation about the origin You can appl algebraic operations to the coordinates (, ) of a point to describe a 180 rotation about the origin: R((, )) (, ) Eample Draw the triangle with vertices R( 2, 2), S(3, 1), and T(2, 3) on a coordinate plane. Then draw its image after a 180 rotation about the origin. J(( 2, 2)) (2, 2) J((3, 1)) ( 3, 1) J((2, 3)) ( 2, 3) Use the transformation J(, ) (, ) to rotate each figure 180 about the origin U G S' T' R O R' T S T O V F O H 14 Reteaching 1.7 Geometr

22 NSWERS Reteaching hapter 1 Lesson N and JN 2. Sample: KN and NJ 3. Sample:, N, Q 4. Sample: K, N,, Q 5. Q, S, SR, QR, R, QS, T, TQ, TR, TS 6. T, TQ, TR, TS 7. TQ, QT 8. S and ST or SQ 9. point 10. R FE Lesson ,, D, and D; and ; D and 8. Q and RS; U and TS; QU and RT; QR and UT 9. NK and NM; JK and JM; LK and LM 10. JM el to atrice: 205 feet; atrice to Zaleika: 95 feet Lesson FD FDE Lesson k and n appear to be perpendicular it is perpendicular to the other line. 3. heck students drawings; the perpendicular bisector of D contains on the perpendicular bisector of the segment. 5. heck students drawings; m appears to bisect D bisects the other angle in the linear pair. Lesson heck students drawings. Lesson heck students drawings. Lesson horizontal move right 1 unit K 2. horizontal move 2 units left, vertical move 1 unit down Q' 3. reflection across the -ais X' Y J Q O Z' O K' O J' ' L R' Z L' X Y' R Geometr nswers 161

23 NSWERS 4. reflection across the -ais M ' N' O N M' 4. n n 2 n + 41 rime? Yes Yes Yes Yes. 5. V' T O U T' V ; No. When n 41, the value of the epression is 41 41, which is not prime. 7. onl one case U' Lesson F G, H' O F' 1. figures with four sides rectangles D H, G' D is a figure with four congruent sides. Reteaching hapter 2 Lesson Yes; the proof would be identical. If n is a positive odd number, then the net positive odd number is n If ou take the bottom row of dots and make it a column at the right, the diagram would be nearl square. It would be missing one dot. For the specific case in the diagram, there are si dots and each row and eight dots in each column. If the bottom row is made into a column, the diagram would be a 7 7 square with one dot missing. That is, Since and are even numbers, let 2n and 2m, where n and m are whole numbers. Then 2n 2m 2(n m), which is an even number, since n m is a whole number. 2. a line intersects a segment at its midpoint 3. the line bisects the segment 4. If a line bisects a segment, then the line intersects the segment at its midpoint. True. 5. If cows have wings, then the can fl. If an animal can fl, then it is a bird. If I am a bird, then I can sing. If I can sing, then I will be a star. If cows have wings, then I will be a star. 162 nswers Geometr

24 NME LSS DTE Reteaching 2.1 n Introduction to roofs Skill Investigating informal proofs ositive even numbers are 2, 4, 6,... and positive odd numbers are 1, 3, 5,... Eample If ou know the square of a positive even number, ou can easil find the square of the net even number b adding 4 plus 4 times the original number. In algebraic terms, this can be epressed as: For an positive even number n, (n 2) 2 n 2 4n 4. Draw a diagram and prove informall that the statement is true. The diagram shows a rectangle whose dimensions are n 2 and n 2. n 2 The areas of the four regions are: n n n 2 2 n 2n n 2 2n n The sum of the areas is n 2 2n 2n 4. Then (n 2) 2 n 2 4n 4. 2 Write informal proofs. 1. ould ou use the method in the solution to prove that for an positive odd number n, (n 2) 2 n 2 4n 4? Eplain. 2. The diagram at right represents the product (n 1)(n 1). Eplain how ou could use the diagram shown to prove that for ever positive integer n, (n 1)(n 1) n rove that the sum of two even numbers is even. (Use the fact that an even number is a number that can be written as 2n, where n is a whole number. Show that if and are both even, then is even.) Geometr Reteaching

25 NME LSS DTE Skill Investigating conjecture and proof prime number is a number that is divisible b eactl two numbers, itself and 1. The first ten prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29. Eample onsider the numbers in the table. Given a whole number n, what appears to be true of the epression n 2 n 41? n n 2 n n n 2 n For each whole number n from 1 to 10, the epression n 2 n 41 gives a prime number. Refer to the eample above. 4. omplete the table. n n 2 n 41 rime? Do the results of our table prove that n 2 n 41 gives a prime number? 5. Find the value of n 2 n 41 for n 41. Show that n 2 n 41 is not prime. 6. Is the conjecture true that n 2 n 41 gives a prime number? Eplain our response. 7. conjecture cannot be proven true b demonstrating that it is true for specific cases, no matter how man are presented. How man cases must be presented to show that a conjecture is not true? 16 Reteaching 2.1 Geometr

26 NSWERS 4. reflection across the -ais M ' N' O N M' 4. n n 2 n + 41 rime? Yes Yes Yes Yes. 5. V' T O U T' V ; No. When n 41, the value of the epression is 41 41, which is not prime. 7. onl one case U' Lesson F G, H' O F' 1. figures with four sides rectangles D H, G' D is a figure with four congruent sides. Reteaching hapter 2 Lesson Yes; the proof would be identical. If n is a positive odd number, then the net positive odd number is n If ou take the bottom row of dots and make it a column at the right, the diagram would be nearl square. It would be missing one dot. For the specific case in the diagram, there are si dots and each row and eight dots in each column. If the bottom row is made into a column, the diagram would be a 7 7 square with one dot missing. That is, Since and are even numbers, let 2n and 2m, where n and m are whole numbers. Then 2n 2m 2(n m), which is an even number, since n m is a whole number. 2. a line intersects a segment at its midpoint 3. the line bisects the segment 4. If a line bisects a segment, then the line intersects the segment at its midpoint. True. 5. If cows have wings, then the can fl. If an animal can fl, then it is a bird. If I am a bird, then I can sing. If I can sing, then I will be a star. If cows have wings, then I will be a star. 162 nswers Geometr

27 NME LSS DTE Reteaching 2.2 n Introduction to Logic Skill Drawing conclusions from conditionals conditional is a statement that can be written in If-then form. The part following if is called the hpothesis. The part following then is called the conclusion. The statement ll cats are mammals is a conditional statement. Mammals If-then form: If an animal is a cat, then it is a mammal. ats hpothesis conclusion The Euler (or Venn) diagram at the right shows that all cats are mammals. This demonstrates that the given conditional statement is true. The process of deductive reasoning, or deduction, involves making logical arguments to reach logicall certain conclusions. Eample a. Draw an Euler diagram that conves the following information. If a figure is a square, then it has four congruent sides. D is a square. b. What conclusion can be drawn about D? a. Figures with b. D has four congruent sides. 4 sides Squares D Use the statements If a figure is a rectangle, then it has four sides and QRS is a rectangle in Eercises 1 and Draw an Euler diagram to represent the information. What conclusion can be drawn from the diagram? Skill Forming the converse of a conditional In logical notation, letters are used to represent the hpothesis and the conclusion of a conditional statement. The conditional If p then q is written p q. This ma also be read p implies q. The converse of a conditional is formed b interchanging the hpothesis and the conclusion. onditional: p q onverse: q p Geometr Reteaching

28 NME LSS DTE The converse of a true conditional statement ma be true or false. conditional statement is false if ou can find one eample, called a countereample, for which the hpothesis is true and the conclusion is false. Eample The statement If a person lives in San ntonio, then he or she lives in Teas is true. Write the hpothesis, the conclusion, and the converse of the statement. If the converse is false, give a countereample. hpothesis: person lives in San ntonio conclusion: he or she lives in Teas converse: If a person lives in Teas, then he or she lives in San ntonio. The converse is false. For eample, the person ma live in Dallas. The statement If a line intersects a segment at its midpoint, then the line bisects the segment is true. 2. Write the hpothesis. 3. Write the conclusion. 4. Write the converse. Is it true? Skill Writing and using logical chains onditional statements can sometimes be linked together in a logical chain. conditional can be derived from the logical chain using the If-Then Transitive ropert: When ou are given If, then and If, then, ou can conclude If, then. This can also be written If and, then. Eample rrange the statements in a logical chain. Write the conditional that follows. If it is cold, I can go ice skating. If it is March, then the wind is blowing. If the wind is blowing, then it is cold. Use letters to represent the hpothesis and conclusion of each statement. : it is cold : I can go ice skating : it is March D: the wind is blowing Write the statements in logical notation and look for a chain. D D The chain is D, D, and. The conditional that follows is, that is, if it is March, then I can go ice skating. rrange the statements in a logical chain and write the conditional that follows. 5. If I can sing, then I will be a star. If I am a bird, then I can sing. If cows have wings, then the can fl. If an animal can fl, then it is a bird. 18 Reteaching 2.2 Geometr

29 NSWERS 4. reflection across the -ais M ' N' O N M' 4. n n 2 n + 41 rime? Yes Yes Yes Yes. 5. V' T O U T' V ; No. When n 41, the value of the epression is 41 41, which is not prime. 7. onl one case U' Lesson F G, H' O F' 1. figures with four sides rectangles D H, G' D is a figure with four congruent sides. Reteaching hapter 2 Lesson Yes; the proof would be identical. If n is a positive odd number, then the net positive odd number is n If ou take the bottom row of dots and make it a column at the right, the diagram would be nearl square. It would be missing one dot. For the specific case in the diagram, there are si dots and each row and eight dots in each column. If the bottom row is made into a column, the diagram would be a 7 7 square with one dot missing. That is, Since and are even numbers, let 2n and 2m, where n and m are whole numbers. Then 2n 2m 2(n m), which is an even number, since n m is a whole number. 2. a line intersects a segment at its midpoint 3. the line bisects the segment 4. If a line bisects a segment, then the line intersects the segment at its midpoint. True. 5. If cows have wings, then the can fl. If an animal can fl, then it is a bird. If I am a bird, then I can sing. If I can sing, then I will be a star. If cows have wings, then I will be a star. 162 nswers Geometr

30 NME LSS DTE Reteaching 2.3 Definitions Skill Using Euler diagrams to stud definitions When a conditional statement and its converse are combined, the result is a conditional statement known as a biconditional, or an if-and-onl-if statement. In logical notation, if ou are given p q and q p, ou can combine them to form the biconditional statement p q, which is read p if and onl q. biconditional can be represented b a pair of Euler diagrams, one representing one conditional statement and one representing its converse. Ever definition can be written as a biconditional. For eample, the definition of adjacent angles can be written as follows. Two angles in the same plane are adjacent angles if and onl if the share a verte and a common side, but no interior points. Eample Determine whether the following sentence is a definition. erpendicular lines are lines that meet to form right angles. If the sentence is a definition, write it as a biconditional and represent it with a pair of Euler diagrams. If two lines are perpendicular, then the meet to form right angles. If two lines meet to form right angles, then the are perpendicular. The sentence is a definition. Two lines are perpendicular if and onl if the meet to form right angles. lines that form right angles perpendicular lines perpendicular lines lines that form right angles Determine whether the sentence is a definition. If so, write it as a biconditional and represent it with a pair of Euler diagrams. If not, eplain wh not. 1. square is a rectangle with four congruent sides. 2. n acute angle is an angle that is not obtuse. Geometr Reteaching

31 NME LSS DTE Skill Using the principles of logic to formulate definitions Ever definition can be written as a biconditional. Eample Suppose, with no information other than that given in the figures below, ou are asked to define the word frisbend. These figures are frisbends. These figures are not frisbends. Use the figures above to write a definition of the word frisbend. You need to determine what frisbends have in common that is not true of objects that are not frisbends. Each of the objects in the first group is a closed figure with at least four sides, including one that etends beond a verte. Some objects in the second group have fewer than four sides. Some have no etended sides, or more than one. Definition: figure is a frisbend if and onl if it is a closed figure with at least four sides, eactl one of which etends beond a verte. Use the given information to determine which figures are greems. 3. These figures are greems. These figures are not greems. Tell whether each figure is a greem.. Use the given information to write a definition of the phrase rectangular solid.. 4. These figures are rectangular solids. These figures are not rectangular solids.. space figure is a rectangular solid if and onl if. 20 Reteaching 2.3 Geometr

32 NSWERS Lesson Yes; a figure is a square if and onl if it is a rectangle with four congruent sides. rectangles with four congruent sides 2. No; if an angle is acute, then it is not obtuse. However an angle that is not obtuse ma be a right angle or a straight angle. 3. a. Yes; b. no; c. es. 4. Sample: all of its surfaces are rectangles Lesson (Given); 3 36 (ddition ropert of Equalit); 12 (Division ropert of Equalit) 2. ddition ropert of Equalit 3. Segment ddition ostulate 4. R QS 5. Given squares squares rectangles with four congruent sides 6. ddition ropert of Equalit 7. ngle ddition ostulate 8. m D 9. m m D Lesson Definition of perpendicular lines 14. Transitive (or Substitution) ropert of Equalit 15. ngle ddition ostulate 16. Transitive (or Substitution) ropert of Equalit 17. Subtraction ropert of Equalit Transitive Reteaching hapter 3 Lesson none Geometr nswers 163

33 NME LSS DTE Reteaching 2.4 uilding a Sstem of Geometr Knowledge Skill Using the properties of equalit and congruence The lgebraic and Equivalence roperties of Equalit which ou are alread familiar are often used in geometr. Since the measures of segments and angles are real numbers, the algebraic properties can be applied to such measures. The equivalence properties also etend to geometric figures. The properties are summarized below. The lgebraic roperties of Equalit For all real numbers a, b, and c, if a b, then: a c b c (ddition) a c b c (Subtraction) a ac bc (Multiplication ) (Division) c b (c 0) c a and b ma be substituted for each other in an equation or inequalit. (Substitution) The Equivalence roperties Equalit ongruence For all real numbers a, b, and c: For geometric figures,, and : Refleive a a Smmetric If a b, then b a. If, then. Transitive If a b and b c, then a c. If and, then. Eample In the figure,. Write and solve an equation to find the value of. Write the steps of the solution at the left. Write the reason that justifies each step to the right of the step. Given m m Definition of congruence Substitution ropert of Equalit Subtraction ropert of Equalit 5 35 Subtraction 7 Division ropert of Equalit In each triangle congruent sides are indicated. Write and solve an equation to find the value of. Write the steps of the solution at the left side of the blanks and the reasons at the right (5 + 12) 29 Geometr Reteaching

34 NME LSS DTE Skill roviding justifications in proofs In the proof of a theorem, each step (unless it is given) must be justified b a definition, a propert, a postulate, or a theorem. Eample Given: m 1 m rove: Statements Reasons 1. m 1 m Given 2. m 2 m Linear air ropert 3. m 1 m 3 m 2 m 3 3. Substitution ropert of Equalit 4. m 1 m 2 4. Subtraction ropert of Equalit Definition of congruence 2 3 omplete each proof. Given: T QT, TR TS rove: R QS S T Q R Statements T QT, TR TS T TR QT TS 2. T TR R, QT TS QS 3. Given Reasons 4. Substitution ropert of Equalit Given: m 1 m 3, m 2 m 4 rove: m m D Statements m 1 m 3, m 2 m 4 5. m 1 m 2 m 3 m 4 6. m 1 m 2 m 7. Reasons 8. m 3 m 4 m ngle ddition ostulate 9. Substitution ropert of Equalit D 22 Reteaching 2.4 Geometr

35 NSWERS Lesson Yes; a figure is a square if and onl if it is a rectangle with four congruent sides. rectangles with four congruent sides 2. No; if an angle is acute, then it is not obtuse. However an angle that is not obtuse ma be a right angle or a straight angle. 3. a. Yes; b. no; c. es. 4. Sample: all of its surfaces are rectangles Lesson (Given); 3 36 (ddition ropert of Equalit); 12 (Division ropert of Equalit) 2. ddition ropert of Equalit 3. Segment ddition ostulate 4. R QS 5. Given squares squares rectangles with four congruent sides 6. ddition ropert of Equalit 7. ngle ddition ostulate 8. m D 9. m m D Lesson Definition of perpendicular lines 14. Transitive (or Substitution) ropert of Equalit 15. ngle ddition ostulate 16. Transitive (or Substitution) ropert of Equalit 17. Subtraction ropert of Equalit Transitive Reteaching hapter 3 Lesson none Geometr nswers 163

36 NME LSS DTE Reteaching 2.5 onjectures That Lead to Theorems Skill Using the Vertical ngles Theorem When two lines intersect, the form two pairs of vertical angles. ccording to the Vertical ngles Theorem, if two angles form a pair of vertical angles, then the are congruent. In the figure at the right, 1 and 2 are vertical angles, so 1 2. lso, 3 and 4 are vertical angles, so Eample omplete: a. KMN and? are vertical angles. J N b. If m JML 77, then m NMQ?. M 45 c. If m NM 45, and m KML (4 3), K then?. L Q a. KMN is formed b intersecting lines K and LN, so KMN and LM are vertical angles. b. JML and NMQ are vertical angles, so m NMQ m JML 77. c. NM and KML are vertical angles. Then their measures are equal and , so 4 48, and 12. Identif the vertical angles in the figure and 2. 2 and 3. 3 and 4. 4 and 5. 5 and 6. 6 and Find the measure of Find the value of (5 + 4) (6 4) (5 5) (3 + 19) Geometr Reteaching

37 NME LSS DTE Skill Using two-column and paragraph forms of proof Eample Given: m 1 m 2 rove: is perpendicular to D. 1 2 D Statements Reasons 1. m 1 m 2 1. Given 2. m 1 m Linear air ropert 3. m 1 m ; 3. Substitution ropert of Equalit 2m m Division ropert of Equalit 5. is perpendicular to D. 5. Definition of perpendicular lines omplete each proof. Given: is perpendicular to D 1 4 rove: 2 3 D Statements is perpendicular to D. Given Reasons m D 90 ; m D m D m D 14. m 1 m 2 m D; 15. m 3 m 4 m D m 1 m 2 m 3 m m 2 m 3 or Given: 2 3 rove: 1 4 roof: 1 and 2 are vertical angles, so (18.) Theorem. It is given that 2 3, so 1 3 b the (19.) ongruence. 3 and (20.) are also vertical angles, so (21.) b the Vertical ngles ropert of as well. Then, b the Transitive ropert of ongruence, (22.). 24 Reteaching 2.5 Geometr

38 NSWERS Lesson Yes; a figure is a square if and onl if it is a rectangle with four congruent sides. rectangles with four congruent sides 2. No; if an angle is acute, then it is not obtuse. However an angle that is not obtuse ma be a right angle or a straight angle. 3. a. Yes; b. no; c. es. 4. Sample: all of its surfaces are rectangles Lesson (Given); 3 36 (ddition ropert of Equalit); 12 (Division ropert of Equalit) 2. ddition ropert of Equalit 3. Segment ddition ostulate 4. R QS 5. Given squares squares rectangles with four congruent sides 6. ddition ropert of Equalit 7. ngle ddition ostulate 8. m D 9. m m D Lesson Definition of perpendicular lines 14. Transitive (or Substitution) ropert of Equalit 15. ngle ddition ostulate 16. Transitive (or Substitution) ropert of Equalit 17. Subtraction ropert of Equalit Transitive Reteaching hapter 3 Lesson none Geometr nswers 163

39 NME LSS DTE Reteaching 3.1 Smmetr in olgons Skill Identifing and drawing figures with reflectional smmetr figure has reflectional smmetr if there is a line, called an ais of smmetr, that divides the figure into two parts that are mirror images of each other. figure ma have no ais of smmetr, or it ma have one or more. If ou fold a figure along an ais of smmetr, the two parts of the figure will coincide. Eample 1 Tell how man aes of smmetr the figure has. Draw them all. a. b. c. a. There is one. b. There are none. c. There are two. Eample 2 omplete the figure so that the dashed line is a line of smmetr. Draw all the aes of smmetr of each figure. If there are none, write none omplete the figure so that each dashed line is an ais of smmetr Geometr Reteaching

40 NME LSS DTE Skill Identifing and drawing figures with rotational smmetr figure has rotational smmetr of about a point if, when the figure is rotated about, the figure and its image coincide. regular polgon is a polgon with all sides congruent and all angles congruent. Its center is the point that is equidistant from the sides. 360 The measure of the central angle of a regular polgon with n sides has measure. n The polgon has n rotational smmetries about its center. Eample 1 Describe the rotational smmetries of a regular heagon. regular heagon has si sides, so it has si rotational smmetries about its 360 center. The measure of a central angle is 60, so the measures of the angles 6 of rotation for the smmetries are multiples of 60 less than or equal to 360 : 60, 120, 180, 240, 300, and 360. Eample 2 Describe the rotational smmetries of the figure. The points of the figure are the vertices of a square. The figure has rotational smmetries of 90, 180, 270, and 360 about the center. Describe the rotational smmetries of the figure. 8. an equilateral triangle with center 9. a regular pentagon with center omplete the drawing so that the figure has 180 rotational smmetr about the indicated point. 26 Reteaching 3.1 Geometr

41 NSWERS Lesson Yes; a figure is a square if and onl if it is a rectangle with four congruent sides. rectangles with four congruent sides 2. No; if an angle is acute, then it is not obtuse. However an angle that is not obtuse ma be a right angle or a straight angle. 3. a. Yes; b. no; c. es. 4. Sample: all of its surfaces are rectangles Lesson (Given); 3 36 (ddition ropert of Equalit); 12 (Division ropert of Equalit) 2. ddition ropert of Equalit 3. Segment ddition ostulate 4. R QS 5. Given squares squares rectangles with four congruent sides 6. ddition ropert of Equalit 7. ngle ddition ostulate 8. m D 9. m m D Lesson Definition of perpendicular lines 14. Transitive (or Substitution) ropert of Equalit 15. ngle ddition ostulate 16. Transitive (or Substitution) ropert of Equalit 17. Subtraction ropert of Equalit Transitive Reteaching hapter 3 Lesson none Geometr nswers 163

42 NSWERS Lesson and 3; 2 and 4; 5 and 7; 6 and and 6; 3 and and 5; 4 and and 3; 6 and Justifications ma var. 4 (Vertical angles are congruent.); 8 (If 2 lines are int. b a trans., corr. s are.); 2 (If 2 lines are int. b a trans., alt. et. s are.) 8. rotational smmetries of 120, 240, and 360 about the center 9. rotational smmetries of 72, 144, 216, 288, and rotational smmetres of 180 and 360 about the center 11. rotational smmetries of 120, 240, and 360 about the center 12. Lesson rhombus, parallelogram 2. square, rhombus, rectangle, parallelogram 3. trapezoid 4. parallelogram a. 143 b. 37 c. 143 d. 37 e. 143 f a. 121 b. 51 c. 59 d a. 24 b. 55 c. 125 Lesson j k; if 2 lines are int. b a trans. and alt. int. s are, the lines are. 2. m n; if 2 lines are int. b a trans. and alt. et. s are, the lines are. 3. none 4. m n; if 2 lints are int. b a trans. and same-side int. s are supp., the lines are. 5. Given 6. MQ 7. N 8. Given 9. MN 10. Q 11. If 2 lines are int. b a trans. and alt. int. s are, the lines are. 12. MNQ is a parallelogram. Lesson ; 70 ; 71 ; ; 64 ; 46 ; nswers Geometr

43 NME LSS DTE Reteaching 3.2 roperties of Quadrilaterals Skill Identifing special quadrilaterals The table summarizes the definitions of special quadrilaterals. Trapezoid arallelogram Rhombus Rectangle Square Eactl one pair of opposite sides are parallel. oth pairs of opposite sides are parallel. ll four sides are congruent. ll four angles are right angles. (ll four angles are congruent.) ll four sides are congruent; all four angles are right angles. The Euler diagram shows how special quadrilaterals are related. Quadrilaterals Trapezoids arallelograms rhombuses squares rectangles Eample In quadrilateral WXYZ, is perpendicular to both and. YZ WX WZ W XY is also perpendicular to both WZ and XY. Identif the Z special tpe(s) of quadrilateral that WXYZ must be. ll four angles of WXYZ are right angles, so WXYZ is a rectangle. ccording to the Euler diagram, all rectangles are parallelograms, so WXYZ is also a parallelogram. Identif the special tpe(s) of quadrilateral that WXYZ must be. 1. WX, XY, YZ, and WZ are congruent. 2. WXYZ is both equilateral and equiangular. 3. WX and ZY are parallel; 4. WX and ZY are parallel; WZ and XY are not parallel. WZ and XY are parallel. Geometr Reteaching

44 NME LSS DTE Skill Using the properties of special quadrilaterals Since rhombuses, rectangles, and squares are all parallelograms, the all have the properties of parallelograms. However, the have special properties of their own as well. The properties are summarized in the table. ll parallelograms Rhombuses Rectangles Squares Opposite sides are congruent. Opposite angles are congruent. The diagonals bisect each other. onsecutive angles are supplementar. The diagonals are perpendicular. The diagonals are congruent. The diagonals are perpendicular and congruent. The diagonals of a square are both perpendicular and congruent because a square is both a rhombus and a rectangle. Eample In parallelogram D, 4, D 3, D 6 and m D 120. Find D,, DE, and m D. E D is a parallelogram. Opposite sides are congruent, so D 4 and D 3. 1 The diagonals of D bisect each other, so DE D 3 2 D D and D are supplementar, so m D In rectangle QRS, S 24, Q 18, and QS 30. omplete. 5. SR 6. QR 7. R 8. T 9. QT 10. m QS In rhombus JKLM, m JML 45, ML 20, JL 15, and MN omplete. 11. KL 12. JK 13. JN 14. m JNM 15. m MJK 16. m JKL In square DE, D 34 and F 24. omplete. 17. E 18. FD 19. D 20. E 21. m E 22. m F Q M T J H F L S R K E D 28 Reteaching 3.2 Geometr

45 NSWERS Lesson and 3; 2 and 4; 5 and 7; 6 and and 6; 3 and and 5; 4 and and 3; 6 and Justifications ma var. 4 (Vertical angles are congruent.); 8 (If 2 lines are int. b a trans., corr. s are.); 2 (If 2 lines are int. b a trans., alt. et. s are.) 8. rotational smmetries of 120, 240, and 360 about the center 9. rotational smmetries of 72, 144, 216, 288, and rotational smmetres of 180 and 360 about the center 11. rotational smmetries of 120, 240, and 360 about the center 12. Lesson rhombus, parallelogram 2. square, rhombus, rectangle, parallelogram 3. trapezoid 4. parallelogram a. 143 b. 37 c. 143 d. 37 e. 143 f a. 121 b. 51 c. 59 d a. 24 b. 55 c. 125 Lesson j k; if 2 lines are int. b a trans. and alt. int. s are, the lines are. 2. m n; if 2 lines are int. b a trans. and alt. et. s are, the lines are. 3. none 4. m n; if 2 lints are int. b a trans. and same-side int. s are supp., the lines are. 5. Given 6. MQ 7. N 8. Given 9. MN 10. Q 11. If 2 lines are int. b a trans. and alt. int. s are, the lines are. 12. MNQ is a parallelogram. Lesson ; 70 ; 71 ; ; 64 ; 46 ; nswers Geometr

46 NME LSS DTE Reteaching 3.3 arallel Lines and Transversals Skill Identifing related pairs of angles transversal of two lines is a line that intersects the given lines in two different points. When a transversal t intersects two lines m and n, the pairs of angles formed have special names. lternate eterior angles: a pair of angles outside m and n on opposite sides of t orresponding angles: a pair of angles in corresponding m n positions relative to m and n lternate interior angles: a pair of angles between m and n on t opposite sides of t Same-side interior angles: a pair of consecutive angles between m and n on the same side of t Eample Indicate whether the pairs of angles are corresponding 1 8 angles, alternate interior angles, alternate eterior angles, 2 7 or same-side interior angles. 3 6 a. 3 and 7 b. 8 and c. 4 and 8 d. 2 and 3 a. 3 and 7 are alternate interior angles. b. 8 and 6 are corresponding angles. c. 4 and 8 are alternate eterior angles. d. 2 and 3 are same-side interior angles. Using the figure in the above eample, identif all pairs of angles of the tpe indicated. 1. corresponding angles 2. alternate interior angles 3. alternate eterior angles 4. same-side interior angles Skill Using the postulates and properties of parallel lines When two parallel lines are intersected b a transversal: orresponding angles are congruent. lternate interior angles are congruent. lternate eterior angles are congruent. Same-side interior angles are supplementar. Geometr Reteaching

47 NME LSS DTE Eample 1 Lines j and k are parallel. Name all the angles that are congruent to 1. Justif our answer j k 7; when two parallel lines are intersected b a transversal, corresponding angles are congruent. 5; when two parallel lines are intersected b a transversal, alternate eterior angles are congruent. 3; vertical angles are congruent. Eample 2 Refer to the figure in Eample 1 above. If m find m 3. When two parallel lines are intersected b a transversal, same-side interior angles are supplementar, so m m Refer to the figure in Eample 1 above. 5. Name all the angles that are congruent to 6. Justif our answer. 6. If m 1 37, find the measure of each numbered angle. a. m 2 b. m 3 c. m 4 d. m 5 e. m 6 f. m 7 For Eercises 7 and 8, refer to the figure at the right, in which Q is parallel to SR and S is parallel to QR. 7. If m S 59 and m RS 51, find the measure of each angle. a. m SQ b. m QR c. m Q d. m RQ 8. If m S (2 7) and m SQ 5( 1), find each value. S R Q a. b. m S c. m SQ 30 Reteaching 3.3 Geometr

48 NSWERS Lesson and 3; 2 and 4; 5 and 7; 6 and and 6; 3 and and 5; 4 and and 3; 6 and Justifications ma var. 4 (Vertical angles are congruent.); 8 (If 2 lines are int. b a trans., corr. s are.); 2 (If 2 lines are int. b a trans., alt. et. s are.) 8. rotational smmetries of 120, 240, and 360 about the center 9. rotational smmetries of 72, 144, 216, 288, and rotational smmetres of 180 and 360 about the center 11. rotational smmetries of 120, 240, and 360 about the center 12. Lesson rhombus, parallelogram 2. square, rhombus, rectangle, parallelogram 3. trapezoid 4. parallelogram a. 143 b. 37 c. 143 d. 37 e. 143 f a. 121 b. 51 c. 59 d a. 24 b. 55 c. 125 Lesson j k; if 2 lines are int. b a trans. and alt. int. s are, the lines are. 2. m n; if 2 lines are int. b a trans. and alt. et. s are, the lines are. 3. none 4. m n; if 2 lints are int. b a trans. and same-side int. s are supp., the lines are. 5. Given 6. MQ 7. N 8. Given 9. MN 10. Q 11. If 2 lines are int. b a trans. and alt. int. s are, the lines are. 12. MNQ is a parallelogram. Lesson ; 70 ; 71 ; ; 64 ; 46 ; nswers Geometr

49 NME LSS DTE Reteaching 3.4 roving That Lines are arallel Skill Identifing parallel lines You have now learned si was to prove that two lines are parallel. Two lines are parallel if, when the are intersected b a transversal: corresponding angles are congruent. alternate interior angles are congruent. alternate eterior angles are congruent. same-side interior angles are supplementar. Two coplanar lines that are perpendicular to the same line are parallel. Two lines that are parallel to the same line are parallel to each other. Eample ased on the given information, tell which lines, if j k an, must be parallel. Eplain wh the lines are m parallel a. 4 5 b n c. 7 and 11 are supplementar a. m and n; if two lines are intersected b a transversal and alternate eterior angles are congruent, then the lines are parallel. b. None. c. j and k; if two lines are intersected b a transversal and same-side interior angles are supplementar, the lines are parallel. Refer to the figure in the eample above. Determine which lines, if an must be parallel. Eplain wh the lines are parallel and 13 are supplementar and 15 are supplementar. Geometr Reteaching

50 NME LSS DTE Skill Using theorems about parallel lines to complete proofs Eample Given: 1 2; and D are supplementar. rove: D is a parallelogram. 2 D 1 roof: Statements Reasons Given 2. D If two lines are intersected b a transversal and alt. et. s are, the lines are. 3. and D are supplementar. Given 4. D If two lines are intersected b a transversal and same-side int. s are supp., the lines are. 5. D is a parallelogram. Definition omplete the two-column proof. Given: 1 2; 3 4 rove: MNQ is a parallelogram. M 1 4 N roof: 2 3 Q Statements Reasons and (7.) are. If two lines are intersected b a transversal and alt. int. s are, the lines are and (10.) are Definition 32 Reteaching 3.4 Geometr

51 NSWERS Lesson and 3; 2 and 4; 5 and 7; 6 and and 6; 3 and and 5; 4 and and 3; 6 and Justifications ma var. 4 (Vertical angles are congruent.); 8 (If 2 lines are int. b a trans., corr. s are.); 2 (If 2 lines are int. b a trans., alt. et. s are.) 8. rotational smmetries of 120, 240, and 360 about the center 9. rotational smmetries of 72, 144, 216, 288, and rotational smmetres of 180 and 360 about the center 11. rotational smmetries of 120, 240, and 360 about the center 12. Lesson rhombus, parallelogram 2. square, rhombus, rectangle, parallelogram 3. trapezoid 4. parallelogram a. 143 b. 37 c. 143 d. 37 e. 143 f a. 121 b. 51 c. 59 d a. 24 b. 55 c. 125 Lesson j k; if 2 lines are int. b a trans. and alt. int. s are, the lines are. 2. m n; if 2 lines are int. b a trans. and alt. et. s are, the lines are. 3. none 4. m n; if 2 lints are int. b a trans. and same-side int. s are supp., the lines are. 5. Given 6. MQ 7. N 8. Given 9. MN 10. Q 11. If 2 lines are int. b a trans. and alt. int. s are, the lines are. 12. MNQ is a parallelogram. Lesson ; 70 ; 71 ; ; 64 ; 46 ; nswers Geometr

52 NME LSS DTE Reteaching 3.5 The Triangle Sum Theorem Skill Using the Triangle Sum Theorem to find angle measures The Triangle Sum Theorem: The sum of the measures of the angles of a triangle is 180. Eample 1 In XYZ, m X 41 and m Y 66. Find m Z. the Triangle Sum Theorem, m X m Y m Z m Z 180 ; m Z 73 Eample 2 In JKL, m J ( 2 10), m K (4 10), and m L (5 10). Find the value of and the measure of each angle of JKL. the Triangle Sum Theorem, m J m K m L ( 10)( 19) 0 10 or 19 If 19, 4 10 and 5 10 are negative, so 10. m J ( 2 10) (100 10) 90 m K (4 10) (40 10) 50 m L (5 10) (50 10) 40 The measures of two angles of a triangle are given. Find the measure of the third angle. 1. m W 92 and m X 65 ; m Y 2. m R 50 and m S 50 ; m T 3. m D 22 and m F 104 ; m E 4. m 78 and m D 34 ; m The measures of the angles of a triangle are given. Find the value of and the measure of each angle. 5. m ( 2 3), m Q (10 1), m R (5 4) m m Q m R 6. m K ( 2 ), m L (5 6), m M (7 14) m K m L m M Geometr Reteaching

53 NME LSS DTE Skill Using the Triangle Sum Theorem to complete proofs Eample Given: with m 2 (m ) and m 3 (m ) rove: is a right triangle. roof: Statements Reasons 1. m 2 (m ), m 3 (m ) Given 2. m m m 180 Triangle Sum Theorem 3. m 2 (m ) 3 (m ) 180 Substitution ropert of Equalit 4. 6 (m ) 180 ; m 30 Division ropert of Equit 5. m 90 Substitution ropert of Equalit 6. is a right angle; Definition is a right triangle. omplete the two-column proof. Given: D is a right angle. rove: E and F are complementar. E roof: F D Statements D is a right angle. m D m D m E m F m E m F m E m F E and F are complementar. 11. Given Reasons 34 Reteaching 3.5 Geometr

54 NSWERS Lesson and 3; 2 and 4; 5 and 7; 6 and and 6; 3 and and 5; 4 and and 3; 6 and Justifications ma var. 4 (Vertical angles are congruent.); 8 (If 2 lines are int. b a trans., corr. s are.); 2 (If 2 lines are int. b a trans., alt. et. s are.) 8. rotational smmetries of 120, 240, and 360 about the center 9. rotational smmetries of 72, 144, 216, 288, and rotational smmetres of 180 and 360 about the center 11. rotational smmetries of 120, 240, and 360 about the center 12. Lesson rhombus, parallelogram 2. square, rhombus, rectangle, parallelogram 3. trapezoid 4. parallelogram a. 143 b. 37 c. 143 d. 37 e. 143 f a. 121 b. 51 c. 59 d a. 24 b. 55 c. 125 Lesson j k; if 2 lines are int. b a trans. and alt. int. s are, the lines are. 2. m n; if 2 lines are int. b a trans. and alt. et. s are, the lines are. 3. none 4. m n; if 2 lints are int. b a trans. and same-side int. s are supp., the lines are. 5. Given 6. MQ 7. N 8. Given 9. MN 10. Q 11. If 2 lines are int. b a trans. and alt. int. s are, the lines are. 12. MNQ is a parallelogram. Lesson ; 70 ; 71 ; ; 64 ; 46 ; nswers Geometr

55 NSWERS 7. Def Triangle Sum Theorem: The sum of the measures of the angles of a triangle is 180. K 9. Substitution rop. of 10. Subtraction. rop. of 11. Def. Lesson ; ; ; ; sides sides sides sides J L O M 1 slope JK slope ML, so JK ML; 2 slope JM slope KL 2, so JM KL. Then JKLM is a parallelogram. Moreover, JM is perpendicular to both JK and ML (The product of their slopes is 1) and KL is perpendicular to both JK and ML. Then JKLM is a rectangle. 7. ( 3, 4) 8. (8, 5) 9. (8, 2.5) sides sides sides sides Lesson Use the method of Eample 2. Draw JL. Midsegments WX of JKL and ZY of JLM are both parallel to JL, so WX ZY. Since E. 2 showed that WZ XY, WXYZ is a parallelogram Lesson parallel 5. perpendicular D O 1 slope slope D, so D; slope 2 D slope 3, so D ; D is a parallelogram; midpoint of (2.5, 3.5) midpoint of D; so the diagonals of D bisect each other. Reteaching hapter 4 Lesson U 4. GN 5. QU 6. Yes; MN RTS. 7. Yes; JKLMN DEF. 8. Yes; DEF LKJM. 9. Given 10. olgon ongruence ostulate 11. Given 12. olgon ongruence ostulate Geometr nswers 165

56 NME LSS DTE Reteaching 3.6 ngles in olgons Skill Determining the measures of interior angles of polgons The sum, S, of the measures of the interior angles of a polgon with n sides is given b S (n 2)180. For an polgon, the sum of the measures of the eterior angles, one at each verte, is 360. Eample a. Find the value of. b. Find the value of a. The sum of the measures of b. The sum of the measures of the interior angles of a the eterior angles of a polgon quadrilateral is 360. is Find the value of The sum of the measures of the interior angles of a 14-sided polgon is. 8. The sum of the measures of the eterior angles of a 7-sided polgon is. Geometr Reteaching

57 NME LSS DTE Skill Determining the measures of interior and eterior angles of a regular polgon The measure, m, of an interior angle of a regular polgon with n sides is given b (n 2)180 m. n The measure, m, of an eterior angle of a regular polgon with n sides is given b 360 m. It is sometimes useful to solve this equation for n and use the equivalent n 360 epression n. n Eample 1 Find the measure of each angle. a. an interior angle of a regular polgon with 20 sides b. an eterior angle of a regular polgon with 16 sides (n 2) a. n 20; m 162 n 20 The measure of an interior angle of a regular 20-sided polgon is 162. b. n 16; m The measure of an eterior angle of a regular 16-sided polgon is Eample 2 How man sides does the regular polgon described have? a. The measure of an interior angle is 160. b. The measure of an eterior angle is 15. a. If the measure of an interior angle is 160, the measure of an eterior angle is n ; n m 20 The regular polgon has 18 sides. 360 b. n ; n m 15 The regular polgon has 24 sides. Find the measure of an interior angle and an eterior angle of a regular polgon with the given number of sides The measure of an interior angle of a regular polgon is given. How man sides does the polgon have? The measure of an eterior angle of a regular polgon is given. How man sides does the polgon have? Reteaching 3.6 Geometr

58 NSWERS 7. Def Triangle Sum Theorem: The sum of the measures of the angles of a triangle is 180. K 9. Substitution rop. of 10. Subtraction. rop. of 11. Def. Lesson ; ; ; ; sides sides sides sides J L O M 1 slope JK slope ML, so JK ML; 2 slope JM slope KL 2, so JM KL. Then JKLM is a parallelogram. Moreover, JM is perpendicular to both JK and ML (The product of their slopes is 1) and KL is perpendicular to both JK and ML. Then JKLM is a rectangle. 7. ( 3, 4) 8. (8, 5) 9. (8, 2.5) sides sides sides sides Lesson Use the method of Eample 2. Draw JL. Midsegments WX of JKL and ZY of JLM are both parallel to JL, so WX ZY. Since E. 2 showed that WZ XY, WXYZ is a parallelogram Lesson parallel 5. perpendicular D O 1 slope slope D, so D; slope 2 D slope 3, so D ; D is a parallelogram; midpoint of (2.5, 3.5) midpoint of D; so the diagonals of D bisect each other. Reteaching hapter 4 Lesson U 4. GN 5. QU 6. Yes; MN RTS. 7. Yes; JKLMN DEF. 8. Yes; DEF LKJM. 9. Given 10. olgon ongruence ostulate 11. Given 12. olgon ongruence ostulate Geometr nswers 165

59 NME LSS DTE Reteaching 3.7 Midsegments of Triangles and Trapezoids Skill Using midsegments of triangles midsegment of a triangle is a segment whose endpoints are midpoints of two sides of the triangle. midsegment is parallel to the third side of the triangle and half as long as that side. Eample 1 In QR, L, M, and N are the midpoints of Q, QR, and R, respectivel. a. Name each midsegment of QR and the side of QR to which the midsegment is parallel. b. If Q 10, QR 8, and R 12, find the length of each midsegment. a. LM R, MN Q, and LN QR b. MN Q 5, LN QR 4, and LM R Eample 2 In quadrilateral JKLM, W, X, Y, and Z are the midpoints of JK, KL, LM, and JM, respectivel. Show that WZ is parallel to XY. Draw KM. (Through an two points there is one and onl line.) W and Z are the midpoints of JK and JM so WZ is a midsegment of JKM and is parallel to KM. Similarl, XY is a midsegment of KML and is parallel to KM. Two lines parallel to a third line are parallel to each other, so WZ XY. Find the indicated measure UV W K X L J L N Y Q Z 66 M R M Geometr Reteaching

60 NME LSS DTE 4. W, X, Y, and Z are the midpoints of the sides of JKLM. Eample 2 showed that WZ XY. Eplain how ou would show that WXYZ is a parallelogram. W K X L J Y Z M Skill Using the midsegment of a trapezoid The midsegment of a trapezoid is the segment joining the endpoints of the nonparallel sides. The midsegment of a trapezoid is parallel to the bases. Its length is half the sum of the lengths of the bases. Eample and D are the midpoints of QT and RS, respectivel. Find the value of and the lengths of QR, TS, and D. Q R D 1 D is the midsegment of QRST, so D (QR ST) ( ) ( 3) 0 0 or 3 If 0, D 0 and QR is negative. Since cannot be 0, 3. Then QR 6(3) 1 17, ST 2(3) 2 4(3) 1 31, 1 and D 8(3) 24. s a check, notice that D (17 31) Find the indicated measure M N 26 D 52 Q 24 XY Q J 10. D W X Y Q Z K J K M T L D Q M N S WZ MN 38 Reteaching 3.7 Geometr

61 NSWERS 7. Def Triangle Sum Theorem: The sum of the measures of the angles of a triangle is 180. K 9. Substitution rop. of 10. Subtraction. rop. of 11. Def. Lesson ; ; ; ; sides sides sides sides J L O M 1 slope JK slope ML, so JK ML; 2 slope JM slope KL 2, so JM KL. Then JKLM is a parallelogram. Moreover, JM is perpendicular to both JK and ML (The product of their slopes is 1) and KL is perpendicular to both JK and ML. Then JKLM is a rectangle. 7. ( 3, 4) 8. (8, 5) 9. (8, 2.5) sides sides sides sides Lesson Use the method of Eample 2. Draw JL. Midsegments WX of JKL and ZY of JLM are both parallel to JL, so WX ZY. Since E. 2 showed that WZ XY, WXYZ is a parallelogram Lesson parallel 5. perpendicular D O 1 slope slope D, so D; slope 2 D slope 3, so D ; D is a parallelogram; midpoint of (2.5, 3.5) midpoint of D; so the diagonals of D bisect each other. Reteaching hapter 4 Lesson U 4. GN 5. QU 6. Yes; MN RTS. 7. Yes; JKLMN DEF. 8. Yes; DEF LKJM. 9. Given 10. olgon ongruence ostulate 11. Given 12. olgon ongruence ostulate Geometr nswers 165

62 NME LSS DTE Reteaching 3.8 nalzing olgons Using oordinates Skill Finding the slope of a line in the coordinate plane The slope, m, of a nonvertical line that contains the points ( 1, 1 ) and ( 2, 2 ) is given b m The slope of a horizontal line is 0. The slope of a vertical line is undefined. Eample Find the slope of the line that contains the points (5, 2) and (12, 1). m Determine the slope, m, of the line that contains the given points. 1. (5, 3) and (2, 6) 2. (5, 8) and ( 3, 2) 3. (14, 8) and ( 2, 6) m m m Skill Using slope to determine whether lines are parallel or perpendicular Two nonvertical lines are parallel if and onl if their slopes are equal. ll vertical lines in the coordinate plane are parallel. Two nonvertical lines are perpendicular if and onl if the product of their slopes is 1. Ever vertical line in the coordinate plane is perpendicular to ever horizontal line in the plane. Eample Draw the triangle with vertices ( 2, 3), (2, 0), and ( 1, 4). Determine whether is a right triangle. Since and appear to be perpendicular, check their slopes. 0 3 slope of 2 ( 2) slope of O Since 3,. Then is a right angle and is a right triangle. Geometr Reteaching

63 NME LSS DTE Determine whether and D are parallel, perpendicular, or neither. 4. ( 3, 1), (4, 2), (0, 2), D(7, 5) 5. ( 2, 2), (3, 2), (0, 2), D(4, 3) 6. Sketch the quadrilateral with vertices J(0, 1), K(4, 3), L(5, 1), and M(1, 1). Show that the quadrilateral is a rectangle. Skill Using the Midpoint Formula The midpoint of the segment with endpoints ( 1, 1 ) and ( 2, 2 ) has coordinates ( 1 2, ) Eample triangle has vertices X( 2, 2), Y(2, 4), and Z(4, 0). Let Q be the midsegment of XYZ that is parallel to XY. Determine the coordinates of and Q. The endpoints of Q are the midpoints of XZ and YZ. Let be the midpoint of XZ and Q the midpoint of YZ. Y ( has coordinates,. 2 2 ) (1, 1) Q Q has coordinates ( 2 4 2, 4 0 Z 2 ) (3, 2) O X Determine the coordinates of M, the midpoint of. 7. ( 13,8); (7, 0) 8. (5, 9); (11, 1) 9. (7, 3); (9, 2) M M M 10. Draw the quadrilateral with vertices (0, 4), (4, 6), (5, 3), and D(1, 1). Show that D is a parallelogram and that the diagonals bisect each other, that is, have the same midpoint. 40 Reteaching 3.8 Geometr

64 NSWERS 7. Def Triangle Sum Theorem: The sum of the measures of the angles of a triangle is 180. K 9. Substitution rop. of 10. Subtraction. rop. of 11. Def. Lesson ; ; ; ; sides sides sides sides J L O M 1 slope JK slope ML, so JK ML; 2 slope JM slope KL 2, so JM KL. Then JKLM is a parallelogram. Moreover, JM is perpendicular to both JK and ML (The product of their slopes is 1) and KL is perpendicular to both JK and ML. Then JKLM is a rectangle. 7. ( 3, 4) 8. (8, 5) 9. (8, 2.5) sides sides sides sides Lesson Use the method of Eample 2. Draw JL. Midsegments WX of JKL and ZY of JLM are both parallel to JL, so WX ZY. Since E. 2 showed that WZ XY, WXYZ is a parallelogram Lesson parallel 5. perpendicular D O 1 slope slope D, so D; slope 2 D slope 3, so D ; D is a parallelogram; midpoint of (2.5, 3.5) midpoint of D; so the diagonals of D bisect each other. Reteaching hapter 4 Lesson U 4. GN 5. QU 6. Yes; MN RTS. 7. Yes; JKLMN DEF. 8. Yes; DEF LKJM. 9. Given 10. olgon ongruence ostulate 11. Given 12. olgon ongruence ostulate Geometr nswers 165

65 NME LSS DTE Reteaching 4.1 ongruent olgons Skill Identifing congruent polgons and congruent parts of congruent polgons polgon is named using consecutive vertices in order, either clockwise or counterclockwise around the polgon. Two polgons are congruent if and onl if there is a correspondence between their sides and angles so that: 1. Each pair of corresponding angles is congruent. 2. Each pair of corresponding sides is congruent. In a congruence statement, a statement identifing congruent polgons, the corresponding vertices of the two polgons are named in the same order. Eample Determine whether the quadrilaterals are congruent. If so, write a congruence statement. Q R S K L M J J, Q M, R L, S K Q JM, QR ML, RS LK, S JK Since the quadrilaterals have four pairs of congruent angles and four corresponding pairs of congruent sides, the quadrilaterals are congruent. Three of the man possible congruence statements are: QRS JMLK SRQ JKLM SQR KJML QUD and ONG are quadrilaterals and QUD ONG. omplete each statement. 1. Q 2. N 3. ON 4. D 5. O Determine whether the polgons are congruent. If so, write a congruence statement. 6. R S 7. M L 8. F N D Q U M O J N G N M T J K F E D D E L K Geometr Reteaching

66 NME LSS DTE Skill Using the olgon ongruence ostulate The olgon ongruence ostulate can be written as a a pair of conditional statements: If all corresponding parts of two polgons are congruent, then the polgons are congruent. If two polgons are congruent, then all corresponding parts are congruent. oth statements are useful in proofs. Eample Given that QR SRQ, QT RT, and T TS, write a paragraph proof that QT RTS. Q T R Since QR SRQ, Q RS and QR RSQ b the olgon ongruence ostulate. QT RTS b the Vertical ngles Theorem. It is given that QT RT and T ST. Since each pair of corresponding angles of the two triangles is congruent and each pair of corresponding sides is congruent, QT RTS b the olgon ongruence ostulate. S Given: E E and ED ED rove: D D E roof: D Statements Reasons E E 9., E E, E E 10. ED ED 11. D D, DE DE, DE DE 12. m E m E, m DE m DE 13. m E m DE m E m DE 14. m D m D or D D 15. D D 16. D D Reteaching 4.1 Geometr

67 NSWERS 7. Def Triangle Sum Theorem: The sum of the measures of the angles of a triangle is 180. K 9. Substitution rop. of 10. Subtraction. rop. of 11. Def. Lesson ; ; ; ; sides sides sides sides J L O M 1 slope JK slope ML, so JK ML; 2 slope JM slope KL 2, so JM KL. Then JKLM is a parallelogram. Moreover, JM is perpendicular to both JK and ML (The product of their slopes is 1) and KL is perpendicular to both JK and ML. Then JKLM is a rectangle. 7. ( 3, 4) 8. (8, 5) 9. (8, 2.5) sides sides sides sides Lesson Use the method of Eample 2. Draw JL. Midsegments WX of JKL and ZY of JLM are both parallel to JL, so WX ZY. Since E. 2 showed that WZ XY, WXYZ is a parallelogram Lesson parallel 5. perpendicular D O 1 slope slope D, so D; slope 2 D slope 3, so D ; D is a parallelogram; midpoint of (2.5, 3.5) midpoint of D; so the diagonals of D bisect each other. Reteaching hapter 4 Lesson U 4. GN 5. QU 6. Yes; MN RTS. 7. Yes; JKLMN DEF. 8. Yes; DEF LKJM. 9. Given 10. olgon ongruence ostulate 11. Given 12. olgon ongruence ostulate Geometr nswers 165

68 NSWERS 13. ngle ongruence ostulate 14. dd. rop. of Equalit 15. Substitution rop., ngle ongruence ostulate 16. Refleive rop. of ongruence 17. olgon ongruence ostulate Lesson Yes; QR XZY; S. 2. Yes; RST FED; SSS. 3. no 4. Yes; GHI RTS; SS. Lesson Yes; QR XWY; SS. 2. Yes; NQM NQ; HL. 3. Yes; HJK HJI; SSS. 4. Yes; TUV MN; S. 5. Yes; DEG FEG; S. 6. no 7. es 15 9 HL 5. es 8. es S 6. no 7. es SSS 8. es SS S 9. es SS 10. es S Lesson ED E (Given); D, E E or E E (T); Segment ongruence ostulate); ED E or ED E (T); Segment ongruence ostulate); E E E ED (Segment ddition ostulate); D or D (Substitution rop.; Segment ongruence ostulate); D D (Refleive rop. of ongruence); D D (SSS ostulate) nswers Geometr

69 NME LSS DTE Reteaching 4.2 Triangle ongruence Skill Using the SSS, SS, and S ongruence ostulates for triangles If three sides of one triangle are congruent to three sides of another triangle, then the triangles are congruent. (SSS ostulate) If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent. (SS ostulate) If two angles and the included side of one triangle are congruent to two sides of another triangle, then the triangles are congruent. (S ostulate) Eample Determine whether the triangles can be shown to be congruent using the SSS, SS, or S ostulate. If so, write a congruence statement and identif the postulate. J K ccording to the markings in the figure: JL LK L Therefore, JLK b the SS ostulate. L Determine whether the triangles can be shown to be congruent using the SSS, SS, or S ostulate. If so, write a congruence statement and identif the postulate. 1. Q Y M 4. G D R Z N X S R T D F H I S T R E Geometr Reteaching

70 NME LSS DTE Skill Determining triangle uniqueness based on the SSS, SS, and S congruence postulates You can use congruence postulates to determine whether given conditions determine a unique triangle. Eample 1 Determine whether there is a unique triangle RST with m R 73, RS 24, and RT 18. The lengths of two sides and the measure of the included angle are given. the SS ostulate, an two triangles with the given measures are congruent. Then there is eactl one triangle RST; that is, RST is unique. Eample 2 Determine whether there is a unique triangle JKL with m J 50, JK 12, and KL 10. The lengths of two sides and the measure of an angle are given. The angle is not included between the given sides. None of the three postulates guarantee a unique triangle. The sketches show two such triangles. L 50 L J K J 12 K T R 24 S Determine whether an of the SSS, SS, or S postulates guarantee that with the given measures is unique. If so, use a protractor and a ruler to sketch the triangle and identif the postulate that guarantees uniqueness. If not, sketch two such triangles cm, m 49, m m 38, m 62, m cm, 3.1 cm, 4.4 cm 8. m 39, 2.5 cm, 3.1 cm 44 Reteaching 4.2 Geometr

71 NSWERS 13. ngle ongruence ostulate 14. dd. rop. of Equalit 15. Substitution rop., ngle ongruence ostulate 16. Refleive rop. of ongruence 17. olgon ongruence ostulate Lesson Yes; QR XZY; S. 2. Yes; RST FED; SSS. 3. no 4. Yes; GHI RTS; SS. Lesson Yes; QR XWY; SS. 2. Yes; NQM NQ; HL. 3. Yes; HJK HJI; SSS. 4. Yes; TUV MN; S. 5. Yes; DEG FEG; S. 6. no 7. es 15 9 HL 5. es 8. es S 6. no 7. es SSS 8. es SS S 9. es SS 10. es S Lesson ED E (Given); D, E E or E E (T); Segment ongruence ostulate); ED E or ED E (T); Segment ongruence ostulate); E E E ED (Segment ddition ostulate); D or D (Substitution rop.; Segment ongruence ostulate); D D (Refleive rop. of ongruence); D D (SSS ostulate) nswers Geometr

72 NME LSS DTE Reteaching 4.3 nalzing Triangle ongruence Skill Using postulates and theorems to determine whether triangles are congruent If two angles and a nonincluded side of one triangle are congruent to the corresponding angles and sides of another triangle, then the triangles are congruent. (S Theorem) If the hpotenuse and one leg of a right triangle are congruent to the hpotenuse and the corresponding leg of another right triangle, then the triangles are congruent. (HL Theorem) Eample Determine whether ou could prove that the D triangles are congruent. If so, write a congruence statement and identif the postulate or theorem that justifies the statement. Notice that and ED are vertical angles, so ED. Then a correspondence can be set up between and ED, with ED, D, and ED. Then ED b the S Theorem. E Determine whether ou could prove that the triangles are congruent. If so, write a congruence statement and identif the postulate or theorem ou could use. 1. Q X 2. R W Y 3. H 4. K J I M T N V Q M U N 5. D 6. L D E F G N M E F Geometr Reteaching

73 NME LSS DTE Skill Determining triangle uniqueness based on congruence postulates and theorems You have learned si was of proving that two triangles are congruent or that a given set of conditions determines a unique triangle. The are the olgon ongruence ostulate, the SSS, SS, and S ostulates, and the S and HL Theorems. The HL Theorem is a special case of SS, that is, two sides and the nonincluded angle. Ecept for right triangles, SS cannot be used to prove congruence or uniqueness. Eample Determine whether there is a unique triangle EFG with m E 52, m F 55, and EG 22. The measures of two angles and the length of a side are given. The side is not included b the given angles. the S Theorem, an two triangles with the given measures are congruent. Then there is eactl one triangle EFG; that is, EFG is unique. E F G Determine whether an of the congruence postulates or theorems guarantee that QR with the given measures is unique. If so, use a protractor and a ruler to sketch the triangle and identif the postulate that guarantees uniqueness. If not, sketch two such triangles cm, 2 cm, m m 63, m 90, 1.5 cm 9. 3 cm, 1.5 cm, m m 50, m 35, 1.5 cm 46 Reteaching 4.3 Geometr

74 NSWERS 13. ngle ongruence ostulate 14. dd. rop. of Equalit 15. Substitution rop., ngle ongruence ostulate 16. Refleive rop. of ongruence 17. olgon ongruence ostulate Lesson Yes; QR XZY; S. 2. Yes; RST FED; SSS. 3. no 4. Yes; GHI RTS; SS. Lesson Yes; QR XWY; SS. 2. Yes; NQM NQ; HL. 3. Yes; HJK HJI; SSS. 4. Yes; TUV MN; S. 5. Yes; DEG FEG; S. 6. no 7. es 15 9 HL 5. es 8. es S 6. no 7. es SSS 8. es SS S 9. es SS 10. es S Lesson ED E (Given); D, E E or E E (T); Segment ongruence ostulate); ED E or ED E (T); Segment ongruence ostulate); E E E ED (Segment ddition ostulate); D or D (Substitution rop.; Segment ongruence ostulate); D D (Refleive rop. of ongruence); D D (SSS ostulate) nswers Geometr

75 NME LSS DTE Reteaching 4.4 Using Triangle ongruence Skill Using congruence of triangles to determine congruence of corresponding parts. the olgon ongruence ostulate, corresponding parts of congruent triangles are congruent. This statement, abbreviated T, is often used to justif statements in proofs such as the flowchart proof in the eample below. Eample Given: D D rove: ED E E D lan: Use T to show that D and D D. Then use the Vertical ngles Theorem to show that ED E, and so ED E b the S Theorem. D D Given D D D D T ED E Vertical s Thm. ED E S Refer to polgon D in the figure above. Write a flowchart proof. 1. Given: ED E rove: D D lan: Use T to show that D, E E, and ED E. Then use the Segment ddition ostulate to show that D, or D. Finall, show that, since D D b the Refleive ropert of ongruence, D D b the SSS ostulate. Geometr Reteaching

76 NME LSS DTE Skill Using the Isosceles Triangle Theorem and its converse The Isosceles Triangle Theorem states that if two sides of a triangle are congruent, then the angles opposite those sides are congruent. The converse of the Isosceles Triangle Theorem is true. That is, if two angles of a triangle are congruent, then the sides opposite those angles are congruent. The Isosceles Triangle Theorem has two easil proven corollaries: The measure of each angle of an equilateral triangle is 60. The bisector of the verte angle of an isosceles triangle is the perpendicular bisector of the base. Eample Find the measure of Z. Z X 47 Y XYZ is isosceles, with XY YZ. the Isosceles Triangle Theorem, X Y, or, m X m Y 47. m X m Y m Z m Z 180 m Z 86 Find the indicated value Q m R m J 5. F 6. S R T G H 39 u m F SU N 8. D ( ) L (10) M N 75 4 E R K M T J 12 u L N V 5 48 Reteaching 4.4 Geometr

77 NSWERS 13. ngle ongruence ostulate 14. dd. rop. of Equalit 15. Substitution rop., ngle ongruence ostulate 16. Refleive rop. of ongruence 17. olgon ongruence ostulate Lesson Yes; QR XZY; S. 2. Yes; RST FED; SSS. 3. no 4. Yes; GHI RTS; SS. Lesson Yes; QR XWY; SS. 2. Yes; NQM NQ; HL. 3. Yes; HJK HJI; SSS. 4. Yes; TUV MN; S. 5. Yes; DEG FEG; S. 6. no 7. es 15 9 HL 5. es 8. es S 6. no 7. es SSS 8. es SS S 9. es SS 10. es S Lesson ED E (Given); D, E E or E E (T); Segment ongruence ostulate); ED E or ED E (T); Segment ongruence ostulate); E E E ED (Segment ddition ostulate); D or D (Substitution rop.; Segment ongruence ostulate); D D (Refleive rop. of ongruence); D D (SSS ostulate) nswers Geometr

78 NSWERS Lesson nswers ma var. Sample answer: Since the diagonals of a parallelogram bisect each other, T RT and QT ST. lso, vertical angles TQ and RTS are congruent, so TQ RTS b the SS ostulate. 9. Square; if the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle; if two adjacent sides of a parallelogram are congruent, then the parallelogram is a rhombus. 10. Rectangle; if the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle. Lesson heck students drawings Lesson LM, LMN, LN, MN, LQ, LQM, MQN, QN 1. ' Lesson Yes; if one pair of opposite sides of a quadrilateral are both congruent and parallel, then the quadrilateral is a parallelogram. v ' 2. no 3. Yes; if both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram. 4. Yes; if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. 5. no 6. Rhombus; if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. 7. Rectangle; if one angle of a parallelogram is a right angle, then the parallelogram is a rectangle ' ' t ' 8. Rhombus; if a diagonal of a rhombus bisects a pair of opposite angles, then the parallelogram is a rhombus. ' z 60 Geometr nswers 167

79 NME LSS DTE Reteaching 4.5 roving Quadrilateral roperties Skill Using the properties of parallelograms In the figure at the right, D is a parallelogram. D has the following properties: and D are congruent, as are D and. and are congruent, as are and D. The following angles are supplementar: and and D and and D and D bisect each other. and D each divide D into two congruent triangles. Eample D In parallelogram JKLM, m MLK J Find each indicated value K a. b. m MJK c. m JKL d N M a. Opposite sides of a parallelogram are congruent, so JK ML means that , and 7. b. Opposite angles of a parallelogram are congruent, so m MJK m MLK 60. c. onsecutive angles of a parallelogram are supplementar, so m JKL 180 m MLK d. The diagonals of a parallelogram bisect each other, so JM KL. Then and 3. L QRS is a parallelogram. omplete each statement. 1. If m QS (4 6) and m QR (3 6), then. 2. If S 5 3 and QR 2 6, then. 3. If QT 5z 3 and TS 3z 6, then z. 4. If m QR (3w 2 3) and m SR (14w 8), then w. 5. Write a plan for proving that TQ RTS. S T R Q Geometr Reteaching

80 NME LSS DTE Skill Using the properties of special parallelograms. Rectangles, rhombuses, and squares are special tpes of parallelograms. The Euler diagram depicts how the figures are related. parallelograms rectangles squares rhombuses Rectangles, rhombuses, and squares have all the properties of parallelograms and special properties as well. The diagonals of a rectangle are congruent. The diagonals of a rhombus are perpendicular. square is a rectangle, so its diagonals are congruent. square is a rhombus, so its diagonals are perpendicular bisectors of each other. Eample DEF is a rhombus. Find each indicated value. D a. b c. m DGE d. If m EDG 65, find m DEG. 2 4 G F E 17 a. ll four sides of a rhombus are congruent, so F FE. That is, and 7. b. rhombus is a parallelogram, so the diagonals of a rhombus bisect each other. Then , and 1. c. The diagonals of a rhombus are perpendicular, so m DGE 90. d. m DGE 90 and m EDG 65, so m DEG 180 (90 65 ) 25. Eercises 6 11 refer to square LMN. omplete each statement. 6. If L 2 4 and N + 2, then. 7. If LQ 3 2 and NQ 6 0.5, then. 8. If LN z 2 9 and M z 3, then z. 9. m LQ. 10. m QL. 11. Name all the isosceles right triangles shown in the figure. L Q M N 50 Reteaching 4.5 Geometr

81 NSWERS Lesson nswers ma var. Sample answer: Since the diagonals of a parallelogram bisect each other, T RT and QT ST. lso, vertical angles TQ and RTS are congruent, so TQ RTS b the SS ostulate. 9. Square; if the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle; if two adjacent sides of a parallelogram are congruent, then the parallelogram is a rhombus. 10. Rectangle; if the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle. Lesson heck students drawings Lesson LM, LMN, LN, MN, LQ, LQM, MQN, QN 1. ' Lesson Yes; if one pair of opposite sides of a quadrilateral are both congruent and parallel, then the quadrilateral is a parallelogram. v ' 2. no 3. Yes; if both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram. 4. Yes; if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. 5. no 6. Rhombus; if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. 7. Rectangle; if one angle of a parallelogram is a right angle, then the parallelogram is a rectangle ' ' t ' 8. Rhombus; if a diagonal of a rhombus bisects a pair of opposite angles, then the parallelogram is a rhombus. ' z 60 Geometr nswers 167

82 NME LSS DTE Reteaching 4.6 onditions for Special Quadrilaterals Skill Recognizing conditions that determine that a quadrilateral is a parallelogram Given a parallelogram, ou know that both pairs of opposite sides of the parallelogram are congruent and that the diagonals bisect each other. It is also true that each of these conditions determine a parallelogram. If both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram. If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. third condition determines a parallelogram as well: If one pair of opposite sides of a quadrilateral are both congruent and parallel, then the quadrilateral is a parallelogram. Each quadrilateral below is a parallelogram. Eample QRS is a quadrilateral with Q SR and S QR. Is QRS necessaril a parallelogram? If so, identif the theorem that justifies our answer. QRS has one pair of parallel sides and one pair of congruent sides. The given information is insufficient to prove that QRS is a parallelogram. For eample, consider the sketch at the right. If it were given that Q RS or that S QR, then QRS would necessaril be a parallelogram. VWXY is a quadrilateral with diagonals intersecting at Z. Tell whether VWXY is necessaril a parallelogram. If so, state the theorem that justifies our answer. 1. VY WX and VY WX 2. VY VW and XY XW 3. WX VY and XY WV 4. VZ XZ and WZ YZ S Q R 5. VZ WZ and YZ XZ Geometr Reteaching

83 NME LSS DTE Skill Recognizing conditions that determine that a parallelogram is a rectangle or a rhombus Given a parallelogram D, the following theorems help ou decide whether D is a rectangle or a rhombus. If one angle of D is a right angle, then D is a rectangle. If the diagonals of D are congruent, then D is a rectangle. If two adjacent sides of D are congruent, then D is a rhombus. If the diagonals, and D, of D are perpendicular, then D is a rhombus. If a diagonal of D bisects a pair of opposite angles, then D is a rhombus. Eample DE is a parallelogram. D E and D E. Tell whether DE is necessaril a rhombus, a rectangle, both, or neither. Eplain. D and E are the diagonals of DE. Since D E, DE is a rhombus. Since D E, DE is a rectangle. Since DE is both a rhombus and a rectangle, DE is a square. MNQ is a parallelogram with diagonals intersecting at R. Tell whether MNQ is necessaril a rhombus, a rectangle, both, or neither. If MNQ is one or both, justif our answer b stating a theorem or theorems. 6. m MRQ MQ MN 8. QM NM and QM NM 9. N Q and MQ Q 10. M NQ 52 Reteaching 4.6 Geometr

84 NSWERS Lesson nswers ma var. Sample answer: Since the diagonals of a parallelogram bisect each other, T RT and QT ST. lso, vertical angles TQ and RTS are congruent, so TQ RTS b the SS ostulate. 9. Square; if the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle; if two adjacent sides of a parallelogram are congruent, then the parallelogram is a rhombus. 10. Rectangle; if the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle. Lesson heck students drawings Lesson LM, LMN, LN, MN, LQ, LQM, MQN, QN 1. ' Lesson Yes; if one pair of opposite sides of a quadrilateral are both congruent and parallel, then the quadrilateral is a parallelogram. v ' 2. no 3. Yes; if both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram. 4. Yes; if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. 5. no 6. Rhombus; if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. 7. Rectangle; if one angle of a parallelogram is a right angle, then the parallelogram is a rectangle ' ' t ' 8. Rhombus; if a diagonal of a rhombus bisects a pair of opposite angles, then the parallelogram is a rhombus. ' z 60 Geometr nswers 167

85 NME LSS DTE Reteaching 4.7 ompass and Straightedge onstructions Skill Use a compass and straightedge to duplicate segments and angles and construct figures In compass and straightedge constructions, a compass is used to cop distances. straightedge is used to draw ras and lines. When ou draw an arc with a compass, the center of the arc is indicated b the compass point. The radius is the distance between the points of the pencil and the compass. Eample Given and, use a compass and straightedge to construct each of the following. a. a segment congruent to b. an angle congruent to. a. Draw line and choose a point on. Draw an arc with radius and center intersecting at Q; Q. Q b. Draw a ra with endpoint X. Draw an arc with center intersecting the sides of at points E and F. Using the same radius, draw an arc with center X intersecting the ra at point Y. Net draw an arc with center Y and radius EF intersecting the first arc at point W. Draw XW ; WXY. E W F X F Refer to in the eample above. hoose segments and angles as indicated and construct a triangle congruent to using the indicated postulate. 1. two sides and the included angle (SS) 2. two angles and the included side (S) Geometr Reteaching

86 NME LSS DTE Skill onstructing lines parallel to or perpendicular to a given line Eample Given line, point M on and point not on, construct each of the following. a. a line through parallel to b. a line through perpendicular to c. a line through M perpendicular to M a. hoose a point X on and draw X, forming X with sides X and. onstruct an angle with verte that is congruent to X as shown. Draw ; Q Q. X N Q b. Draw an arc with center intersecting line at points R and S. Using an reasonable radius, draw arcs with centers R and S that intersect at a point Q on the opposite side of from. Draw ; Q Q. R Q S c. Draw two arcs with center M and the same radius intersecting at points J and K. Using a radius greater than MJ, draw two arcs with centers J and K intersecting at point N. Draw ; MN MN. Use a compass and a straightedge to construct each figure. 3. rectangle 4. rhombus 5. parallelogram 6. square J N M K 54 Reteaching 4.7 Geometr

87 NSWERS Lesson nswers ma var. Sample answer: Since the diagonals of a parallelogram bisect each other, T RT and QT ST. lso, vertical angles TQ and RTS are congruent, so TQ RTS b the SS ostulate. 9. Square; if the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle; if two adjacent sides of a parallelogram are congruent, then the parallelogram is a rhombus. 10. Rectangle; if the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle. Lesson heck students drawings Lesson LM, LMN, LN, MN, LQ, LQM, MQN, QN 1. ' Lesson Yes; if one pair of opposite sides of a quadrilateral are both congruent and parallel, then the quadrilateral is a parallelogram. v ' 2. no 3. Yes; if both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram. 4. Yes; if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. 5. no 6. Rhombus; if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. 7. Rectangle; if one angle of a parallelogram is a right angle, then the parallelogram is a rectangle ' ' t ' 8. Rhombus; if a diagonal of a rhombus bisects a pair of opposite angles, then the parallelogram is a rhombus. ' z 60 Geometr nswers 167

88 NME LSS DTE Reteaching 4.8 onstructing Transformations Skill Translating a segment using a compass and straightedge translation vector shows the direction and distance of a translation. To translate a segment, translate its endpoints and connect their images. Eample Use a compass and straightedge to translate the segment as indicated b the translation vector. Let be the length of the translation vector v. onstruct a line l 1 through parallel to v and a line l 2 through parallel to v. Using a compass opening of, construct ' on l 1 and ' on l 2. Draw ''. Then '' is the image of. ' ' v v 1 2 Skill Rotating a segment using a compass and straightedge To rotate a segment, rotate its endpoints and connect their images. Eample Rotate about b m J. 40 J Draw and. onstruct Q and R, both congruent to J. onstruct ' on Q and ' on R. Draw ''. Then '' is the rotation image of. ' R ' Q Geometr Reteaching

89 NME LSS DTE Skill Reflecting a segment using a compass and straightedge To reflect a segment, reflect its endpoints and connect their images. Eample Reflect across. onstruct a perpendicular, 1, to through intersecting at X. onstruct a perpendicular, 2, to through intersecting at Y. onstruct X' X on 1. onstruct Y' Y on 2. Draw ''. Then '' is the reflection of. ' ' 1 2 Use a compass and straightedge to construct each transformation of the given segment,. 1. the translation indicated b vector v 2. the translation indicated b vector t t v 3. rotation about b m Z 4. rotation about M b m W 60 z 5. reflection across n 6. reflection across n n k 56 Reteaching 4.8 Geometr

90 NSWERS Lesson nswers ma var. Sample answer: Since the diagonals of a parallelogram bisect each other, T RT and QT ST. lso, vertical angles TQ and RTS are congruent, so TQ RTS b the SS ostulate. 9. Square; if the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle; if two adjacent sides of a parallelogram are congruent, then the parallelogram is a rhombus. 10. Rectangle; if the diagonals of a parallelogram are congruent, then the parallelogram is a rectangle. Lesson heck students drawings Lesson LM, LMN, LN, MN, LQ, LQM, MQN, QN 1. ' Lesson Yes; if one pair of opposite sides of a quadrilateral are both congruent and parallel, then the quadrilateral is a parallelogram. v ' 2. no 3. Yes; if both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram. 4. Yes; if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. 5. no 6. Rhombus; if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. 7. Rectangle; if one angle of a parallelogram is a right angle, then the parallelogram is a rectangle ' ' t ' 8. Rhombus; if a diagonal of a rhombus bisects a pair of opposite angles, then the parallelogram is a rhombus. ' z 60 Geometr nswers 167

91 NSWERS 4. Lesson 5.2 ' square meters square feet ' square centimeters 5. n ' square inches square meters square units square millimeters ' square feet square centimeters square units square meters ' k square inches square centimeters Reteaching hapter 5 Lesson units units units units units units units units units units square units square units 13. about 8 square units 14. h 21 inches, b 14 inches, 294 square inches 15. b 7 meters, h 15 meters, 44 meters ' square ards square units square millimeters square units square feet square millimeters Lesson nswers were obtained using 3.14 for π unless otherwise noted π units; units 2. 11π inches; inches 3. 8π meters; meters feet; 48 feet meters; 2.4 meters π 168 nswers Geometr

92 NME LSS DTE Reteaching 5.1 erimeter and rea Skill Finding the perimeter of a polgon The perimeter of a polgon is the sum of the length of its sides. The perimeter of a rectangle with base b and height h is 2b 2h. Eample Find the perimeter of each figure. a. Q 5 R b. J 2 2 S 3 3 U 4 T M 12 a units b. JKLM is a rectangle; 2b 2h 2(12) 2(8) 40 units. K 8 L Find the perimeter of each figure a rhombus 82 in. on a side 5. a regular pentagon with sides 9 m long 6. a rectangle with base 12 ft and height 5 ft 7. a rectangle with base 11 in. and height 8 in. 8. a regular octagon with sides 4 3 cm long 9. 6 a decagon with each side units long a dodecagon with each side units long 3 Geometr Reteaching

93 NME LSS DTE Skill Finding areas using the formula for the area of a rectangle To find the area of a rectangle with base b and height h, use the formula bh. To find the area of region that is not rectangular, ou ma be able to divide it into nonoverlapping rectangular regions or approimate its shape with rectangles. Eample 1 Find each area. ngles that appear to be right angles are right angles. a. b. 2 c a. Use the formula bh square units b. Divide the figure into rectangles c. Sample: There are 8 squares One method is shown. completel inside the figure ( ) square units and 12 partl inside the figure ( ) square units XX XXX XXX Eample 2 The height of a rectangle is 3 times its base. The perimeter is 48 inches. Find the area of the rectangle. Use 48 and h 3b to write an equation: 2b 2h 48 2b 2(3b) ; 48 8b; b 6 Then h 18 and bh The area is 108 square inches. Find the area. Estimate if necessar. ngles that appear to be right angles are right angles The perimeter of a rectangle is 70 inches. The base is the height. Find the base, 3 the height, and the area of the rectangle. 15. The area of a rectangle is 105 square meters. The height is 1 more than twice the base. Find the base, the height, and the perimeter of the rectangle. 58 Reteaching 5.1 Geometr

94 NSWERS 4. Lesson 5.2 ' square meters square feet ' square centimeters 5. n ' square inches square meters square units square millimeters ' square feet square centimeters square units square meters ' k square inches square centimeters Reteaching hapter 5 Lesson units units units units units units units units units units square units square units 13. about 8 square units 14. h 21 inches, b 14 inches, 294 square inches 15. b 7 meters, h 15 meters, 44 meters ' square ards square units square millimeters square units square feet square millimeters Lesson nswers were obtained using 3.14 for π unless otherwise noted π units; units 2. 11π inches; inches 3. 8π meters; meters feet; 48 feet meters; 2.4 meters π 168 nswers Geometr

95 NME LSS DTE Reteaching 5.2 reas of Triangles, arallelograms, and Trapezoids Skill Finding the area of a triangle The area of a triangle with base b and height h is 1. 2 bh Eample Find the area of. Since ou know the length of the altitude to be the base. Then b 16 and h 5, so square units., let 1 (1 16) Find the area of the triangle with the given base and height. 1. b 42 m, h 8 m 2. b 16 ft, h 0 ft 3. b 12.8 cm, h 9 cm Find the area of each triangle in. 11 in. 20 m 21 m 14 in. Skill Finding the area of a parallelogram 29 m The area of a parallelogram with base b and height h is bh. Eample Find the area of parallelogram JKLM Since ou know the length of the altitude to JK and LM, let JK be the base. Then b 19 and h 10, so M J L K (19 10) 190 square units. Geometr Reteaching

96 NME LSS DTE Find the area of the parallelogram with the given base and height. 7. b 42 m, h 8 m 8. b 16 ft, h 70 ft 9. b 12.8 cm, h 9 cm Find the area of each parallelogram m in m 5 m 18 in. 7 in. Skill Finding the area of a trapezoid The area of a trapezoid with bases and and height h is 1. 2 (b 1 b 2 ) b 1 b 2 Eample Find the area of trapezoid QRS. 32 Q Q and RS are the bases of QRS, so the height is Then 1 (32)(32 64) 1536 square units. 2 S 64 R Find the area of the trapezoid with the given bases and height. 13. b 1 15 cm, b 2 9 cm, h 7 cm 14. b 1 13 d, b 2 7 d, h 5 d Find the area of each trapezoid m m 6 m 7 m 7 6 m b 1 7 ft, b 2 10 ft, h 8 ft 19. b 1 40 m, b 2 32 m, h 20 m 60 Reteaching 5.2 Geometr

97 NSWERS 4. Lesson 5.2 ' square meters square feet ' square centimeters 5. n ' square inches square meters square units square millimeters ' square feet square centimeters square units square meters ' k square inches square centimeters Reteaching hapter 5 Lesson units units units units units units units units units units square units square units 13. about 8 square units 14. h 21 inches, b 14 inches, 294 square inches 15. b 7 meters, h 15 meters, 44 meters ' square ards square units square millimeters square units square feet square millimeters Lesson nswers were obtained using 3.14 for π unless otherwise noted π units; units 2. 11π inches; inches 3. 8π meters; meters feet; 48 feet meters; 2.4 meters π 168 nswers Geometr

98 NME LSS DTE Reteaching 5.3 ircumferences and reas of ircles Skill Using the formulas for the circumference of a circle The circumference of a circle with radius r and diameter d is 2πr or πd. n approimate value of the circumference can be obtained b substituting one of the 22 values 3.14 or for π. 7 Eample 1 Find the area of each circle. Give an eact answer and an approimation. a. b ft a. 22 r 7, so 2πr 14π; substituting for π, 22 (14) 44 units. 7 7 b. d 13 ft, so 13π ft; substituting 3.14 for π, 13(3.14) feet. Eample 2 circle has circumference 72 centimeters. Find the eact radius and an approimation πr; 72 2πr; r cm and r cm 2π π 3.14 For each circle, find the eact circumference and an approimation. If necessar, round to the nearest hundredth in. For each circle, find the eact radius and an approimation to the nearest tenth π ft m in. 8 m Geometr Reteaching

99 NME LSS DTE Skill Using the formula for the area of a circle The area of a circle with radius r is = πr 2. n approimate value of the area can be 22 obtained b substituting one of the values 3.14 or for π. 7 Eample 1 Find the area of each circle. (In c find the area of the shaded region.) Give an eact answer and an approimation. a. b. c ft 15 km 22 a. r 7, so πr 2 49π; substituting for π, 22 (49) 154 square units 7 7 b. d 13 ft, so r ft and 42.25π ft 2 ; substituting 3.14 for π and 2 rounding to the nearest hundredth, 42.25(3.14) ft 2. c. area of shaded region area of square area of circle π; substituting 3.14 for π, area km 2. Eample 2 circle has an area of 56 square units. Find the eact radius of the circle and an approimation to the nearest tenth. 56 πr 2 ; 56 πr 2 ; r 2 56, so r and r units π 56 π 3.14 For each shaded region, find the eact area and an approimation. If necessar, round to the nearest hundredth d 7.1 m For each circle, find the eact radius and an approimation to the nearest tenth. 5 m 12 in. 18 in π km m in Reteaching 5.3 Geometr

100 NSWERS 4. Lesson 5.2 ' square meters square feet ' square centimeters 5. n ' square inches square meters square units square millimeters ' square feet square centimeters square units square meters ' k square inches square centimeters Reteaching hapter 5 Lesson units units units units units units units units units units square units square units 13. about 8 square units 14. h 21 inches, b 14 inches, 294 square inches 15. b 7 meters, h 15 meters, 44 meters ' square ards square units square millimeters square units square feet square millimeters Lesson nswers were obtained using 3.14 for π unless otherwise noted π units; units 2. 11π inches; inches 3. 8π meters; meters feet; 48 feet meters; 2.4 meters π 168 nswers Geometr

101 NSWERS inches; 17.8 inches π π square ards; 616 square ards ( π 22 7 ) 8. 25π square meters; square meters 9. 45π square inches; square inches kilometers; 7.28 kilometers 11. meters; 9.5 meters 285 π 12. inches; 12.6 inches 498 π Lesson feet 8. es 9. es 10. es 11. no 12. es 13. es 14. no 15. es 16. es 17. acute 18. obtuse 19. obtuse 20. acute 21. obtuse 22. right 23. acute 24. acute 25. obtuse Lesson ; ; 4 3. ; ; ; ; ; ; square units square inches square meters square feet square units square meters square millimeters square inches Lesson ; ; ; ; nswers ma var due to when rounding is done ; Sample: Yes; the value calculated using the formula is Lesson ( a, a), ( a, a), D(a, a) 2. M(0, 0), N(0, a), (a, a) 3. J( a, 0) 4. S( b, c) 5. G(0, 0), E(a b, c) 6. D(0, 0), F(2a, 0) 7. Sample proof: Let, Q, R, and S be as shown in the figure. ( b, c) S( a, 0) Q(b, c) R(a, 0) R (a ( b)) 2 (0 c) 2 (a b) 2 c 2 and QS (b ( a)) 2 (c 0) 2 (a b) 2 c 2, so R QS. Geometr nswers 169

102 NME LSS DTE Reteaching 5.4 The thagorean Theorem Skill Using the thagorean Theorem In an right triangle, the square the length of the hpotenuse is equal to the sum of the squares of the other two sides. For the right triangle shown, c 2 a 2 b 2. Eample The hpotenuse of a right triangle is 9 inches long and one leg is 5 inches long. Find the length of the other leg. Let c 9, a 5, and b the length of the other leg. the thagorean Theorem, a 2 b 2 c 2, so b , and b c b a The lengths of two sides of a right triangle are given. Find the length of the third side. If necessar, write our answer in simplest radical form. 1. a 20, b 21, c 2. a 8, b 15, c 3. a 5, b 8, c 4. b 12, c 19, a 5. a 10, c 20, b 6. a 5, c 11, b 7. rectangular field is 180 feet long and 50 feet wide. Find the length of a diagonal of the field. Give our answer in simplest radical form and as a decimal rounded to the nearest tenth of a foot. Skill Using the converse of the thagorean Theorem If the square of the length of the longest side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle is a right triangle. Eample The sides of a triangle are 9 cm, 40 cm, and 41 cm long. Is the triangle a right triangle? Square the length of each side: dd the squares of the lengths of the two shorter sides: ompare: the converse of the thagorean Theorem, the triangle is a right triangle. Geometr Reteaching

103 NME LSS DTE The lengths of the sides of a triangle are given. Is the triangle a right triangle? 8. 12, 16, in., 12 in., 13 in m, 24 m, 26 m km, 15 km, 16 km 12. 7, 7, cm, 12 cm, 12.5 cm ft, 6 ft, 2 15 ft mm, 9 3 mm, 18 mm , 20, 25 Skill Using the thagorean Inequalities Given an two real numbers and, eactl one of the following is true. Let be a triangle with longest side of length c and shorter sides of lengths a and b. The converse of the thagorean Theorem deals with the case in which c 2 a 2 b 2. The thagorean Inequalities deal with the other two possibilities. If c 2 a 2 b 2, then is an obtuse triangle. If c 2 a 2 b 2, then is an acute triangle. Eample The sides of a triangle are 23 feet, 15 feet, and 12 feet long. Is the triangle right, obtuse, or acute? Square the length of each side: dd the squares of the lengths of the two shorter sides: ompare: The triangle is an obtuse triangle. The lengths of the sides of a triangle are given. Determine whether the triangle is right, acute, or obtuse , 20, ft, 9 ft, 14 ft in., 4 in., 7 in m, 18 m, 22 m , 8, mm, 60 mm, 61 mm 23. 8, 8, ft, 15 ft, 15 ft , 40, Reteaching 5.4 Geometr

104 NSWERS inches; 17.8 inches π π square ards; 616 square ards ( π 22 7 ) 8. 25π square meters; square meters 9. 45π square inches; square inches kilometers; 7.28 kilometers 11. meters; 9.5 meters 285 π 12. inches; 12.6 inches 498 π Lesson feet 8. es 9. es 10. es 11. no 12. es 13. es 14. no 15. es 16. es 17. acute 18. obtuse 19. obtuse 20. acute 21. obtuse 22. right 23. acute 24. acute 25. obtuse Lesson ; ; 4 3. ; ; ; ; ; ; square units square inches square meters square feet square units square meters square millimeters square inches Lesson ; ; ; ; nswers ma var due to when rounding is done ; Sample: Yes; the value calculated using the formula is Lesson ( a, a), ( a, a), D(a, a) 2. M(0, 0), N(0, a), (a, a) 3. J( a, 0) 4. S( b, c) 5. G(0, 0), E(a b, c) 6. D(0, 0), F(2a, 0) 7. Sample proof: Let, Q, R, and S be as shown in the figure. ( b, c) S( a, 0) Q(b, c) R(a, 0) R (a ( b)) 2 (0 c) 2 (a b) 2 c 2 and QS (b ( a)) 2 (c 0) 2 (a b) 2 c 2, so R QS. Geometr nswers 169

105 NME LSS DTE Reteaching 5.5 Special Triangles and reas of Regular olgons Skill Using the Triangle Theorem and the Triangle Theorem triangle is a right triangle in which both acute angles have measure 45. In a triangle, if the length of each leg is, the length of the hpotenuse is 2. If the length of the 2 hpotenuse is, then the length of each leg is or triangle is a right triangle in which the acute angles have measures of 30 and 60. The shorter leg of the triangle is opposite the 30 angle and the longer leg is opposite the 60 angle. In a triangle, if the length of the shorter leg is, then the length of the longer leg is 3, and the length of the hpotenuse is Eample a. The legs of a triangle are each 15 centimeters long. Find the length, h, of the hpotenuse to the nearest tenth of a centimeter. b. The longer leg of a triangle is 5 inches long. Find the length, h, of the hpotenuse in simplest radical form. a. Refer to the triangle in the figure above. Since 15, h centimeters. b. Refer to the triangle in the figure above. Since 3 5, and h inches. The length of one side of a triangle is given. Find the lengths of the other two sides in simplest radical form. 1. d 13, e 2. d 4, e f 3. e 4.5, f 4. f 9 2, d d The length of one side of a triangle is given. Find the lengths of the other two sides in simplest radical form. f e e 45 d f z , , z z , 8. 18, z z Geometr Reteaching

106 NME LSS DTE Find the area of each figure. Round our answers to the nearest tenth if necessar in m 96 6 in in in m Skill Finding the area of a regular polgon The apothem of a regular polgon is the length of a perpendicular segment from the center of the polgon to a side. The apothem can be used to determine the area of the polgon. The area of a regular polgon with apothem a and perimeter p is 1. 2 ap Eample Find the area of a regular heagon with sides that are 8 centimeters long. Use the Triangle Theorem to determine the apothem. In the triangle shown, the length of the shorter leg is 4, so the apothem is 4 3. Since the perimeter is , a Find the area of each regular polgon. Give our answer in simplest radical form a 10 ft 20 ft a regular octagon with sides that are 15 mm long and an apothem of 18.1 mm 16. a regular decagon with sides that are 10 in. long and an apothem of 15.4 in. 6 a 14 m 9.6 m 7 m 66 Reteaching 5.5 Geometr

107 NSWERS inches; 17.8 inches π π square ards; 616 square ards ( π 22 7 ) 8. 25π square meters; square meters 9. 45π square inches; square inches kilometers; 7.28 kilometers 11. meters; 9.5 meters 285 π 12. inches; 12.6 inches 498 π Lesson feet 8. es 9. es 10. es 11. no 12. es 13. es 14. no 15. es 16. es 17. acute 18. obtuse 19. obtuse 20. acute 21. obtuse 22. right 23. acute 24. acute 25. obtuse Lesson ; ; 4 3. ; ; ; ; ; ; square units square inches square meters square feet square units square meters square millimeters square inches Lesson ; ; ; ; nswers ma var due to when rounding is done ; Sample: Yes; the value calculated using the formula is Lesson ( a, a), ( a, a), D(a, a) 2. M(0, 0), N(0, a), (a, a) 3. J( a, 0) 4. S( b, c) 5. G(0, 0), E(a b, c) 6. D(0, 0), F(2a, 0) 7. Sample proof: Let, Q, R, and S be as shown in the figure. ( b, c) S( a, 0) Q(b, c) R(a, 0) R (a ( b)) 2 (0 c) 2 (a b) 2 c 2 and QS (b ( a)) 2 (c 0) 2 (a b) 2 c 2, so R QS. Geometr nswers 169

108 NME LSS DTE Reteaching 5.6 The Distance Formula and the Method of Quadrature Skill Using the distance formula The thagorean Theorem can be used to determine the distance between two points in a coordinate plane. The distance between two points ( 1, 1 ) and ( 2, 2 ) in a coordinate plane is d ( 2 1 ) 2 ( 2 1 ) 2. To determine the length of a segment in the plane, use the distance formula to determine the distance between the endpoints. Eample 1 Find the distance between the points ( 4, 7) and (3, 8). Give the answer in simplest radical form and as a decimal rounded to the nearest hundredth. Let ( 1, 1 ) ( 4, 7) and ( 2, 2 ) (3, 8). d (3 ( 4)) 2 (8 7) d 7.07 Eample 2 triangle has vertices ( 3, 3), Q( 1, 5), and R(6, 0). Find the perimeter of QR to the nearest tenth. Q (5 ( 3)) 2 ( 1 ( 3)) QR (6 ( 1)) 2 (0 5) ( 5) R (6 ( 3)) 2 (0 ( 3)) perimeter of QR Find the distance between the points. Give our answer in simplest radical form and as a decimal rounded to the nearest hundredth. 1. (8, 5) and (3, 7) 2. ( 10, 8) and ( 6, 5) 3. (4, 5) and (7, 10) 4. ( 2, 8) and (8, 2) 5. (7, 7) and ( 3, 2) 6. (12, 5) and (18, 8) The vertices of are given. Find the perimeter of to the nearest tenth. 7. (0, 0), (5, 10), and ( 2, 4) 8. (5, 8), ( 6, 6), and (1, 10) Geometr Reteaching

109 NME LSS DTE 9. ( 4, 4), (3, 7) and (8, 2) 10. (1, 1), (5, 12), and (6, 4) Skill Using quadrature to estimate the area under a curve Quadrature is a technique involving approimating the area of a region of a coordinate plane using the sum of the areas of rectangles. Eample Estimate the area of a quarter circle with radius 3 and center O(0, 0). We will use two different methods and find the average. We will determine E, D, D, and b using the thagorean Theorem. 1. (0, 3) O E D 3 O D 3 Use the given rectangles to estimate the area of the region to the nearest hundredth D E 4 2 O 2 In Figure 1, E is a leg of a right triangle with hpotenuse 3 and base 1, so E Similarl, D 5. In Figure 2, D 2 2 and 5. Then the area from Figure 1 is , or about The area from Figure 2 is 2 2 5, or about The average is about Using the formula for the area of a circle, the area is π , so this is a 4 fairl good estimate. 4 F O D E 6 F 13. verage our estimates from Eercises 11 and 12. Is it a reasonable estimate? 68 Reteaching 5.6 Geometr

110 NSWERS inches; 17.8 inches π π square ards; 616 square ards ( π 22 7 ) 8. 25π square meters; square meters 9. 45π square inches; square inches kilometers; 7.28 kilometers 11. meters; 9.5 meters 285 π 12. inches; 12.6 inches 498 π Lesson feet 8. es 9. es 10. es 11. no 12. es 13. es 14. no 15. es 16. es 17. acute 18. obtuse 19. obtuse 20. acute 21. obtuse 22. right 23. acute 24. acute 25. obtuse Lesson ; ; 4 3. ; ; ; ; ; ; square units square inches square meters square feet square units square meters square millimeters square inches Lesson ; ; ; ; nswers ma var due to when rounding is done ; Sample: Yes; the value calculated using the formula is Lesson ( a, a), ( a, a), D(a, a) 2. M(0, 0), N(0, a), (a, a) 3. J( a, 0) 4. S( b, c) 5. G(0, 0), E(a b, c) 6. D(0, 0), F(2a, 0) 7. Sample proof: Let, Q, R, and S be as shown in the figure. ( b, c) S( a, 0) Q(b, c) R(a, 0) R (a ( b)) 2 (0 c) 2 (a b) 2 c 2 and QS (b ( a)) 2 (c 0) 2 (a b) 2 c 2, so R QS. Geometr nswers 169

111 NME LSS DTE Reteaching 5.7 roofs Using oordinate Geometr Skill Using coordinate geometr to represent geometric properties of figures oordinate proofs are particularl convenient for proving that segments are congruent, that two lines intersect or are parallel or perpendicular, and that a line or segment bisects another segment In writing a coordinate proof, it is helpful to position the aes and assign coordinates in such a wa that the calculations are as simple as possible. If a figure is smmetric, position the aes so that one of them is the line of smmetr. If a figure is not smmetric, position the aes so that as man vertices as possible lie on an ais (and have at least one 0 coordinate). When working with midpoints, assign even coordinates such as 2a and 2b. lwas be careful not to assign coordinates that give the figure properties it does not actuall have. Eample D is a rectangle. Give coordinates of the unknown vertices using onl the variables shown. (a, b) D is a rectangle, so point has the same -coordinate as point and the same -coordinate as point, so ( a, b). oint D has the same ( a, b) D -coordinate as point and the same -coordinate as point, so D (a, b). If possible, give the coordinates of the unknown verte or vertices without introducing an new variables. If not, use another variable. 1. square D 2. square MNQ 3. isosceles triangle JKL (a, a) 4. isosceles trapezoid QRS 5. parallelogram DEFG 6. equilateral triangle XYZ D N M Q(a, 0) J K(0, c) L(a, 0) S R(b, c) F( b, c) E E(a, b) ( a, 0) Q(a, 0) G D(a, 0) D F Geometr Reteaching

112 NME LSS DTE Skill Using coordinate geometr in proofs The Distance Formula and the Midpoint Formula are frequentl used in coordinate proofs, as are the definition of slope and the equation of a line. Eample Given: isosceles triangle ; rove: The medians to the legs of are congruent. (0, 2b) roof: is isosceles so let the -ais be an ais of smmetr for. Since we want to determine the coordinates of midpoints, make all coordinates ( 2a, 0) (2a, 0) multiples of 2, as shown in the figure. the Midpoint Formula, the midpoint, M, of has coordinates ( 2a 0 0 2b,, ( a, b). Similarl, the midpoint of is N(a, b). 2 2 ) M (2a ( a)) 2 (0 b) 2 9a 2 b 2. N (2a a) 2 (0 b) 2 9a 2 b 2. Make a sketch and write a coordinate proof. 7. The diagonals of an isosceles trapezoid are congruent. Given: isosceles trapezoid QRS, S QR rove: R QS roof: 8. The segment joining the midpoints of consecutive sides of a rectangle is a rhombus. Given: rectangle WXYZ, with midpoints M, N,, and Q rove: MNQ is a rhombus. roof: 70 Reteaching 5.7 Geometr

113 NSWERS inches; 17.8 inches π π square ards; 616 square ards ( π 22 7 ) 8. 25π square meters; square meters 9. 45π square inches; square inches kilometers; 7.28 kilometers 11. meters; 9.5 meters 285 π 12. inches; 12.6 inches 498 π Lesson feet 8. es 9. es 10. es 11. no 12. es 13. es 14. no 15. es 16. es 17. acute 18. obtuse 19. obtuse 20. acute 21. obtuse 22. right 23. acute 24. acute 25. obtuse Lesson ; ; 4 3. ; ; ; ; ; ; square units square inches square meters square feet square units square meters square millimeters square inches Lesson ; ; ; ; nswers ma var due to when rounding is done ; Sample: Yes; the value calculated using the formula is Lesson ( a, a), ( a, a), D(a, a) 2. M(0, 0), N(0, a), (a, a) 3. J( a, 0) 4. S( b, c) 5. G(0, 0), E(a b, c) 6. D(0, 0), F(2a, 0) 7. Sample proof: Let, Q, R, and S be as shown in the figure. ( b, c) S( a, 0) Q(b, c) R(a, 0) R (a ( b)) 2 (0 c) 2 (a b) 2 c 2 and QS (b ( a)) 2 (c 0) 2 (a b) 2 c 2, so R QS. Geometr nswers 169

114 NSWERS 8. Sample proof: Let W, X, Y, and Z be as shown. 2. X( a, b) N Y(a, b) D E M 3. W( a, b) Q Z(a, b) D E MNQ is a parallelogram since the midpoint of NQ and M is (0,0). M, N,, and Q lie on the aes, so the diagonals, M and NQ, of MNQ are perpendicular and MNQ is a rhombus (Theorem 4.6.8: If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.) 4. front right back top left bottom Lesson ; 0.375; 37.5% 8 5. front back left 1 2. ; 0.5; 50% 2 right top bottom 1 3. ; 0.125; 12.5% ; 0.25; 25% ; 0.2; 20% ; 0.1; 11 % ; 0.83; % Reteaching hapter 6 Lesson ; 0.16; % D sq. units; 8 cu. units sq. units; 8 cu. units sq. units; 11 cu. units Lesson JKLM, JMRN 2. JKN, MLQR 3. JK, ML, N, RQ 4. ML, RQ, MR, LQ 5. a. FE, E b. 6. a. F, E b. F, DE 7. MLQR 8. JMRN 9. perpendicular 170 nswers Geometr

115 NME LSS DTE Reteaching 5.8 Geometric robabilit Skill Determining the theoretical probabilit of an event The probabilit of an event is a number that indicates how likel it is that the event will occur. The probabilit of an event is alwas between 0 and 1. If the probabilit of an event is 0, the event cannot occur. If the probabilit of an event is 1, the event is sure to happen. probabilit can be epressed as a fraction, a decimal, or a percent. If all possible outcomes are equall likel, the theoretical probabilit of an event is number of favorable outcomes. favorable outcome is one for which the event occurs. number of possible outcomes Eample bag of 20 marbles contains 7 red marbles, 5 green marbles, and 8 ellow marbles. You remove one marble from the bag without looking. Find the probabilit that the marble is ellow. ll of the twent outcomes are equall likel, and eight are favorable. Then %. 5 The spinner at the right is spun once. Find the probabilit that the spinner lands on the number indicated. Give each probabilit as a fraction in lowest terms, as a decimal, and as a percent a number greater than 5 2. an even number 3. a number less than 2 Skill Determining the geometric probabilit of an event Geometric probabilities are based on length or on area. Eample a. point on F is chosen at random. Find the probabilit,, that the point is on. b. point in rectangle D is chosen at random. Find the probabilit,, that the point is inside the circle. length of a. length of F % 5 area of circle b. area of rectangle π % 24 D E F Geometr Reteaching

116 NME LSS DTE In Eercises 4 5, use the given segment to find the indicated probabilit. 4. You are epecting a deliver of a package between 12 noon and 4:00.M. You are called awa at 1:00.M. and return half an hour later. Find the probabilit that ou miss the deliver bus arrives at a stop ever 10 minutes and waits 2 minutes before leaving. Find the probabilit that if ou arrive at a random time, there is a bus waiting In Eercises 6 8, find the probabilit that a point chosen at random in the figure is inside the shaded region In Eercises 9 11, a dart is tossed at random onto the board. Find the probabilit that a dart that hits the board hits the shaded region. Give our answer as a decimal rounded to the nearest hundredth. (Use π 3.14.) m 2 m 2 m 5 m 12 in. 9 m 3 ft 2 ft 72 Reteaching 5.8 Geometr

117 NSWERS 8. Sample proof: Let W, X, Y, and Z be as shown. 2. X( a, b) N Y(a, b) D E M 3. W( a, b) Q Z(a, b) D E MNQ is a parallelogram since the midpoint of NQ and M is (0,0). M, N,, and Q lie on the aes, so the diagonals, M and NQ, of MNQ are perpendicular and MNQ is a rhombus (Theorem 4.6.8: If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.) 4. front right back top left bottom Lesson ; 0.375; 37.5% 8 5. front back left 1 2. ; 0.5; 50% 2 right top bottom 1 3. ; 0.125; 12.5% ; 0.25; 25% ; 0.2; 20% ; 0.1; 11 % ; 0.83; % Reteaching hapter 6 Lesson ; 0.16; % D sq. units; 8 cu. units sq. units; 8 cu. units sq. units; 11 cu. units Lesson JKLM, JMRN 2. JKN, MLQR 3. JK, ML, N, RQ 4. ML, RQ, MR, LQ 5. a. FE, E b. 6. a. F, E b. F, DE 7. MLQR 8. JMRN 9. perpendicular 170 nswers Geometr

118 NME LSS DTE Reteaching 6.1 Solid Shapes Skill Using isometric dot paper to draw three-dimensional figures ll segments drawn on isometric dot paper b connecting two consecutive dots on a vertical or a diagonal line are the same length, or isometric. Eample The figure consists of five unit cubes. ube E, which is not visible, is underneath cube D. Redraw the figure, placing cube D net to E, so that cubes, D, and E are in a row. D op the figure and make erasures to remove cube D. Redraw lines to restore cube E. Then draw the visible edges of cube D in its new position. E D Refer to the top figure in the eample. Redraw the figure with the indicated change. 1. dd a cube on top of. 2. lace on top of. 3. lace in front of. Skill Drawing orthographic projections n isometric drawing appears to be three-dimensional. Orthographic projection, however, produces an image in which a three-dimensional figure appears to be two-dimensional. Eample In the top figure in the eample above, let the face lettered be the front and the faces lettered and be the left side. Draw orthographic projections showing the front, the back, the left, the right, the top, and the bottom. front back left right top bottom Geometr Reteaching

119 NME LSS DTE Draw si orthographic projections of each figure. The front and left faces of each figure are indicated left front left front Skill Determining surface area and volume using unit cubes Each face of a unit cube has area 1 square unit. The volume of a unit cube is 1 cubic unit. To determine the surface area of a figure formed from unit cubes, count all eposed cube faces, that is, faces not covered b another cube. To determine the volume, count the number of unit cubes in the figure. Eample Determine the surface area and volume of the figure. One method of determining surface is area is to consider each unit cube and determine how man of its faces are eposed. For eample, the topmost cube has five eposed faces, while the cube at the right front corner has four. The total number of eposed faces is 28, so the surface area of the figure is 28 square units. There are 7 cubes so the volume is 7 cubic units. Determine the surface area and volume of the figure Reteaching 6.1 Geometr

120 NSWERS 8. Sample proof: Let W, X, Y, and Z be as shown. 2. X( a, b) N Y(a, b) D E M 3. W( a, b) Q Z(a, b) D E MNQ is a parallelogram since the midpoint of NQ and M is (0,0). M, N,, and Q lie on the aes, so the diagonals, M and NQ, of MNQ are perpendicular and MNQ is a rhombus (Theorem 4.6.8: If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.) 4. front right back top left bottom Lesson ; 0.375; 37.5% 8 5. front back left 1 2. ; 0.5; 50% 2 right top bottom 1 3. ; 0.125; 12.5% ; 0.25; 25% ; 0.2; 20% ; 0.1; 11 % ; 0.83; % Reteaching hapter 6 Lesson ; 0.16; % D sq. units; 8 cu. units sq. units; 8 cu. units sq. units; 11 cu. units Lesson JKLM, JMRN 2. JKN, MLQR 3. JK, ML, N, RQ 4. ML, RQ, MR, LQ 5. a. FE, E b. 6. a. F, E b. F, DE 7. MLQR 8. JMRN 9. perpendicular 170 nswers Geometr

121 NME LSS DTE Reteaching 6.2 Spatial Relationships Skill Identifing relationships between lines and planes in space K L The space figure shown is a called a rectangular solid, since each face is rectangular. Such a figure can be used to illustrate relationships between lines and planes in space. J N R M Q line is perpendicular to a plane at point if the line is perpendicular to ever line in the plane that passes through. line is parallel to a plane if the line and the plane do not intersect. Eample Refer to the figure above. Name each of the following. a. a plane perpendicular to b. a plane parallel to JK JK a. lane JMRN is perpendicular to JK at J and plane KLQ is perpendicular to JK at K. b. lanes NQR and MLQR are both parallel to JK. Refer to the figure in the eample above. Name each of the following. 1. all planes shown that are parallel to 2. all planes shown that are perpendicular to 3. all lines shown that are perpendicular to plane KLQ 4. all lines shown that are parallel to plane JKN Eercises 5 and 6 refer to the figure at the right. Two surfaces, and DE, of the space figure are right triangles. Name all lines shown that are: 5. a. parallel to plane D b. parallel to plane DEF 6. a. perpendicular to plane DEF b. perpendicular to plane EF Q KL F D E Geometr Reteaching

122 NME LSS DTE Skill Identifing relationships between planes Two lines in a plane ma be parallel or ma intersect. The same is true of two planes in space. Two planes that do not intersect are parallel. Two planes that intersect form dihedral angles. In a plane, the sides of an angle are ras and their intersection is the verte of the angle. In space, the sides of a dihedral angle are half-planes, called the faces of the angle, and their intersection is a line, called the edge of the angle. The measure of a dihedral angle is the measure of an angle formed b ras on the faces that are perpendicular to the edge. Eample a. Name a plane parallel to plane JKLM. K L b. Identif an angle whose measure ou could use to determine the measure of the dihedral angle J M formed b the intersection of planes JKLM and MLQR. Then give the measure of the dihedral Q angle. a. lane NQR is parallel to plane JKLM. b. JMR or KLQ; 90 N R Refer to the figure above. omplete each statement. 7. lane is parallel to plane JKN. 8. lane KLQ is parallel to plane. 9. Since m JN 90, planes JKN and NQR are. 10. lanes,,, and are perpendicular to plane MLQR. 11. The measure of the dihedral angle formed b the intersection of planes KLRN and NQR is equal to m or m. 12. If JKN and MLQR are squares, then the measure of the dihedral angle formed b the intersection of planes KLRN and NQR is. 76 Reteaching 6.2 Geometr

123 NSWERS 8. Sample proof: Let W, X, Y, and Z be as shown. 2. X( a, b) N Y(a, b) D E M 3. W( a, b) Q Z(a, b) D E MNQ is a parallelogram since the midpoint of NQ and M is (0,0). M, N,, and Q lie on the aes, so the diagonals, M and NQ, of MNQ are perpendicular and MNQ is a rhombus (Theorem 4.6.8: If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.) 4. front right back top left bottom Lesson ; 0.375; 37.5% 8 5. front back left 1 2. ; 0.5; 50% 2 right top bottom 1 3. ; 0.125; 12.5% ; 0.25; 25% ; 0.2; 20% ; 0.1; 11 % ; 0.83; % Reteaching hapter 6 Lesson ; 0.16; % D sq. units; 8 cu. units sq. units; 8 cu. units sq. units; 11 cu. units Lesson JKLM, JMRN 2. JKN, MLQR 3. JK, ML, N, RQ 4. ML, RQ, MR, LQ 5. a. FE, E b. 6. a. F, E b. F, DE 7. MLQR 8. JMRN 9. perpendicular 170 nswers Geometr

124 NSWERS 10. JKLM; JMRN; NQR; KLQ 11. LRQ; KN Lesson No; the figure does not have two parallel faces and onl one face is a parallelogram. 2. Yes; right rectangular prism. 3. Yes; oblique pentagonal prism. 4. right trapezoidal prism 5. D and EFGH D, EF, GH E, F, G 8. FE, DGH 9. angles E, EF, F, FE, F, FG, G, GF, DG, GH, GHD, DH, DE, EH, EHD, DH ; ; ; ; front-right-bottom 10. (2, 0, 2) 11. (0, 0, 2) 12. (2, 4, 2) 13. (0, 4, 0) 14 (2, 4, 0) ; ; ; ; ; ; ; 4.90 Lesson (5, 0, 0); (0, 7.5, 0); (0, 0, 3) 2. (8, 0, 0); (0, 2, 0); (0, 0, 4) 3. (3, 0, 0); (0, 6, 0); (0, 0, 9) 4. z (0, 0, 4) ( 1, 0, 0) (0, 4, 0) ; ; ; ; ; 7.42 Lesson z ( 1, 1, 4) (2, 2, 5) (3, 2, 3) (2, 4, 2) 5. z (0, 0, 3) (0, 4, 0) (3, 0, 0) (0, 5, 1) (4, 0, 0) 7. -ais 8. back-right-bottom Geometr nswers 171

125 NME LSS DTE Reteaching 6.3 risms Skill Identifing prisms and properties of their bases, faces, and edges The bases of a prism are congruent polgons that lie in parallel planes. prism is named for the shape of its bases. For eample, if the bases are pentagons, the prism is called a pentagonal prism. The faces that are not bases are called lateral faces. The segments joining corresponding vertices in the bases are called lateral edges. ll lateral faces of a prism are parallelograms and all lateral edges are congruent. If the lateral faces are rectangles, the prism is called a right prism. If the faces are not rectangles, the prism is oblique. Eample The figure at the right is an oblique triangular prism. R QR and STU are equilateral triangles. a. Name the bases of the prism. b. Name all segments congruent to Q. Q U c. Name all segments congruent to S. d. re an of the lateral faces congruent? S e. re an angles of the prism right angles? T a. QR and STU b. QR and STU are congruent equilateral triangles so, QR, R, ST, TU, and SU are all congruent to Q. c. ll lateral edges of a prism are congruent, so QT and RU are congruent to S. d. Yes; because the bases are equilateral triangles, all three lateral faces are congruent. e. No; the angles of the bases are acute angles. Since the lateral faces are parallelograms that are not rectangles, none of their angles are right angles. Tell whether the figure appears to be a prism. If so, name the tpe of prism. If not, tell wh not Geometr Reteaching

126 NME LSS DTE The figure at the right is a right prism. The bases are isosceles trapezoids. 4. What tpe of prism is the figure? D 5. Name the bases of the prism. E H 6. Name all segments congruent to. 7. Name all segments congruent to DH. 8. Name an congruent lateral faces. 9. Name an right angles in the prism. F G Skill Using the formula for the length of a diagonal of a right rectangular prism diagonal of a polhedron is a segment that joins two points that are vertices of different faces of the polhedron. The thagorean Theorem can be used to derive a formula for the length of a diagonal of a right rectangular prism. If a right rectangular prism has length, width w, and height h, then the length, d, of a diagonal is given b d 2 w 2 h 2. Eample Find the length of a diagonal of 3 m the right rectangular prism. d d 2 w 2 h m Find the length of a diagonal of a right rectangular prism with the given dimensions. Give our answer as a radical in simplest form and as a decimal rounded to the nearest hundredth , w 4, h , w 2, h , w 7, h , w 2, h , w 6, h , w 12, h , w 10, h , w 8, h 4 Find the missing dimension of the right rectangular prism. Give our answer as a radical in simplest form and as a decimal rounded to the nearest hundredth. 18. d 17, w 9, 12, h 19. d 12, w 8, h 5, 5 m 2 m 78 Reteaching 6.3 Geometr

127 NSWERS 10. JKLM; JMRN; NQR; KLQ 11. LRQ; KN Lesson No; the figure does not have two parallel faces and onl one face is a parallelogram. 2. Yes; right rectangular prism. 3. Yes; oblique pentagonal prism. 4. right trapezoidal prism 5. D and EFGH D, EF, GH E, F, G 8. FE, DGH 9. angles E, EF, F, FE, F, FG, G, GF, DG, GH, GHD, DH, DE, EH, EHD, DH ; ; ; ; front-right-bottom 10. (2, 0, 2) 11. (0, 0, 2) 12. (2, 4, 2) 13. (0, 4, 0) 14 (2, 4, 0) ; ; ; ; ; ; ; 4.90 Lesson (5, 0, 0); (0, 7.5, 0); (0, 0, 3) 2. (8, 0, 0); (0, 2, 0); (0, 0, 4) 3. (3, 0, 0); (0, 6, 0); (0, 0, 9) 4. z (0, 0, 4) ( 1, 0, 0) (0, 4, 0) ; ; ; ; ; 7.42 Lesson z ( 1, 1, 4) (2, 2, 5) (3, 2, 3) (2, 4, 2) 5. z (0, 0, 3) (0, 4, 0) (3, 0, 0) (0, 5, 1) (4, 0, 0) 7. -ais 8. back-right-bottom Geometr nswers 171

128 NME LSS DTE Reteaching 6.4 oordinates in Three Dimensions Skill Locating points in three dimensions The position of a point in a plane is determined b an ordered pair (, ). The position of a point is space is determined b an ordered triple (,, z) Each pair of aes determine a plane and the three planes (the -plane, the zplane, and the z-plane) divide space into eight octants. The first octant is the octant in which all coordinates of ever point are positive. The phrases front or back, left or right, and top or bottom describe the position of an octant with respect to the z-plane, the z-plane, and the -plane, respectivel. The first octant could also be referred to as the front-right-top octant. Eample lot the point (3, 1, 2). Determine the octant in which the point is located. egin at the origin and move three units in the z positive direction (forward) on the -ais. Mark the -ais at that point. Move one unit in the negative direction (to the left) parallel to the -ais. Draw a line showing the path. Finall, move two units in the positive (3, 1, 2) direction (up) parallel to the z-ais. gain, draw a line representing our path. Label the point where ou finish (3, 1, 2). Since the point is in front of the z-plane,to the left of the z-plane, and above the -plane, it is in the front-left-top octant. lot each point on the coordinate aes provided. 1. (4, 0, 0) 2. (2, 4, 2) 3. ( 1, 1, 4) 4. (3, 2, 3) 5. (0, 5, 1) 6. (2, 2, 5) Determine the octant in which the point is located. If the point is on an ais, identif the ais. 7. (0, 12, 0) 8. ( 2, 2, 2) 9. (3, 4, 5) Geometr Reteaching

129 NME LSS DTE Use the given coordinates of the vertices of a right rectangular solid to find the coordinates of the following vertices: 10. J z 11. K K L (0, 4, 2) 12. M Q J M O (0, 0, 0) N (2, 0, 0) Q Skill Using the distance formula for three dimensions The distance between two points in space, ( 1, 1, z 1 ) and ( 2, 2, z 2 ), is given b d ( 2 1 ) 2 ( 2 1 ) 2 (z 2 z 1 ) 2 Eample Find the distance between the points (4, 3, 2) and (5, 0, 1). d ( 2 1 ) 2 ( 2 1 ) 2 (z 2 z 1 ) 2 (5 4) 2 (0 3) 2 (1 2) Find the distance between the given points. Give our answer in simplest radical form and as a decimal rounded to the nearest hundredth. 15. (3, 2, 0) and (0, 7, 4) 16. (2, 4, 2) and (3, 5, 2) 17. ( 3, 5, 2) and (4, 1, 1) 18. (5, 5, 5) and ( 2, 1, 5) 19. (8, 10, 2) and ( 2, 0, 7) In the figure above, give the length of each segment in simplest radical form and as a decimal rounded to the nearest hundredth. 20. JK 21. KM 22. M 23. KQ 80 Reteaching 6.4 Geometr

130 NSWERS 10. JKLM; JMRN; NQR; KLQ 11. LRQ; KN Lesson No; the figure does not have two parallel faces and onl one face is a parallelogram. 2. Yes; right rectangular prism. 3. Yes; oblique pentagonal prism. 4. right trapezoidal prism 5. D and EFGH D, EF, GH E, F, G 8. FE, DGH 9. angles E, EF, F, FE, F, FG, G, GF, DG, GH, GHD, DH, DE, EH, EHD, DH ; ; ; ; front-right-bottom 10. (2, 0, 2) 11. (0, 0, 2) 12. (2, 4, 2) 13. (0, 4, 0) 14 (2, 4, 0) ; ; ; ; ; ; ; 4.90 Lesson (5, 0, 0); (0, 7.5, 0); (0, 0, 3) 2. (8, 0, 0); (0, 2, 0); (0, 0, 4) 3. (3, 0, 0); (0, 6, 0); (0, 0, 9) 4. z (0, 0, 4) ( 1, 0, 0) (0, 4, 0) ; ; ; ; ; 7.42 Lesson z ( 1, 1, 4) (2, 2, 5) (3, 2, 3) (2, 4, 2) 5. z (0, 0, 3) (0, 4, 0) (3, 0, 0) (0, 5, 1) (4, 0, 0) 7. -ais 8. back-right-bottom Geometr nswers 171

131 NME LSS DTE Reteaching 6.5 Lines and lanes in Space Skill Sketching planes in space n equation of a plane can be written in the form z D, where,,, and D are real numbers that are not all zero, and is nonnegative. Just as in a plane, ou can graph an equation b locating its intercepts. To determine the - intercept, set and z equal to 0 and solve the equation for. Then plot the point (, 0, 0). Use a similar method to plot the points (0,, 0) and (0, 0, z). Draw the triangle determined b the points and shade the plane. Eample Sketch the plane with equation 6 4 3z 12. Determine the intercepts. 6 4(0) 3(0) 12; 2 6(0) 4 3(0) 12; 3 6(0) 4(0) 3z 12; z 4 lot the points ( 2, 0, 0), (0, 3, 0), and (0, 0, 4). Sketch the triangle determined b the points and shade the plane. (0, 3, 0) z (0, 0, 4) (1, 0, 0) Determine the -, -, and z-intercepts of the graph of the equation z z z 18 Sketch the plane with the given equation z z z 3 z z z Geometr Reteaching

132 NME LSS DTE Skill Sketching lines in space It ma seem that since is an equation of a line in a plane, that z D should be an equation of a line in space. However, ou have seen that the graph of such an equation is a plane, not a line. Then what does the equation of a line in space look like? In our stud of algebra, ou learned about parametric equations, equations that epress the variables in terms of another variable, called the parameter. arametric equations provide one wa to describe an equation of a line in space. Eample Graph the line with the given parametric equations. t 1; t 1; z 2t Find the values of,, and z for given values of t. In space, as in a plane, two points determine a line, so we will find coordinates of three points, using the third as a check. We will make a table, letting t 1, 2, and 3, and find corresponding values of,, and z. t z z (4, 2, 6) (3, 1, 4) (2, 0, 2) The resulting ordered triples are (2, 0, 2), (3, 1, 4), and (4, 2, 6). Graph the line with the given parametric equations. Make a table if needed. 7. t 2 8. t 9. 2t 1 t 1 2 t 3 t 2 z 3 z t 3 z 5 t z z z 82 Reteaching 6.5 Geometr

133 NSWERS 10. JKLM; JMRN; NQR; KLQ 11. LRQ; KN Lesson No; the figure does not have two parallel faces and onl one face is a parallelogram. 2. Yes; right rectangular prism. 3. Yes; oblique pentagonal prism. 4. right trapezoidal prism 5. D and EFGH D, EF, GH E, F, G 8. FE, DGH 9. angles E, EF, F, FE, F, FG, G, GF, DG, GH, GHD, DH, DE, EH, EHD, DH ; ; ; ; front-right-bottom 10. (2, 0, 2) 11. (0, 0, 2) 12. (2, 4, 2) 13. (0, 4, 0) 14 (2, 4, 0) ; ; ; ; ; ; ; 4.90 Lesson (5, 0, 0); (0, 7.5, 0); (0, 0, 3) 2. (8, 0, 0); (0, 2, 0); (0, 0, 4) 3. (3, 0, 0); (0, 6, 0); (0, 0, 9) 4. z (0, 0, 4) ( 1, 0, 0) (0, 4, 0) ; ; ; ; ; 7.42 Lesson z ( 1, 1, 4) (2, 2, 5) (3, 2, 3) (2, 4, 2) 5. z (0, 0, 3) (0, 4, 0) (3, 0, 0) (0, 5, 1) (4, 0, 0) 7. -ais 8. back-right-bottom Geometr nswers 171

134 NSWERS 6. z (0, 0, 3) (0, 3, 0) Reteaching hapter 7 Lesson square units square meters (3, 0, 0) square inches square meters 7. z square units ( 1, 2, 3) (0, 3, 3) (1, 4, 3) square centimeters cubic units cubic units 8. z cubic centimeters cubic meters cubic units cubic feet (1, 1, 2) 9. Lesson 6.6 z (2, 1, 1) (3, 3, 0) (1, 3, 4) (3, 4, 3) (5, 5, 2) 13. ; the area of the front face is 96 square inches, which is 1.6 times larger than the area of the front face of. 14. a cube that is 240, or approimatel 7.83 inches on a side 15. The first package has greater surface area and would be more epensive to produce. The second, however, would be ver difficult to hold in one hand for pouring. Lesson square millimeters square feet square meters square inches cubic feet 1 6. heck drawings cubic meters cubic inches cubic feet cubic meters cubic meters 172 nswers Geometr

135 NME LSS DTE Reteaching 6.6 erspective Drawing Skill Using one-point perspective In a drawing with one-point perspective, parallel lines appear to intersect at a single point called the vanishing point. Eample Draw a rectangular solid in one-point perspective. Let the rectangle be the front face of the solid and the given point be the vanishing point. In the drawing at the left below, labels have been added for clarit. Lightl draw,,, and D. hoose a point, E, on and draw EF parallel to, FG parallel to, HG parallel to D, and EH parallel to D. Erase all segments that are not edges of the rectangular solid and make hidden lines dashed. E H F G D Draw a rectangular solid in one-point perspective that has the given rectangle as its front face and the given point as the vanishing point Geometr Reteaching

136 NME LSS DTE Skill Using two-point perspective drawing done in two-point perspective has two vanishing points. Eample Draw a rectangular solid in two-point perspective. Let the given vertical segment be the front edge of the solid, the horizontal line be the horizon, and the given points be the vanishing points. In the drawing at the left below, labels have been added for clarit. Lightl draw segments from to and W and from Q to and W. hoose point D on and point on Q and draw vertical segments DZ and X as shown. Net, draw segments from to and X and from Q to D and Z. Let be the intersection of and QD and Y be the intersection of X and QZ. Draw a dashed segment from to Y. Erase all segments that are not edges of the rectangular solid and make hidden lines dashed. Z Y X Q W Draw a rectangular solid in two-point perspective Let the vertical segment be the front edge of the solid. Let the horizontal line be the horizon and the two given points the vanishing points Reteaching 6.6 Geometr

137 NSWERS 6. z (0, 0, 3) (0, 3, 0) Reteaching hapter 7 Lesson square units square meters (3, 0, 0) square inches square meters 7. z square units ( 1, 2, 3) (0, 3, 3) (1, 4, 3) square centimeters cubic units cubic units 8. z cubic centimeters cubic meters cubic units cubic feet (1, 1, 2) 9. Lesson 6.6 z (2, 1, 1) (3, 3, 0) (1, 3, 4) (3, 4, 3) (5, 5, 2) 13. ; the area of the front face is 96 square inches, which is 1.6 times larger than the area of the front face of. 14. a cube that is 240, or approimatel 7.83 inches on a side 15. The first package has greater surface area and would be more epensive to produce. The second, however, would be ver difficult to hold in one hand for pouring. Lesson square millimeters square feet square meters square inches cubic feet 1 6. heck drawings cubic meters cubic inches cubic feet cubic meters cubic meters 172 nswers Geometr

138 NME LSS DTE Reteaching 7.1 Surface rea and Volume Skill Finding the surface area of a right rectangular prism The surface area, S, of a right rectangular prism with length, width w, and height h is S 2 w 2 h 2wh. Eample Find the surface area of the right rectangular prism. 15, w 7, and h 8, so S 2(15)(7) 2(15)(8) 2(7)(8) 772. The surface area is 772 square units Find the surface area of the right rectangular prism with the given dimensions. 1. 6, w 6, h m, w 5.5 m, in., w 12 in., h 6 m h 4 in m, w 2 m, 5. 9, w 6.5, h cm, w 2 cm, h 4 m h 1.5 cm Skill Finding the volume of a right rectangular prism The volume, V, of a right rectangular prism with length, width w, and height h is V w h. Eample Refer to the right rectangular prism in the figure above. Find the volume of the prism. V w h The volume is 840 cubic units. Find the volume of the right rectangular prism with the given dimensions , w 6.2, h , w 6, h cm, w 7 cm, h 10 cm m, w 4.2 m, h 8 m 11. 8, w 7, h ft, w 2.5 ft, h 5 ft Geometr Reteaching

139 NME LSS DTE Skill omparing the surface areas and volumes of right rectangular prisms For a right rectangular prism with volume, V, the prism with minimum surface area 3 is a cube with sides of length V. Eample manufacturer is considering two packages for a new laundr detergent. ompare the surface areas and volumes 8 in. of the boes. ased on cost of materials 5 in. alone, which would be a better package? 12 in. 7.5 in. ackage : S V in. 8 in. ackage : S V The packages will contain the same amount of detergent. However, ackage has a smaller surface area. Thus, it can be produced using a lesser amount of material. ased on cost alone, ackage would be a better package than ackage. Refer to packages and in the eample above. 13. Suppose the packages were places side-b-side on a supermarket shelf. Which would be more noticeable? Eplain. 14. Suppose ou wanted to produce another size package with half the volume of those above. Describe the dimensions of a rectangular package with the given volume and the smallest possible surface area. 15. cereal bo will have an opening for pouring at the top. ompare the advantages and disadvantages of two packages, one that is 8.5 inches long, 3 inches deep, and 12 inches high and one that is a cube 6.7 inches on a side. 86 Reteaching 7.1 Geometr

140 NSWERS 6. z (0, 0, 3) (0, 3, 0) Reteaching hapter 7 Lesson square units square meters (3, 0, 0) square inches square meters 7. z square units ( 1, 2, 3) (0, 3, 3) (1, 4, 3) square centimeters cubic units cubic units 8. z cubic centimeters cubic meters cubic units cubic feet (1, 1, 2) 9. Lesson 6.6 z (2, 1, 1) (3, 3, 0) (1, 3, 4) (3, 4, 3) (5, 5, 2) 13. ; the area of the front face is 96 square inches, which is 1.6 times larger than the area of the front face of. 14. a cube that is 240, or approimatel 7.83 inches on a side 15. The first package has greater surface area and would be more epensive to produce. The second, however, would be ver difficult to hold in one hand for pouring. Lesson square millimeters square feet square meters square inches cubic feet 1 6. heck drawings cubic meters cubic inches cubic feet cubic meters cubic meters 172 nswers Geometr

141 NME LSS DTE Reteaching 7.2 Surface rea and Volume of risms Skill Finding the surface area of a right prism The surface area of an space figure is the sum of the areas of its surfaces. The surface area of an prism is equal to the sum of its lateral area and the areas of both bases, that is, S L 2. For a right prism, if the perimeter is p, the height is h, and the area of a base is, the surface area is S hp 2. Note that if the base of a prism is a regular polgon with apothem a, ou can use the area formula ( 1 2 ap ) for such a polgon to determine that 2 ap and S hp ap. Eample The figure shows a net for a right triangular prism. Find the surface area of the prism. 3 in. 4 in. 5 in. 4 in. 1 Solution 2 h 4; p h 4; L 3(4) 5(4) 4(4) (3)(4) 6 (3)(4) S hp 2 4(12) 2(6) 60 S L The surface area of the prism is 60 square inches. Find the surface area of the prism with the given dimensions. 1. L 124 mm 2, 60 mm 2 2. h 15 ft, p 32 ft, 100 ft 2 Each figure shows a net for a right prism. Find the surface area of the prism m 5 m 4. 8 in. 6 m 4 m 3 m 5.5 in. 4 in. Geometr Reteaching

142 NME LSS DTE Skill Finding the volume of a prism The volume of a prism with height h and area of a base is V h. This formula is true for all prisms, whether the are right or oblique. Eample 1 The bases of a triangular prism are right triangles with sides of length 5 inches, 12 inches, and 13 inches. The prism is 9 inches high. Find the volume of the prism. 1 (5)(12) 30 2 V h 30(9) 270 The volume of the prism is 270 cubic inches. Eample 2 The base of the prism is a regular heagon with apothem 45 centimeters. Find the volume of the prism. 52 cm 60 cm 1 1 ap (45)(6)(52) V h (7020)(60) 421,200 The volume of the prism is 421,200 cubic centimeters. Find the volume of each prism ft 6. 6 m 7. 6 ft 4 m 8 ft 8 ft 3 m 5 m 8. The area of a base is 80 square feet and the height is 10 feet. 9. Each base of the right prism is a rectangle that is 7 meters long and 10 in. 10 in. 7 in. 4 meters wide and the height of the prism is 9 meters. 10. Each base of the oblique prism is a trapezoid with height 5 inches and bases that are 4 inches and 10 inches. The height of the prism is 12 inches. 88 Reteaching 7.2 Geometr

143 NSWERS 6. z (0, 0, 3) (0, 3, 0) Reteaching hapter 7 Lesson square units square meters (3, 0, 0) square inches square meters 7. z square units ( 1, 2, 3) (0, 3, 3) (1, 4, 3) square centimeters cubic units cubic units 8. z cubic centimeters cubic meters cubic units cubic feet (1, 1, 2) 9. Lesson 6.6 z (2, 1, 1) (3, 3, 0) (1, 3, 4) (3, 4, 3) (5, 5, 2) 13. ; the area of the front face is 96 square inches, which is 1.6 times larger than the area of the front face of. 14. a cube that is 240, or approimatel 7.83 inches on a side 15. The first package has greater surface area and would be more epensive to produce. The second, however, would be ver difficult to hold in one hand for pouring. Lesson square millimeters square feet square meters square inches cubic feet 1 6. heck drawings cubic meters cubic inches cubic feet cubic meters cubic meters 172 nswers Geometr

144 NME LSS DTE Reteaching 7.3 Surface rea and Volume of ramids Skill Finding the surface area of a regular pramid The surface area, S, of a regular pramid with lateral area L and base area is S L. If the slant height of the pramid is and the perimeter is p, the surface 1 area is S p. lso, 1, where a is the apothem. 2 2 ap Eample Find the surface area of the regular square pramid. 12 The base is a square with sides that are 10 units long, so 100. The perimeter of the base is 40 units. Net use the thagorean Theorem to find the slant height, Then S p (13)(40) The surface area of the pramid is 360 square units Find the surface area of each regular pramid = Find the surface area of each pramid. 4. a regular pramid whose base is a pentagon with sides 4 inches long and area 9 a = = 22.5 a = square inches, and slant height 10 inches 5. a regular pramid whose base is a heagon with sides 9 meters long and apothem 7.8 meters, and slant height 12 meters Geometr Reteaching

145 NME LSS DTE 6. a regular pramid whose base is an equilateral triangle with sides 7 feet long and apothem 2 feet, and slant height 4 feet Skill Finding the volume of a pramid 1 The volume, V, of a pramid with base area and height h is V h. 3 Eample Find the volume of a regular square pramid The area of the base is 100 square units and the height is 12 units. 1 V h 3 1 (100)(12) The volume is 400 cubic units. Find the volume of each pramid d = 29 in. 7 d 5.3 m 10 d 4 m 3 m 40 in. 20 in. 10. an octagonal pramid with base area 412 square meters and height 15 meters 11. a pentagonal pramid with base area 108 square inches and height 7.5 inches 12. a rectangular pramid whose base is 20 centimeters long and 16 centimeters wide, and whose height is 10 centimeters 13. a pramid whose base is a right triangle with sides of length 9 feet, 12 feet, and 15 feet and whose height is 10 feet 90 Reteaching 7.3 Geometr

146 NSWERS Lesson square units square units square units square inches square meters square feet cubic ards cubic meters 9. 11,200 cubic inches cubic meters cubic inches cubic centimeters cubic feet Lesson π square inches; square inches 2. 12π square feet; square feet π square centimeters; square centimeters 3π 4. square millimeters; 4.71 square 2 millimeters meters feet ds cubic centimeters 9. 4π cubic inches; 12.6 cubic inches π cubic meters; cubic meters π cubic centimeters; cubic centimeters π cubic feet; cubic feet π cubic units; cubic units meters π inches; 44.0 inches π cubic meters; cubic meters π cubic meters; cubic meters Lesson π square inches 2. 33π square feet 3. 49π 7π 149 square units 4. 85π square inches; square inches 5. 16π square millimeters; 50.3 square millimeters 6. 7π π square feet; square feet 7. 11π π square millimeters; square millimeters 80π 8. cubic meters π cubic feet π cubic inches π cubic inches; cubic inches π cubic feet; cubic feet π cubic meters; 26.2 cubic meters feet feet π cubic inches π cubic millimeters Lesson π square inches; square inches π square millimeters; square millimeters π square centimeters; square centimeters Geometr nswers 173

147 NME LSS DTE Reteaching 7.4 Surface rea and Volume of linders Skill Finding the surface area of a right clinder The surface area, S, of a clinder is the sum of its lateral area and the area of both bases, that is, S L 2. For a right clinder, if the radius of a base is r and the height of the clinder is h, the surface area, S, is S 2πrh 2πr 2. Eample 1 9 in. The radius of a base of a right clinder is 9 inches. The clinder is 6 inches high. Find the eact surface area 6 in. of the clinder and the surface area rounded to the nearest tenth. r 9 and h 6, so S 2π(9)(6) 2π(9 2 ) 270π. The surface area is 270π square inches, or approimatel square inches. Eample 2 right clinder that is 15 meters high has surface area 1400π square meters. Find the eact circumference of a base of the clinder. To find the circumference, first find the radius. S 2πrh 2πr π 2πr(15) 2πr r r 2 Solve the quadratic equation: r 2 15r 700 0; r 35 or r 20 Since r must be positive, r 20. Then 2πr 40π. The circumference of a base is 40π meters. The dimensions of a right clinder are given. Find the eact surface area of the clinder and the surface area rounded to the nearest hundredth. 1. radius of a base 12 in., height 12 in. 2. radius of a base 2 ft, height 1 ft 3. radius of a base 30 cm, height 8 cm 4. diameter of a base 1 mm, height 1 mm Find the unknown value for each clinder. Give the eact value as well as the value rounded to the nearest tenth. 5. surface area 92π m 2, radius of a base 5 m, height 6. surface area 507π ft 2, height 12 ft, diameter of a base 7. surface area 7500π d 2, height 12 d, circumference of a base Geometr Reteaching

148 NME LSS DTE Skill Finding the volume of a clinder If the area of a base of a clinder is and the height of the clinder is h, then the volume, V, of the clinder is V h. If the radius of the clinder is r, then the volume is V πr 2 h. Eample The diameter of a base of an oblique clinder is 46 inches. The height of the clinder is 10 inches. Find the eact volume of the clinder and the volume rounded to the nearest tenth. V πr 2 h Since d 46, r 23. Then, since h 10, V π(23) 2 (10) 5290π. The volume of the clinder is 5290π cubic units, or approimatel 16,619.0 cubic inches. 46 in. 10 in. Find the eact volume of each clinder. Then give the volume rounded to the nearest tenth. 8. area of a base 68 cm 2, height 12 cm 9. radius of a base 1 in., height 4 in. 10. diameter of a base 20 m, height 8 m 11. radius of a base 2.4 cm, height 10 cm 12. radius of a base 5 ft, height 4 ft 13. circumference of a base 24π, height 6 Find the unknown value for each clinder. Give the eact value as well as the value rounded to the nearest tenth. In Eercises 16 17, the clinder is a right clinder. 14. volume 128π cm 3, diameter of a base 8 m, height 15. volume 245π in. 3, height 5 in., circumference of a base 16. surface area 80π m 2, height 3 m, volume 17. surface area 250π m 2, height 20 m, volume 92 Reteaching 7.4 Geometr

149 NSWERS Lesson square units square units square units square inches square meters square feet cubic ards cubic meters 9. 11,200 cubic inches cubic meters cubic inches cubic centimeters cubic feet Lesson π square inches; square inches 2. 12π square feet; square feet π square centimeters; square centimeters 3π 4. square millimeters; 4.71 square 2 millimeters meters feet ds cubic centimeters 9. 4π cubic inches; 12.6 cubic inches π cubic meters; cubic meters π cubic centimeters; cubic centimeters π cubic feet; cubic feet π cubic units; cubic units meters π inches; 44.0 inches π cubic meters; cubic meters π cubic meters; cubic meters Lesson π square inches 2. 33π square feet 3. 49π 7π 149 square units 4. 85π square inches; square inches 5. 16π square millimeters; 50.3 square millimeters 6. 7π π square feet; square feet 7. 11π π square millimeters; square millimeters 80π 8. cubic meters π cubic feet π cubic inches π cubic inches; cubic inches π cubic feet; cubic feet π cubic meters; 26.2 cubic meters feet feet π cubic inches π cubic millimeters Lesson π square inches; square inches π square millimeters; square millimeters π square centimeters; square centimeters Geometr nswers 173

150 NME LSS DTE Reteaching 7.5 Surface rea and Volume of ones Skill Finding the surface area of a right cone The surface area, S, of a cone is the sum of its lateral area and the area of its base, that is, S L. For a right cone, if the radius of the base is r and the slant height of the cone is, the surface area, S, is S πr πr 2. Eample 1 Find the eact surface area of the right cone and the surface area rounded to the nearest tenth. 8 m 5 m First, use the thagorean Theorem to find the slant height of the cone ; 89 S π(5) ( 89 ) π(5 2 ) 5π 89 25π The surface area of the cone is 5π 89 25π square meters, or approimatel square meters. Find the eact surface area of each right cone in in. 8 ft Find the eact surface area of each right cone. Then find the surface area to the nearest tenth. 6 ft 4. radius of the base 5 in., slant height 12 in. 5. radius of the base 2 mm, slant height 6 mm radius of the base 7 ft, height 8 ft 7. diameter of the base 22 mm, height 16 mm Geometr Reteaching

151 NME LSS DTE Skill Finding the volume of a cone If the area of the base of a cone is and the slant height of the cone is, then the 1 volume, V, of the cone is V h. If the radius of the cone is r, then the volume is 3 1 V πr 2 h. 3 Eample Find the eact volume of the cone. Then find the volume to the nearest hundredth. 8 m 1 V πr 2 1 h π( π )(8) m 200π The volume of the cone is cubic meters, 3 or approimatel cubic meters. Find the eact volume of each cone ft m 7 ft 4 m 6 in. 6 in. Find the eact volume of each cone. Then find the volume rounded to the nearest tenth. 11. radius of the base 18 in., height 6 in. 12. radius of the base 11 ft, height 4 ft 13. radius of the base 2.5 m, height 4 m Find the eact unknown value for each cone. In Eercises 16 and 17, the cones are right cones. 14. volume 432π ft 2, height 16 ft, radius of the base 15. volume 784π m 2, diameter of the base 28 ft, height 16. diameter of the base 16 in., slant height 17 in., volume 17. radius of the base 9 mm, slant height 41 mm, volume 94 Reteaching 7.5 Geometr

152 NSWERS Lesson square units square units square units square inches square meters square feet cubic ards cubic meters 9. 11,200 cubic inches cubic meters cubic inches cubic centimeters cubic feet Lesson π square inches; square inches 2. 12π square feet; square feet π square centimeters; square centimeters 3π 4. square millimeters; 4.71 square 2 millimeters meters feet ds cubic centimeters 9. 4π cubic inches; 12.6 cubic inches π cubic meters; cubic meters π cubic centimeters; cubic centimeters π cubic feet; cubic feet π cubic units; cubic units meters π inches; 44.0 inches π cubic meters; cubic meters π cubic meters; cubic meters Lesson π square inches 2. 33π square feet 3. 49π 7π 149 square units 4. 85π square inches; square inches 5. 16π square millimeters; 50.3 square millimeters 6. 7π π square feet; square feet 7. 11π π square millimeters; square millimeters 80π 8. cubic meters π cubic feet π cubic inches π cubic inches; cubic inches π cubic feet; cubic feet π cubic meters; 26.2 cubic meters feet feet π cubic inches π cubic millimeters Lesson π square inches; square inches π square millimeters; square millimeters π square centimeters; square centimeters Geometr nswers 173

153 NME LSS DTE Reteaching 7.6 Surface rea and Volume of Spheres Skill Finding the surface area of a sphere In the figure at the right, O is the center of the sphere. O and O are radii and is a diameter. The radius of the sphere is O. The diameter is. The surface area, S, of a sphere with radius r is S 4πr 2. O Eample sphere has radius 10 meters. Find the eact surface area of the sphere and the surface area rounded to the nearest tenth. Since r 10, S 4π(10) 2 400π. The surface area of the sphere is 400π square meters, or approimatel square meters. Find the eact surface area of each sphere and the surface area rounded to the nearest hundredth. 1. radius 3 in. 2. radius 12 mm 3. radius 6.2 cm 4. diameter 16 d 5. diameter 7 ft 6. diameter 4.6 in. Skill Finding the volume of a sphere 4 The volume, V, of a sphere with radius r is V πr 3. 3 Eample Find the eact volume of the sphere in the eample above. Then find the volume to the nearest tenth of a cubic meter. 4 r 10, so V π( π ) π The volume is cubic meters, or approimatel cubic meters. 3 Find the eact volume of each sphere and the volume rounded to the nearest tenth. 7. radius 24 d 8. radius 15 mm 9. radius 11 m Geometr Reteaching

154 NME LSS DTE 10. diameter 18 in. 11. diameter 17 d 12. diameter 13 cm Skill Solving problems using the formula for the volume of a sphere Eample The sphere is inscribed in the cube. The volume of the cube is 4096 cubic feet. Find the volume of the sphere. volume of the cube 4096 s 3 3 ; s radius of the sphere s volume of the sphere π(8) π The volume of the sphere is cubic feet, or approimatel square feet. 3 Find the eact volume of each space figure m 3 in. 16. spherical scoop of ice cream has radius 1.25 inches. The scoop is placed on top of a cone with radius 1 inch and height 5 inches. Is the cone large enough to hold all the ice cream if it melts? Eplain. 9 m r = 2 ft 5 ft 96 Reteaching 7.6 Geometr

155 NSWERS Lesson square units square units square units square inches square meters square feet cubic ards cubic meters 9. 11,200 cubic inches cubic meters cubic inches cubic centimeters cubic feet Lesson π square inches; square inches 2. 12π square feet; square feet π square centimeters; square centimeters 3π 4. square millimeters; 4.71 square 2 millimeters meters feet ds cubic centimeters 9. 4π cubic inches; 12.6 cubic inches π cubic meters; cubic meters π cubic centimeters; cubic centimeters π cubic feet; cubic feet π cubic units; cubic units meters π inches; 44.0 inches π cubic meters; cubic meters π cubic meters; cubic meters Lesson π square inches 2. 33π square feet 3. 49π 7π 149 square units 4. 85π square inches; square inches 5. 16π square millimeters; 50.3 square millimeters 6. 7π π square feet; square feet 7. 11π π square millimeters; square millimeters 80π 8. cubic meters π cubic feet π cubic inches π cubic inches; cubic inches π cubic feet; cubic feet π cubic meters; 26.2 cubic meters feet feet π cubic inches π cubic millimeters Lesson π square inches; square inches π square millimeters; square millimeters π square centimeters; square centimeters Geometr nswers 173

156 NSWERS π square ards; square ards 5. 49π square feet; square feet π square inches; square inches π cubic ards; 57,905.8 cubic ards π cubic millimeters; 14,137.2 cubic millimeters 5324π 9. cubic meters; cubic meters π cubic inches; cubic inches 4913π 11. cubic ards; cubic ards π 12. cubic centimeters; cubic 6 centimeters π cubic inches π cubic meters 92π 15. cubic feet No; the volume of the ice cream is about 8.2 cubic inches, while the volume of the cone is onl about 5.2 cubic inches. 9. a clinder with base radius 12 in. and height 10 in. 10. a clinder with base radius 10 in. and height 12 in. 11. a clinder with base radius 5 in. and height 12 in. Reteaching hapter 8 Lesson (0, 10) 2. ( 4,2) 3. (8, 4) (12, 8) 6. (2, 12) ( 5 2, 6 ) Q' Q ' Lesson (2, 4, 1) 2. (2, 3, 4) 3. (5, 0, 2) 4. '(0, 2, 1); '(3, 3, 4) 5. '( 2, 2, 2); '(4, 3, 1) 6. (0, 4, 2); (3, 5, 3) z (0, 3, 2) (0, 3, 0) 9. R R' S S' 8. z 174 nswers Geometr

157 NME LSS DTE Reteaching 7.7 Three-Dimensional Smmetr Skill Reflecting points and segments across a plane in space Reflecting the point (,, z) across one of the planes defined b a pair of aes in space produces a change in the coordinates according to the plane across which is reflected. The image of reflected across the z-plane is (,, z). The image of reflected across the z-plane is (,, z). The image of reflected across the -plane is (,, z). Eample The segment with endpoints (3, 1, 2) and (1, 2, 3) is reflected across the zplane. Identif the coordinates of the endpoints of the image. The endpoints of the image are '(3, 1, 2) and '(1, 2, 3). Give the coordinates of the image of the given point after reflection across the indicated plane. 1. (2, 4, 1); -plane 2. ( 2, 3, 4); z-plane 3. (5, 0, 2); z-plane Given and, find the coordinates of the endpoints of the image of after reflection across the indicated plane. 4. (0, 2, 1) and ( 3, 3, 4) 5. ( 2, 2, 2) and (4, 3, 1) 6. (0, 4, 2) and (3, 5, 3) z-plane z-plane -plane Skill Rotating segments about an ais in space When a segment in space is rotated about an ais to which it is perpendicular, the figure formed ma be a circle or a donut-shaped region called an annulus. If the segment is parallel to the ais of rotation, the image is the lateral surface of a right clinder whose ais is the ais of rotation. Eample Draw the image of the segment with endpoints (3, 0, 2) and (1, 0, 2) rotated (a) about the z-ais and (b) about the -ais. Geometr Reteaching

158 NME LSS DTE a. z (3, 0, 2) b. (1, 0, 2) z (3, 0, 2) (1, 0, 2) has endpoints (0, 3, 2) and (0, 3, 0). Sketch the figure formed b rotating about the indicated ais. 7. -ais 8. z-ais z z Skill Visualizing and describing solids of revolution closed plane figure that is rotated about an ais in space produces a solid figure called a solid of revolution. Eample Describe the solid of revolution formed when the right triangle is rotated about (a) its vertical leg and (b) its horizontal leg. icture the figure rotating about the indicated ais. oth rotations produce cones. Since the cones have different heights and base radii, the have different volumes. Describe the solid of revolution formed when the rectangle is rotated about the indicated ais. Give the dimensions of the solid of revolution in. D 12 in. 11. the line through the midpoints of and D 98 Reteaching 7.7 Geometr

159 NSWERS π square ards; square ards 5. 49π square feet; square feet π square inches; square inches π cubic ards; 57,905.8 cubic ards π cubic millimeters; 14,137.2 cubic millimeters 5324π 9. cubic meters; cubic meters π cubic inches; cubic inches 4913π 11. cubic ards; cubic ards π 12. cubic centimeters; cubic 6 centimeters π cubic inches π cubic meters 92π 15. cubic feet No; the volume of the ice cream is about 8.2 cubic inches, while the volume of the cone is onl about 5.2 cubic inches. 9. a clinder with base radius 12 in. and height 10 in. 10. a clinder with base radius 10 in. and height 12 in. 11. a clinder with base radius 5 in. and height 12 in. Reteaching hapter 8 Lesson (0, 10) 2. ( 4,2) 3. (8, 4) (12, 8) 6. (2, 12) ( 5 2, 6 ) Q' Q ' Lesson (2, 4, 1) 2. (2, 3, 4) 3. (5, 0, 2) 4. '(0, 2, 1); '(3, 3, 4) 5. '( 2, 2, 2); '(4, 3, 1) 6. (0, 4, 2); (3, 5, 3) z (0, 3, 2) (0, 3, 0) 9. R R' S S' 8. z 174 nswers Geometr

160 NME LSS DTE Reteaching 8.1 Dilations and Scale Factors Skill Determining the coordinates of the image of a point under a dilation Let be a point in the coordinate plane and n a real number, n not 0, and let D be the dilation with center and scale factor n. The image of an point Q under D is the point on OQ (on the ra opposite OQ Q if n is negative) such that R' = nq OQ n OQ. If n 1, the dilation is called an epansion. If n 0, the dilation is called a contraction. Q Q' If the center of the dilation is the origin and Q has O coordinates (, ), then Q has coordinates (n, n). Eample Let D be a dilation with center (0, 0) and scale factor 2 and let ( 2, 1). Determine the coordinates of. '(4, 2) Since the origin is the center of the dilation and the scale factor is 2, D( 2, 1) has coordinates ( 2( 2), 2( 1)) (4, 2). O ( 2, 1) For each point, find the coordinates of, the image of under the dilation with center (0, 0) and scale factor n. 1. n 2.5; (0, 4) 2. n 1; (4, 2) 3. n 4; ( 2, 1) 4. n 1 ; ( 5, 12) 5. n 2 ; (18, 12) 6. n 3; The dilation D has center (0, 0) and scale factor 1 1. Draw the segment 2 with endpoints ( 2, 0) and Q(2, 2) and its image, Q, under D. ( 2 3, 4 ) 8. The dilation E has center (0, 0) and scale factor 5. Draw the segment with endpoints R(0, 1) and S(1, 0) and its image, R S, under E. O O Geometr Reteaching

161 NME LSS DTE Skill Drawing the image of a plane figure under a dilation To draw the image of a plane figure under a dilation with center and scale factor n, first draw the images of the vertices of the figure. If n 0 and point is a verte of the figure to be dilated, draw. Measure and locate on so that. is the image of. Use the same method to locate the images of the other vertices and connect them. If n 0, proceed as described above, but locate on the ra opposite instead. Eample Draw the image of JKL under a dilation with 3 center L and scale factor. 4 K Draw J, K, and L. J 36 cm, K 24 cm, and 20 cm J L L So J , K , and 4 4 L Locate J on J, K on K, and L on L. Finall, draw J K, K L, and J L. J K L is the image of JKL. J K J' K' L' L Draw the image of the figure under the dilation with center and the indicated scale factor. 9. n n n n n n Reteaching 8.1 Geometr

162 NSWERS π square ards; square ards 5. 49π square feet; square feet π square inches; square inches π cubic ards; 57,905.8 cubic ards π cubic millimeters; 14,137.2 cubic millimeters 5324π 9. cubic meters; cubic meters π cubic inches; cubic inches 4913π 11. cubic ards; cubic ards π 12. cubic centimeters; cubic 6 centimeters π cubic inches π cubic meters 92π 15. cubic feet No; the volume of the ice cream is about 8.2 cubic inches, while the volume of the cone is onl about 5.2 cubic inches. 9. a clinder with base radius 12 in. and height 10 in. 10. a clinder with base radius 10 in. and height 12 in. 11. a clinder with base radius 5 in. and height 12 in. Reteaching hapter 8 Lesson (0, 10) 2. ( 4,2) 3. (8, 4) (12, 8) 6. (2, 12) ( 5 2, 6 ) Q' Q ' Lesson (2, 4, 1) 2. (2, 3, 4) 3. (5, 0, 2) 4. '(0, 2, 1); '(3, 3, 4) 5. '( 2, 2, 2); '(4, 3, 1) 6. (0, 4, 2); (3, 5, 3) z (0, 3, 2) (0, 3, 0) 9. R R' S S' 8. z 174 nswers Geometr

163 NSWERS Yes; UVW ZXY; SS 4. Yes; FGH LMK or FGH MLK; SSS 5. Yes; RST MN; Yes; ZXY; SS feet feet Lesson Lesson feet feet feet feet feet 6. 6 meters 14. Lesson Yes; EFD 2. no 3. Yes; DEFG LMJK inches Lesson Yes; JKL STR; SSS 2. Yes; E STR; Lesson ; 9 25 ; ; ; cm π square ards; 4500π cubic ards 8. 6 inches; 10 square inches 9. 16π meters; 256π cubic meters Reteaching hapter 9 Lesson Q, Q, Q, QD, D 3. D and D, and 4., D, D, D Geometr nswers 175

164 NME LSS DTE Reteaching 8.2 Similar olgons Skill Using the olgon Similarit ostulate to determine whether two polgons are similar The olgon Similarit ostulate: Two polgons are similar if and onl if there is a wa of setting up a correspondence between their sides and angles so that: (1) ll corresponding angles are congruent. (2) ll corresponding sides are proportional. Eample Determine whether the triangles are similar. If so, write a similarit statement. 46 K J L the Triangle Sum Theorem: m L 180 (70 80 ) 30 m Q 180 (70 30 ) 80 Then J, K Q, and L R. Then JK and Q are corresponding sides, as are KL and QR, and JL and R. JK 46 Q Q 129 KL 86 QR JL 90 R The ratios of the lengths of corresponding sides are equal, so corresponding sides are in proportion. the olgon Similarit ostulate, JKL QR. Determine whether the polgons are similar. If so, write a similarit statement S V W Z T 6 U X 4 Y D E F 15 R D G 8 E 127 J 4 K 8 5 F L M 7 Geometr Reteaching

165 NME LSS DTE Skill Using properties of proportions and scale factors to solve problems involving similar polgons proportion is a statement that two ratios are equal. Given two similar polgons, then, ou can write proportions comparing the ratios of an two pairs of corresponding sides. Given some of the side lengths of two smaller figures, this method allows ou to determine unknown lengths. Suppose that XYZ. Then. the Echange ropert of roportions, it is also true that XY YZ XY. That is, the ratio of the lengths of two sides of one of a pair of similar YZ figures is equal to the ratio of the lengths of the corresponding sides of the other figure. Eample MNQ RSTU. Find the values of and. N 8 S 10 T M 4 Q R 5 U 7.5 N Q 8 Since the quadrilaterals are similar, or. 10 ST TU 7.5 Then and 6. MQ M MQ RU lso,, so b the Echange ropert of roportions,. RU ST N ST Then and omplete. 4. E 5. T U D 12 F 10 D S 12 V I DEF HIJ, so D STUV, so WXYZ JKLM, so H 3 J n architect is making plans for a rectangular office building that is 840 feet long and 252 feet wide. blueprint of the floor plan for the first floor is 15 inches long. How wide is the blueprint? 16 W Z X Y M 15 J L 10 K 102 Reteaching 8.2 Geometr

166 NSWERS Yes; UVW ZXY; SS 4. Yes; FGH LMK or FGH MLK; SSS 5. Yes; RST MN; Yes; ZXY; SS feet feet Lesson Lesson feet feet feet feet feet 6. 6 meters 14. Lesson Yes; EFD 2. no 3. Yes; DEFG LMJK inches Lesson Yes; JKL STR; SSS 2. Yes; E STR; Lesson ; 9 25 ; ; ; cm π square ards; 4500π cubic ards 8. 6 inches; 10 square inches 9. 16π meters; 256π cubic meters Reteaching hapter 9 Lesson Q, Q, Q, QD, D 3. D and D, and 4., D, D, D Geometr nswers 175

167 NME LSS DTE Reteaching 8.3 Triangle Similarit ostulates Skill Using similarit postulates and theorems to determine whether two triangles are similar Similarit ostulate: If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. SSS Similarit ostulate: If the three sides of one triangle are proportional to the three sides of another triangle, the triangles are similar. SS Similarit ostulate: If two sides of one triangle are proportional to two sides of another triangle, and their included angles are congruent, then the triangles are similar. Eample re the triangles similar? Q R 13 D F 9 E m Q 180 (m m R) 180 (40 72 ) 68 Then Q E and R F, so QR DEF b the Similarit ostulate. re the triangles similar? If so, write a similarit statement and identif the postulate or theorem that justifies our answer. 1. K R S T J 18 L 46 E S R 44 T V Y Z u 8 W X F 15 G H K L 9 M Geometr Reteaching

168 NME LSS DTE 5. R 6 6. M N S T Y X Z Skill Using triangle similarit to solve problems If ou can show that two triangles are similar, then ou can use the proportionalit of the sides to solve problems. Eample park is in the shape of an isosceles triangle with legs 60 feet long. The base is 104 feet and is marked b a 60 ft 60 ft picket fence. If each of the congruent sides of the park is cut back to a length of 45 feet, how long a fence will 104 ft be needed for the base? Make a sketch. the SS Similarit Theorem, the new triangle is similar to the original triangle. Let be the base of the new triangle. The ratio of the lengths of the 60 4 legs is, so the ratio of the bases is also and ft 45 ft Then 78; a fence 78 feet long will be needed. Solve each problem. Identif the postulate or theorem that justifies our solution. 7. portion of an eterior wall of a house is marked off b the edges of the roof and a horizontal piece of trim 50 feet long. Decorative vertical strips are spaced evenl as shown. If the middle strip is 10 feet long, how long are each of the shortest strips? 8. Students plan to make a large replica of a school banner to displa at a football game. The banner is triangular as shown in the figure. If the shortest side of the replica will be 6 1 feet long, 4 how man feet of trim will be needed to edge the replica? 10 in. 24 in. 24 in. 104 Reteaching 8.3 Geometr

169 NSWERS Yes; UVW ZXY; SS 4. Yes; FGH LMK or FGH MLK; SSS 5. Yes; RST MN; Yes; ZXY; SS feet feet Lesson Lesson feet feet feet feet feet 6. 6 meters 14. Lesson Yes; EFD 2. no 3. Yes; DEFG LMJK inches Lesson Yes; JKL STR; SSS 2. Yes; E STR; Lesson ; 9 25 ; ; ; cm π square ards; 4500π cubic ards 8. 6 inches; 10 square inches 9. 16π meters; 256π cubic meters Reteaching hapter 9 Lesson Q, Q, Q, QD, D 3. D and D, and 4., D, D, D Geometr nswers 175

170 NME LSS DTE Reteaching 8.4 The Side-Splitting Theorem Skill Using the Side-Splitting Theorem to solve problems The Side-Splitting Theorem states that a line parallel to one side of a triangle divides the other two sides proportionall. In the figure below, the dashed segment is parallel to a side of the triangle. The Side-Splitting Theorem can be a d used to justif an proportion equivalent to the first f c proportion listed. Several eamples are given. a c d e a b c d a c b d b c d f To use the Side-Splitting Theorem to find an unknown length in a triangle, find a proportion in which is isolated as the numerator or denominator of one of the ratios. Eample 2 Find the value of b e Find the value of Geometr Reteaching

171 NME LSS DTE Skill Using the Two-Transversal roportionalit orollar to solve problems corollar of a theorem is a theorem that can be easil proven using the given theorem. The Two-Transversal roportionalit orollar is a corollar of the Side-Splitting Theorem: If three or more parallel lines intersect two transversals, the divide them proportionall. Eample In the figure, j, k, and are parallel. Find the value of. j k 5 In the figure, lines m, n, p, and q are parallel. Use the Two- Transversal roportionalit orollar and similar triangles to find each value a b c d 14. e Since j k, the transversals are divided proportionall. That is,. 5 Then d a 5 e 10 4 b c m n p q 106 Reteaching 8.4 Geometr

172 NSWERS Yes; UVW ZXY; SS 4. Yes; FGH LMK or FGH MLK; SSS 5. Yes; RST MN; Yes; ZXY; SS feet feet Lesson Lesson feet feet feet feet feet 6. 6 meters 14. Lesson Yes; EFD 2. no 3. Yes; DEFG LMJK inches Lesson Yes; JKL STR; SSS 2. Yes; E STR; Lesson ; 9 25 ; ; ; cm π square ards; 4500π cubic ards 8. 6 inches; 10 square inches 9. 16π meters; 256π cubic meters Reteaching hapter 9 Lesson Q, Q, Q, QD, D 3. D and D, and 4., D, D, D Geometr nswers 175

173 NME LSS DTE Reteaching 8.5 Indirect Measurement and dditional Similarit Theorems Skill Using similar triangles to measure distances indirectl Similar triangles ma be used to determine unknown lengths that ma be difficult or even impossible to measure using ordinar methods. Eample rvid wanted to determine the height of a tree in his ard. He stood in the shadow of the tree and positioned himself carefull so that his shadow ended at the same point as the tree s shadow. He then had his sister measure his shadow 6 ft (7.5 feet) and the sun s shadow (30 feet). If rvid is 6 feet tall, E D 7.5 ft how tall is the tree? 30 ft D D D 30 E D b the Similarit ostulate. or so D 24. E E The tree is 24 feet high. Find the value of ft 10 ft 60 ft ft 48 ft 16 ft ft 80 ft ft 10 ft 72 ft 88 ft 50 ft 200 ft 40 ft 152 cm 5.25 m 133 cm Geometr Reteaching

174 NME LSS DTE Skill Using the roportional ltitudes, Medians, and ngle isector Theorems If two triangles are similar, their sides are proportional. s the following theorems indicate, other lines related to the triangles are proportional as well. roportional ltitudes Theorem: If two triangles are similar, then the ratio of the lengths of two corresponding altitudes is the same as the ratio of the lengths of two corresponding sides. roportional Medians Theorem: If two triangles are similar, then the ratio of the lengths of two corresponding medians is the same as the ratio of the lengths of two corresponding sides. roportional ngle isectors Theorem: If two triangles are similar, then the ratio of the lengths of two corresponding angle bisectors is the same as the ratio of the lengths of two corresponding sides. Eample If JKL MN, find the length of JX. N J K 5 X L M 9 12 Y Since X is the midpoint of KL and Y is the midpoint of N, JX and MY are medians. the roportional Medians Theorem, JX JL JX. Then and JX MY M 12 The triangles in each pair are similar. Find the value of Reteaching 8.5 Geometr

175 NSWERS Yes; UVW ZXY; SS 4. Yes; FGH LMK or FGH MLK; SSS 5. Yes; RST MN; Yes; ZXY; SS feet feet Lesson Lesson feet feet feet feet feet 6. 6 meters 14. Lesson Yes; EFD 2. no 3. Yes; DEFG LMJK inches Lesson Yes; JKL STR; SSS 2. Yes; E STR; Lesson ; 9 25 ; ; ; cm π square ards; 4500π cubic ards 8. 6 inches; 10 square inches 9. 16π meters; 256π cubic meters Reteaching hapter 9 Lesson Q, Q, Q, QD, D 3. D and D, and 4., D, D, D Geometr nswers 175

176 NME LSS DTE Reteaching 8.6 rea and Volume Ratios Skill Determining the ratio of the perimeters and the areas of similar plane figures and the ratio of the volumes of similar solids a If the scale factor of two similar plane figures is, then the ratio of their perimeters b a a is and the ratio of their areas is 2. b b 2 a If the scale factor of two similar solids is, then the ratio of corresponding areas is b a 2 a and the ratio of their volumes is 3. b 2 b 3 Eample 1 3 The scale factor of two rectangular pramids is. Find the ratio of the perimeters 7 of their bases, the ratio of the areas of their bases, and the ratio of their volumes. 3 3 The scale factor is, so the ratio of the perimeters of their bases is The ratio of the areas of their bases is 2. The ratio of the volumes is Find the indicated ratio for the similar figures rectangles, scale factor ; 11 ratio of areas: ; ratio of perimeters: 1 2. pramids, scale factor ; 5 ratio of areas of bases: ; ratio of volumes: 3 3. spheres, scale factor ; 2 ratio of volumes: ; ratio of surface areas: QR and FGH, ratio of areas ; 9 QR ratio of perimeters: ; : GH square prisms, ratio of areas of bases ; 25 ratio of volumes: ; ratio of perimeters: Geometr Reteaching

177 NME LSS DTE Skill Using ratios for perimeters, areas, and volumes of similar figures The scale factor of two similar figures is related to the ratios of other linear dimensions as well as to the ratios of areas and, for space figures, volumes. Then ou can use information about an one of the ratios to solve problems involving the others. Eample The scale factor of the similar rectangular 8 prisms is. If the volume of the smaller prism is cubic meters, find the volume of the larger prism Let V be the volume of the larger prism. Since the scale factor is, the ratio of the volumes is. Then and V V 3375 The volume of the larger prism is cubic meters. omplete The ratio of the areas of two similar triangles is. If the perimeter of the smaller triangle is centimeters, then the perimeter of the larger is The scale factor of two similar spheres is. If the surface area of the larger sphere is 900π square 2 ards, the surface area of the smaller sphere is. If the volume of the smaller sphere is 288π cubic ards, then the volume of the larger sphere is. 8. Square pramids and are similar. The ratio of the volume of pramid to the volume of 8 pramid is. If the height of pramid is 4 inches, then the height of pramid is 27. If the area of a lateral face of pramid is 22.5 square inches, then the area of a lateral face of pramid is. 9. linder X is 16 meters high. similar clinder, Y, is 28 meters high. If the area of the base of clinder Y is 49π square meters, then the area of a base of clinder X is. If the volume of clinder Y is 1372π cubic meters, then the volume of clinder X is. 110 Reteaching 8.6 Geometr

178 NSWERS Yes; UVW ZXY; SS 4. Yes; FGH LMK or FGH MLK; SSS 5. Yes; RST MN; Yes; ZXY; SS feet feet Lesson Lesson feet feet feet feet feet 6. 6 meters 14. Lesson Yes; EFD 2. no 3. Yes; DEFG LMJK inches Lesson Yes; JKL STR; SSS 2. Yes; E STR; Lesson ; 9 25 ; ; ; cm π square ards; 4500π cubic ards 8. 6 inches; 10 square inches 9. 16π meters; 256π cubic meters Reteaching hapter 9 Lesson Q, Q, Q, QD, D 3. D and D, and 4., D, D, D Geometr nswers 175

179 NME LSS DTE Reteaching 9.1 hords and rcs Skill Identifing parts of circles circle,, is the set of all points in a plane that are equidistant from a point, called the center of the circle. radius of is a segment with one endpoint and the other endpoint on the circle. chord is a segment with both endpoints on the circle. chord that contains is a diameter of. Two points on a circle determine a chord and two arcs. If the points are the endpoints of a diameter, both arcs are semicircles. If the two points are not endpoints of a diameter, the determine a minor arc (shorter than a semicircle) and a major arc (longer than a semicircle). Three points on an arc must be used to denote a semicircle or a major arc. Eample Given circle, name each of the following. a. the center b. a radius or radii c. a diameter d. a major arc e. a semicircle f. a minor arc g. a chord that is not a diameter and the arc of the chord S R Q a. b. R,, and Q c. Q d. RQS or QRS e. RQ or SQ f. R, S, SQ, RQ, or RS g. R, R Given Q, name all of the following: 1. radii 2. diameters 3. chords that are not diameters and their arcs 4. semicircles D Q 5. minor arcs 6. major arcs Geometr Reteaching

180 NME LSS DTE Skill Finding the degree measure of an arc of a circle central angle of O is an angle with verte O. In the figure, and are the arcs intercepted b O, and O is the central angle of arcs O and. The measure of a minor arc is the measure of its central angle. The measure of a major arc is 360 minus the measure of its central angle. The measure of a semicircle is 180. In a circle, or in congruent circles, arcs of congruent chords are congruent. Eample Find the measure of each arc. a. b. D c. D 55 R a. m m R 55 b. md D 360 m R 305 c. D, so md m 55. Find the measure of each arc. 7. Q SQ 10. QR SR S 35 T Q 11. QT 12. RS R Skill Finding the length of an arc of a circle If is an arc of circle with radius r and m M, then the length, L, of is M given b L (2πr). 360 Eample Find the length of arc with measure 40 in a circle with radius 18 cm. M 40 L (2πr) (2)(π)(18) 4π Find the length of an arc with the given measure in a circle with the given radius. Give our answer to the nearest hundredth ; r 25 m ; r 100 in ; r 15 m ; r 9.5 cm 112 Reteaching 9.1 Geometr

181 NSWERS Yes; UVW ZXY; SS 4. Yes; FGH LMK or FGH MLK; SSS 5. Yes; RST MN; Yes; ZXY; SS feet feet Lesson Lesson feet feet feet feet feet 6. 6 meters 14. Lesson Yes; EFD 2. no 3. Yes; DEFG LMJK inches Lesson Yes; JKL STR; SSS 2. Yes; E STR; Lesson ; 9 25 ; ; ; cm π square ards; 4500π cubic ards 8. 6 inches; 10 square inches 9. 16π meters; 256π cubic meters Reteaching hapter 9 Lesson Q, Q, Q, QD, D 3. D and D, and 4., D, D, D Geometr nswers 175

182 NSWERS 5.,, D, D meters inches meters centimeters Lesson onsider Q. 2 3, Q radius of 3 1, and Q 12, 2 so (Q) 2 (Q) 2 () 2. the converse of the thagorean Theorem, Q is a right triangle and Q. Then b the converse of the Tangent Theorem, is tangent to. Similarl, and are tangent to Lesson No; NQM is not an inscribed angle. 8. No; the measure of the intercepted arc cannot be determined from the figure D,, D, D, R; Q, QR Lesson Lesson Lesson ; 1.5; (0, 0); ; 12, 6; (0, 3); , 17.85; 10, 14; (5, 2); , 7.74; none; ( 4, 1); ( 5) ( 2) 2 ( 2) ( 3) 2 ( 6) ( 5) 2 ( 5) ( 6) 2 ( 4) ( 1) ( 3) 2 ( 3) ( 2) 2 ( 2) 2 16 Reteaching hapter 10 Lesson ; ; ; ; ; nswers Geometr

183 NME LSS DTE Reteaching 9.2 Tangents to ircles Skill Using the Tangent Theorem and its converse In each figure, nearest tenth. The Tangent Theorem: If a line is tangent to a circle, then the line is perpendicular to a radius of the circle drawn to the point of tangenc. The onverse of the Tangent Theorem: If a line is perpendicular to a radius of a circle at its endpoint on the circle, then the line is tangent to the circle. Eample 1 Q is tangent to O. Find the value of. 15 Q Q O 9 is tangent to O. the Tangent Theorem, Q is perpendicular to radius O. the thagorean Theorem, (OQ) 2 (Q) 2 (O) 2 ( 9) Eample 2 Show that 12 is tangent to. 5 8 D is a radius of, so D 5. Then D D () 2 () 2 () the converse of the thagorean Theorem, is a right triangle and is perpendicular to radius. the converse of the Tangent Theorem, is tangent to. is tangent to O. Find the value of to the O O 20 8 O Geometr Reteaching

184 NME LSS DTE 4. In the figure, is an equilateral triangle with 6. M, N, and Q are the midpoints of the sides of The radius of is 3 and 2 3. Show that is inscribed in, that is, show that each side of is tangent to. M N Q Skill Using the Radius and hord Theorem The Radius and hord Theorem: radius that is perpendicular to a chord of a circle bisects the chord. Eample The radius of N is 16 and 10. Find the value of to the nearest tenth. Radius NE is perpendicular to chord, so NE bisects b the Radius and hord Theorem. Then D 5. the thagorean Theorem, (N) , and E D N Eercises 5 8 refer to. Give our answers to the nearest tenth, if necessar. 5. Name two pairs of congruent segments. 6. If R 8, find Q. 7. If Q 10 and R 24, find the radius of. 8. If the radius of is 9 and QS 1, find R. Find the indicated measure. Round our answers to the nearest tenth if necessar R Q L K J Q S R S R 18, QS 44, Q M JM 24, KM 42, JL 114 Reteaching 9.2 Geometr

185 NSWERS 5.,, D, D meters inches meters centimeters Lesson onsider Q. 2 3, Q radius of 3 1, and Q 12, 2 so (Q) 2 (Q) 2 () 2. the converse of the thagorean Theorem, Q is a right triangle and Q. Then b the converse of the Tangent Theorem, is tangent to. Similarl, and are tangent to Lesson No; NQM is not an inscribed angle. 8. No; the measure of the intercepted arc cannot be determined from the figure D,, D, D, R; Q, QR Lesson Lesson Lesson ; 1.5; (0, 0); ; 12, 6; (0, 3); , 17.85; 10, 14; (5, 2); , 7.74; none; ( 4, 1); ( 5) ( 2) 2 ( 2) ( 3) 2 ( 6) ( 5) 2 ( 5) ( 6) 2 ( 4) ( 1) ( 3) 2 ( 3) ( 2) 2 ( 2) 2 16 Reteaching hapter 10 Lesson ; ; ; ; ; nswers Geometr

186 NME LSS DTE Reteaching 9.3 Inscribed ngles and rcs Skill Using the Inscribed ngle Theorem n inscribed angle of a circle is an angle whose verte lies on the circle and whose sides are chords of the circle. You can use the Inscribed ngle Theorem to determine the measure of an inscribed angle of a circle. The Inscribed ngle Theorem: The measure of an angle inscribed in a circle is equal to one-half the measure of the intercepted arc. Eample Find the measure of QR. 74 O R Q the Inscribed ngle Theorem, 1 1 m QR mr (74 ) Find the measure of each inscribed angle of. 1. JKN 2. NKM 3. JKM 4. NLM 5. KNL 6. KML Refer to the figure for Eercises 1 6. an ou use the Inscribed ngle Theorem to determine the measure of the given angle? Eplain wh or wh not. K J 52 Q 46 L 30 M R 7. NQM 8. MLR 9. omplete: is inscribed in and intersects. If m 102, then m. Geometr Reteaching

187 NME LSS DTE Skill Using the corollaries of the Inscribed ngle Theorem The following corollaries follow directl from the Inscribed ngle Theorem. The Right-ngle orollar: If an inscribed angle of a circle intercepts a semicircle, then the angle is a right angle. The rc-intercept orollar: If two inscribed angles intercept the same arc, then the angles have the same measure. Eample X Y 64 XZ is a diameter of '. Find the measure of each angle. W 24 a. XYZ b. XWY c. YWZ Z d. WYZ a. XYZ intercepts XWZ, which is a semicircle, so XYZ is a right angle and m XYZ 90. b. XWY intercepts the same arc as XZY, so m XWY m XZY 24. c. YWZ intercepts YZ. XY and YZ are adjacent arcs, that is, the share onl one point. So mxy myz m XYZ. Since XZY intercepts XY and m XZY 24, mxy 48 b the Inscribed ngle Theorem. Substituting, 48 myz 180 and myz 132. gain b the Inscribed ngle Theorem, m YWZ 66. d. mxw mwz mxwz 180, so mwz Then b the Inscribed ngle Theorem, m WYZ 58. Find the measure of each angle or arc in O with diameters D and E. 10. E 11. D 12. E 13. E E Find the measure of each angle or arc in J with diameter UW. 16. VXW 17. UXW 18. UV 19. UXV 20. VWX 21. VUX 27 U O E V J X D 38 W Reteaching 9.3 Geometr

188 NSWERS 5.,, D, D meters inches meters centimeters Lesson onsider Q. 2 3, Q radius of 3 1, and Q 12, 2 so (Q) 2 (Q) 2 () 2. the converse of the thagorean Theorem, Q is a right triangle and Q. Then b the converse of the Tangent Theorem, is tangent to. Similarl, and are tangent to Lesson No; NQM is not an inscribed angle. 8. No; the measure of the intercepted arc cannot be determined from the figure D,, D, D, R; Q, QR Lesson Lesson Lesson ; 1.5; (0, 0); ; 12, 6; (0, 3); , 17.85; 10, 14; (5, 2); , 7.74; none; ( 4, 1); ( 5) ( 2) 2 ( 2) ( 3) 2 ( 6) ( 5) 2 ( 5) ( 6) 2 ( 4) ( 1) ( 3) 2 ( 3) ( 2) 2 ( 2) 2 16 Reteaching hapter 10 Lesson ; ; ; ; ; nswers Geometr

189 NME LSS DTE Reteaching 9.4 ngles Formed b Secants and Tangents Skill Finding the measure of an angle formed b a secant and a tangent of a circle that intersect on the circle Theorem: If a tangent and a secant or chord of a circle intersect on the circle at the point of tangenc, the measure of the angle formed is half the measure of the intercepted arc. Eample is tangent to O at. Find m. m 1 1 m O 140 Q is tangent to at. Find the measure of QR R Q Q R 216 Q 150 R Skill Finding the measure of an angle formed b two secants or chords of a circle that intersect inside the circle The measure of an angle formed b two secants or chords that intersect in the interior of a circle is equal to half the sum of the measures of the intercepted arcs. Eample Find m JNK m JNK (50 62 ) 56 2 (mjk ML ) K 2 Find the measure of each angle. 4. F D Q T R H E S G J O N D M L m FHG m STR m DE Geometr Reteaching

190 NME LSS DTE Skill Finding the measure of an angle formed b two secants, two tangents, or a secant and a tangent of a circle that intersect outside the circle The measure of an angle formed b two secants, two tangents, or a secant and a tangent of a circle that intersect outside the circle is one-half the difference of the measures of the intercepted arcs. Eample X is tangent to at X. Y is tangent to at Y. Find m XY. X Y 252 Z The measure of XY is one-half the difference of the measures of the intercepted arcs, XZY and XY. Since mxzy mxy 360, 1 1 m 108. Then m XY ( ) 72 2 (XZY XY XY ) 2 Find the measure of the each angle. 7. N m NM M T m STU 60 S U m QRS Q R T S 118 Reteaching 9.4 Geometr

191 NSWERS 5.,, D, D meters inches meters centimeters Lesson onsider Q. 2 3, Q radius of 3 1, and Q 12, 2 so (Q) 2 (Q) 2 () 2. the converse of the thagorean Theorem, Q is a right triangle and Q. Then b the converse of the Tangent Theorem, is tangent to. Similarl, and are tangent to Lesson No; NQM is not an inscribed angle. 8. No; the measure of the intercepted arc cannot be determined from the figure D,, D, D, R; Q, QR Lesson Lesson Lesson ; 1.5; (0, 0); ; 12, 6; (0, 3); , 17.85; 10, 14; (5, 2); , 7.74; none; ( 4, 1); ( 5) ( 2) 2 ( 2) ( 3) 2 ( 6) ( 5) 2 ( 5) ( 6) 2 ( 4) ( 1) ( 3) 2 ( 3) ( 2) 2 ( 2) 2 16 Reteaching hapter 10 Lesson ; ; ; ; ; nswers Geometr

192 NME LSS DTE Reteaching 9.5 Segments of Tangents, Secants, and hords Skill Finding lengths of segments determined b two tangents or two secants of a circle that intersect outside the circle The figures below illustrate the terms tangent segment, secant segment, and eternal secant segment. X and XQ are tangents of O. K and M are secants of. O X L J K Q M X and XQ are tangent segments. K and M are secant segments. J and L are eternal secant segments. In the figures, X XQ and K J M L. These relationships are formalized in the following theorems. If two tangents are drawn to a circle from the same eternal point, the tangent segments are congruent. If two secants are drawn to a circle from the same eternal point, the product of the lengths of one secant segment and its eternal segment is equal to the product of the lengths of the other segment and its eternal segment. Eample Refer to O and above. omplete each statement. a. In O, if XQ 7.2, then X _?. b. In, if J 5, K 15, and M 13.6, then L _?. a. X and XQ are tangents to O from the same eternal point, so X XQ 7.2. b. Secants K and M intersect at point outside Q, so K J M L. K J 5 15 Then L 5.5. M 13.6 Find the measure of the indicated length D E 3. F Q G Q D R S T H E RS 5, RT 18, RQ 8 EF 8, DF 18, FH 16 3, 12, D 4 R FG E Geometr Reteaching

193 NME LSS DTE Skill Finding lengths of segments determined b a tangent and a secant of a circle that intersect outside the circle or b two chords that intersect If a secant and a tangent to a circle intersect at a point outside the circle, then the product of the lengths of the secant segment and its eternal segment is equal to the square of the length of the tangent segment. If two chords of a circle intersect, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other. Eample is tangent to. a. If 6 and D 10, find to the nearest hundredth. b. If EF 4, FD 8, and F 4.5, find FG to the nearest hundredth. a. Tangent and secant D intersect at point outside of. Then D () 2 and D b. ED and G are chords of. Then EF FD F FG EF D and FG F E F G D In Eercises 4 9, Q is tangent to O. Find each length to the nearest hundredth X Y O Q O Q O R M N S Q 8, QR 5 QM 5, QN 16 QY 3, QX 13 QS Q Q W 9. G V X Z O E H D Y F D E Q E 5, E 1, ED 1.5 XZ 7, VX 8, XY 9 DH 9, HF 4, HE 5 E WX GH 120 Reteaching 9.5 Geometr

194 NSWERS 5.,, D, D meters inches meters centimeters Lesson onsider Q. 2 3, Q radius of 3 1, and Q 12, 2 so (Q) 2 (Q) 2 () 2. the converse of the thagorean Theorem, Q is a right triangle and Q. Then b the converse of the Tangent Theorem, is tangent to. Similarl, and are tangent to Lesson No; NQM is not an inscribed angle. 8. No; the measure of the intercepted arc cannot be determined from the figure D,, D, D, R; Q, QR Lesson Lesson Lesson ; 1.5; (0, 0); ; 12, 6; (0, 3); , 17.85; 10, 14; (5, 2); , 7.74; none; ( 4, 1); ( 5) ( 2) 2 ( 2) ( 3) 2 ( 6) ( 5) 2 ( 5) ( 6) 2 ( 4) ( 1) ( 3) 2 ( 3) ( 2) 2 ( 2) 2 16 Reteaching hapter 10 Lesson ; ; ; ; ; nswers Geometr

195 NME LSS DTE Reteaching 9.6 ircles in the oordinate lane Skill Using an equation of a circle to determine the intercept(s), the center, and the radius of the circle The equation ( k) 2 ( k) 2 r 2 is the standard form of the equation of a circle with center (h, k) and radius r. If the center is the origin, then h 0 and k 0 and the equation can be written 2 2 r 2. Eample circle has equation ( 1) Find: a. the -intercept(s) and the -intercept(s). b. the center and radius of the circle. a. (1) Let 0 and solve for. (2) Let 0 and solve for : ( 1) The -intercepts are approimatel 8.28 and The -intercepts are approimatel 7.21 and b. First write the equation in standard form. ( k) 2 ( k) 2 r 2 ( ( 1)) 2 ( 0) 2 ( 53) h 1, k 0, and r 53 The center is ( 1, 0) and the radius is Find the -intercept(s), the -intercept(s), the center, and the radius of the circle with the given equation. If necessar, round to the nearest hundredth. 1. equation: intercept(s): -intercepts: center: radius: 2. equation: 2 ( 3) intercept(s): -intercepts: center: radius: 3. equation: ( 5) 2 ( 2) intercept(s): -intercepts: center: radius: 4. equation: ( 4) 2 ( 1) intercept(s): -intercepts: center: radius: Geometr Reteaching

196 NME LSS DTE Skill Writing an equation of a circle Given the coordinates of the center of a circle and the radius of a circle, ou can write an equation of the circle. Eample Write an equation in standard form for each circle. a. the circle with center ( 7, 3) and radius 11 b., shown at the right a. Substitute 7 for h, 3 for k, and 11 for r in the standard equation. ( ( 7) 2 ( 3) ( 7) 2 ( 3) b. From the graph, ou can determine that the center of the circle is ( 1,1) and the radius is 3. Substitute 1 for h, 1 for k, and 3 for r in the standard equation. ( ( 1) 2 ( 1) ( 1) 2 ( 1) 2 9 ( 1, 1) O ( 1, 2) Write an equation in standard form for each circle. 5. the circle with center (0, 0) and radius the circle with center (5, 0) and radius the circle with center ( 2, 2) and radius the circle with center ( 3, 6) and radius the circle with center (5, 5) and radius the circle with center ( 6, 4) and radius ( 2, 2) O ( 1, 0) (1, 0) (3, 3) 4 O 4 ( 2, 2) O (3, 1) Reteaching 9.6 Geometr

197 NSWERS 5.,, D, D meters inches meters centimeters Lesson onsider Q. 2 3, Q radius of 3 1, and Q 12, 2 so (Q) 2 (Q) 2 () 2. the converse of the thagorean Theorem, Q is a right triangle and Q. Then b the converse of the Tangent Theorem, is tangent to. Similarl, and are tangent to Lesson No; NQM is not an inscribed angle. 8. No; the measure of the intercepted arc cannot be determined from the figure D,, D, D, R; Q, QR Lesson Lesson Lesson ; 1.5; (0, 0); ; 12, 6; (0, 3); , 17.85; 10, 14; (5, 2); , 7.74; none; ( 4, 1); ( 5) ( 2) 2 ( 2) ( 3) 2 ( 6) ( 5) 2 ( 5) ( 6) 2 ( 4) ( 1) ( 3) 2 ( 3) ( 2) 2 ( 2) 2 16 Reteaching hapter 10 Lesson ; ; ; ; ; nswers Geometr

198 NME LSS DTE Reteaching 10.1 Tangent Ratios Skill Finding the tangent of an acute angle If is an acute angle of a right triangle, the tangent of, written tan or tan, is the ratio of the length of the leg opposite to the length of the leg adjacent to. This can be abbreviated tan opposite. If the measure in degrees of is, adjacent the ratio ma also be referred to as tan. The tangent of an angle can be determined using a scientific or graphics calculator, or a table of trigonometric ratios, such as the one given on page 870 of our tetbook. Eample 1 Find tan. Give our answer to the nearest hundredth. 2.5 cm and 3.4 cm, so tan 2.5 opposite adjacent cm 2.5 cm Eample 2 In JKL, m L 33. Use a calculator to find tan L. Round our answer to the nearest hundredth. J e sure the calculator is set in degree mode. sample ke sequence is shown. K 33 L TN 3 3 ENTER To the nearest tenth, tan Find tan as a fraction and as a decimal rounded to the nearest hundredth Geometr Reteaching

199 NME LSS DTE Use a scientific or graphics calculator to find the tangent of the angle with the given measure. Round our answer to the nearest hundredth Skill Finding the measure of an acute angle given its tangent To find the measure of an acute angle with a given tangent r 0, ou could write r a as a fraction, then construct a right triangle with legs of length a and b. You could b then measure the angle opposite the side with length a. Since measurements ma be inaccurate, it is better to use the inverse tangent function, tan 1, of a scientific or graphics calculator. For r 0, tan 1 (r), where 0 90 and tan r. Eample is an acute angle with tan 5. Find m to the nearest degree. 8 e sure the calculator is set in degree mode. sample ke sequence is shown. 2nd TN 5 8 ) ENTER To the nearest degree, tan The tangent of an acute angle is given. Use a calculator to find the measure of the angle to the nearest degree Reteaching 10.1 Geometr

200 NSWERS 5.,, D, D meters inches meters centimeters Lesson onsider Q. 2 3, Q radius of 3 1, and Q 12, 2 so (Q) 2 (Q) 2 () 2. the converse of the thagorean Theorem, Q is a right triangle and Q. Then b the converse of the Tangent Theorem, is tangent to. Similarl, and are tangent to Lesson No; NQM is not an inscribed angle. 8. No; the measure of the intercepted arc cannot be determined from the figure D,, D, D, R; Q, QR Lesson Lesson Lesson ; 1.5; (0, 0); ; 12, 6; (0, 3); , 17.85; 10, 14; (5, 2); , 7.74; none; ( 4, 1); ( 5) ( 2) 2 ( 2) ( 3) 2 ( 6) ( 5) 2 ( 5) ( 6) 2 ( 4) ( 1) ( 3) 2 ( 3) ( 2) 2 ( 2) 2 16 Reteaching hapter 10 Lesson ; ; ; ; ; nswers Geometr

201 NSWERS Lesson ; ; 9 5 ; 3 29 ; ; ; ft ft Lesson (1, 0) O ' ( , ); ; ' (1, 0) O ( , ); ; (1, 0) ' (0.4226, ); ; ; ; ( , ) ; ; ( , ) ; ; (0.9397, ) 7. 12, , , , , Lesson nswers ma var slightl due to rounding. 5. m 97, b 11.4, c m L 66, k 14.8, m 65, b 11.0, c or no solution or nswers ma var slightl due to rounding. 12. m K 40, m L 48, l no solution 14. (1) m 63, m 76, a 16.3 (2) m 117, m 22, a m Q 43, m R 27, r 2.7 Lesson nswers ma var slightl due to rounding or to difference in solution methods. 7. m 80, m 57, m , m X 29, m Z m S 73, m R 57, m Q c 34.4, m 33, m 37 Geometr nswers 177

202 NME LSS DTE Reteaching 10.2 Sines and osines Skill Finding the sine and the cosine of an acute angle Two other important trigonometric ratios are the sine of the angle and the cosine of the angle. The terms sine and cosine are abbreviated sin and cos. sin cos The sine of an angle with measure can be referred to as sin, sin, or sin. Similar epressions can be used to identif the cosine of an angle. The sine and cosine can be determined using a scientific or graphics calculator, or a table of trigonometric ratios, such as the one given on page 870 of our tetbook. Eample a. Epress sin as a fraction. 5 b. Use a scientific or graphics calculator to find cos to the nearest hundredth. 4 a. sin opposite hpotenuse length of side opposite length of hpotenuse 3 5 b. e sure the calculator is set in degree mode. sample ke sequence is shown. OS 5 9 ENTER To the nearest tenth, cos or length of side adjacent to length of hpotenuse Write sin and cos as fractions in simplest form opposite hpotenuse or adjacent hpotenuse sin : cos : sin : cos : sin : cos : Use a scientific or graphics calculator to find sin and cos to the nearest hundredth m m m 27 sin : cos : sin : cos : sin : cos : Geometr Reteaching

203 NME LSS DTE Skill Finding the measure of an acute angle given its sine or cosine You can use a scientific or graphics calculator to find the measure of an acute angle given its sine or cosine. Eample The cosine of an acute angle is Find m to the nearest degree. Use the cos 1 function: cos m 22 Find the measure of to the nearest degree. 7. sin 0.15, m 8. cos 0.69, m 9. sin 0.71, m 10. cos 0.55, m 11. sin 0.05, m 12. sin 0.31, m Skill Using the sine and cosine ratios to solve problems You can use the sine and cosine ratios to solve problems. Eample 20 foot ladder placed against a building makes an angle of 72 with the ground. Find how high up the building the ladder reaches. opposite sin 72, hpotenuse 20 so 20(sin 72 ) 20(0.95) 19. The ladder reaches 19 feet up the building. The figure shows a tower that is 96 feet tall and has two attached gu wires. 13. The wire on the left meets the ground at an angle of 64. Find the length of the wire to the nearest foot ft ft 14. The wire on the right meets the tower at an angle of 48. Find the length of the wire to the nearest foot. 126 Reteaching 10.2 Geometr

204 NSWERS Lesson ; ; 9 5 ; 3 29 ; ; ; ft ft Lesson (1, 0) O ' ( , ); ; ' (1, 0) O ( , ); ; (1, 0) ' (0.4226, ); ; ; ; ( , ) ; ; ( , ) ; ; (0.9397, ) 7. 12, , , , , Lesson nswers ma var slightl due to rounding. 5. m 97, b 11.4, c m L 66, k 14.8, m 65, b 11.0, c or no solution or nswers ma var slightl due to rounding. 12. m K 40, m L 48, l no solution 14. (1) m 63, m 76, a 16.3 (2) m 117, m 22, a m Q 43, m R 27, r 2.7 Lesson nswers ma var slightl due to rounding or to difference in solution methods. 7. m 80, m 57, m , m X 29, m Z m S 73, m R 57, m Q c 34.4, m 33, m 37 Geometr nswers 177

205 NME LSS DTE Reteaching 10.3 Etending the Trigonometric Ratios Skill Using the unit circle to find the sine and cosine of an angle ra with endpoint O lies along the positive -ais and intersects the unit circle at (0, 1). If O is rotated about the origin, angles of rotation are produced. Suppose is rotated through an angle. Let be the point where the terminal O side of O intersects the unit circle. Then the -coordinate of is cos and the -coordinate of is sin or (cos, sin ). Eample For 217, find the coordinates of to four decimal places and use them to give sin 217 and cos (1, 0) First note that since is in the third quadrant, both Q coordinates of are negative. Draw the unit circle and an O angle of rotation of 217. ' Draw Q to form a right triangle with one side on the -ais. m QO cos 37 OQ O sin 37 Q O OQ Q 1 1 OQ Q ( , ), so cos and sin Use the unit circle to sketch O, the result of rotating O through an angle with measure. Find the coordinates of to four decimal places. Then find cos and sin O (1, 0) O (1, 0) O (1, 0) cos 199 cos 130 cos 295 sin 199 sin 130 sin 295 Geometr Reteaching

206 NME LSS DTE Use a scientific calculator to find the sine and cosine of each angle to four decimal places. Then give the - and -coordinates of the point where the terminal side of the given angle intersects the unit circle. 4. cos 105 sin sin 200 cos sin 340 cos 340 Skill Finding the measures of angles given their sines or cosines Since we will be using sines and cosines to work with triangles, we will be mainl interested in angles whose measures are between 0 and 180. The figures below show the graphs of sin and cos for = sin Note the following from the graphs. For 0 180, sin is between 0 and 1. lso, if ou draw a horizontal line through the graph of sin, ou will note that for an value z between 0 and 1, there are two angles with measures between 0 and 180 whose sines are equal to z. If ou use a calculator to find sin 1 z, the calculator will give ou an angle with measure between 0 and 90. The other angle has measure 180. For 0 180, cos is between 1 and 1. For an value z between 1 and 1, there is onl one angle, for which cos z. Eample Use a scientific calculator to find all values of, 0 180, for which sin Round our answer(s) to the nearest degree. 26 and Find all values of between 0 and 180 for which the given statement is true. Round our answer(s) to the nearest degree. 7. sin cos cos cos sin sin sin sin cos Reteaching 10.3 Geometr

207 NSWERS Lesson ; ; 9 5 ; 3 29 ; ; ; ft ft Lesson (1, 0) O ' ( , ); ; ' (1, 0) O ( , ); ; (1, 0) ' (0.4226, ); ; ; ; ( , ) ; ; ( , ) ; ; (0.9397, ) 7. 12, , , , , Lesson nswers ma var slightl due to rounding. 5. m 97, b 11.4, c m L 66, k 14.8, m 65, b 11.0, c or no solution or nswers ma var slightl due to rounding. 12. m K 40, m L 48, l no solution 14. (1) m 63, m 76, a 16.3 (2) m 117, m 22, a m Q 43, m R 27, r 2.7 Lesson nswers ma var slightl due to rounding or to difference in solution methods. 7. m 80, m 57, m , m X 29, m Z m S 73, m R 57, m Q c 34.4, m 33, m 37 Geometr nswers 177

208 NME LSS DTE Reteaching 10.4 The Law of Sines Skill Using the Law of Sines to solve a triangle given the measures of two angles and a side The Law of Sines states that for an triangle with sides a, b, and c, sin sin sin. In some cases, the Law of Sines can be used to solve triangles. a b c Solving a triangle means using given lengths and angle measures to determine all possible values of all other measurements. Given the measures of two angles and a side of a triangle, there is eactl one solution of the triangle. Eample Solve, in which m 61, m 38, and b 15. First note that m 180 (61 38 ) 81. Net use the Law of Sines to find a and c. sin a sin b a b sin sin 15 sin 81 sin sin c sin b c b sin sin 15 sin 38 sin Find the indicated measure in. Give lengths to the nearest tenth. 1. m 42, m 70, c 25, b 2. m 54, m 45, a 15, c 3. m 105, m 40, b 12, a 4. m 22, m 62, c 14, b Solve each triangle. Give lengths to the nearest tenth and angle measures to the nearest degree. 5. : m 50, m 33, a JKL: m J 70, m K 44, j QR: m 95, m Q 20, r 10 Geometr Reteaching

209 NME LSS DTE Skill Using the Law of Sines to solve a triangle given the measures of two sides and an angle that is opposite one of the sides If ou are given the lengths of two sides of a triangle and the measure of an angle that is opposite one of the sides, it is possible that there is no solution, one solution, or two solutions. It is sometimes helpful to make a sketch that is roughl to scale. Eample Solve each triangle. a. m 65, a 6, b 8 b. m 47, a 8, b 9 sin sin 65 a. the Law of Sines, sin. Then sin, and a b sin 65 sin Since the sine of ever angle is between 1 and 1, 6 there is no such angle, so there is no such triangle. sin sin 47 b. the Law of Sines, sin. Then sin, and a b sin 47 sin Two possible angles are 55 and If m 55, then m 78 and c If m 125, then m 8 and c 1.5. Find the measure of the indicated angle of to the nearest degree. If there is no solution, write no solution. If there is more than one solution, give both. 8. m 34, c 25, b 30, m 9. a 6, c 12, m 30, m 10. m 29, a 16, c 7, m 11. m 33, b 13, c 18, m Solve each triangle. Give lengths to the nearest tenth and angle measures to the nearest degree. If there is no solution, write no solution. If there is more than one solution, give both. 12. JKL: m J 92, j 7, k DEF: m D 96, d 5, f : m 41, b 11, c QR: p 5.5, q 4, m Reteaching 10.4 Geometr

210 NSWERS Lesson ; ; 9 5 ; 3 29 ; ; ; ft ft Lesson (1, 0) O ' ( , ); ; ' (1, 0) O ( , ); ; (1, 0) ' (0.4226, ); ; ; ; ( , ) ; ; ( , ) ; ; (0.9397, ) 7. 12, , , , , Lesson nswers ma var slightl due to rounding. 5. m 97, b 11.4, c m L 66, k 14.8, m 65, b 11.0, c or no solution or nswers ma var slightl due to rounding. 12. m K 40, m L 48, l no solution 14. (1) m 63, m 76, a 16.3 (2) m 117, m 22, a m Q 43, m R 27, r 2.7 Lesson nswers ma var slightl due to rounding or to difference in solution methods. 7. m 80, m 57, m , m X 29, m Z m S 73, m R 57, m Q c 34.4, m 33, m 37 Geometr nswers 177

211 NME LSS DTE Reteaching 10.5 The Law of osines Skill Using the Law of osines to find the measure of an angle or side of a triangle The Law of osines states that for an triangle with sides a, b, and c, each of the following are true: a 2 b 2 c 2 2bc cos c a b 2 a 2 c 2 2ac cos c 2 a 2 b 2 2ab cos b The Law of osines can be used in solving triangles where the measures of two sides and the included angle are known or where the lengths of all three sides are known. The following theorem ma simplif our work when solving such triangles. Theorem: If one side of a triangle is longer than a second side, then the angle opposite the first side is larger than the angle opposite the second. It follows from the theorem that the largest angle of a triangle is opposite the longest side. It is also helpful to recall that, for an numbers, 1 1, there is onl value of between 0 and 180 for which cos. Eample a. Find. Z Y b. Find m. = 5 85 z = 8 c = 8 a = 10 X a. the Law of osines, 2 2 z 2 2z cos X (40) (cos 85 ) 82. Then b. the Law of osines, c 2 a 2 b 2 2ab cos, so cos c2 a 2 b ab 2(12)(15) Then m cos Find each indicated measure. Give lengths to the nearest tenth and angles to the nearest degree. 1. N 2. S 3. M p = 30 n = 20 m = 18 t = 5 R 70 s = 7 r T b = 42 b = 12 c = a m N r a Geometr Reteaching

212 NME LSS DTE 4. DEF: m D 25, e 12.5, f 11, d 5. MN: m 18, n 12, p 7.5, m M 6. STU: s 30, t 35, u 32, m U Skill Using the Law of osines along with the Law of Sines to solve triangles The Law of osines ma be used together with the Law of Sines to solve triangles. For eample, consider in the eample on the preceding page. It is possible to use the Law of osines to determine the measure of each angle of the triangle. However, once ou have found the measure of one angle, it is much simpler to use the Law of Sines to find the measure of a second and the Triangle Sum Theorem to find the measure of the third. Eample Solve HJK. H the Law of osines, k 2 h 2 j 2 2hj cosk (16)(14) cos Then k j = K h = 16 k J sin J sin sin 54 the Law of Sines,, so sin J Since J is not the largest angle, J must be acute, so m J sin Solve each triangle. Give lengths to the nearest tenth and angle measures to the nearest degree z = 5 9. X Y 125 c = 100 a = 125 = 5.5 b = : m 110, a 20, b FGH: m G 85, f 12, h 15 Z R q = 8 S s = 10 r = 8.5 Q 12. VWX: v 73, w 40, Reteaching 10.5 Geometr

213 NSWERS Lesson ; ; 9 5 ; 3 29 ; ; ; ft ft Lesson (1, 0) O ' ( , ); ; ' (1, 0) O ( , ); ; (1, 0) ' (0.4226, ); ; ; ; ( , ) ; ; ( , ) ; ; (0.9397, ) 7. 12, , , , , Lesson nswers ma var slightl due to rounding. 5. m 97, b 11.4, c m L 66, k 14.8, m 65, b 11.0, c or no solution or nswers ma var slightl due to rounding. 12. m K 40, m L 48, l no solution 14. (1) m 63, m 76, a 16.3 (2) m 117, m 22, a m Q 43, m R 27, r 2.7 Lesson nswers ma var slightl due to rounding or to difference in solution methods. 7. m 80, m 57, m , m X 29, m Z m S 73, m R 57, m Q c 34.4, m 33, m 37 Geometr nswers 177

214 NSWERS 11. g 18.4, m F 41, m H m V 143, m W 19, m X 18 Lesson (0.71, 2.12), (4.95, 0.71), ( 2.12, 4.95) 11. (4.19, 2.73), (1.71, 4.70), (4.95, 3.93) 12. (7.31, 6.82), ( 7.80, 0.44), ( 7.90, 3.26) 13. ( 4.69, 9.33), ( 1.22, 10.37), ( 7.88, 1.39) 14. (0.62, 7.04), ( 3.31, 5.39), ( 8.95, 4.90) (1.22, 0.71), (2.26, 4.57), (6.57, 1.34), (7.61, 5.21) Reteaching hapter 11 Lesson ft d m mm ft in mi/h; mi/h; 20 Lesson (0, 2) 2. ( 2.60, 1.5) 3. ( 0.64, 1.26) 4. (3.57, 0.52) 5. ( 2.66, 3.59) 6. ( 2.94, 1.16) 7. (1, 2) 8. (2, 1) 9. ( 2, 2) TU QT heck drawings. Lesson heck drawings ; ; ; ; ; ; O 178 nswers Geometr

215 NME LSS DTE Reteaching 10.6 Vectors in Geometr Skill Drawing the sum of vectors using the head-to-head method and the parallelogram method vector is a quantit that has both magnitude, or size, and direction. It ma be represented b an arrow. The length of the arrow represents the magnitude of the vector. To describe the direction of the vector, picture superimposing a coordinate grid on the vector with the origin at the tail of the vector. The direction is the angle the vector makes with the positive -ais. The figure at the right shows v. The head of the vector is indicated b the arrow. The tail is the other end. The magnitude of v is denoted v v. Two vectors are equivalent if the have the same direction and magnitude. For an point in the plane, a vector equivalent to v can be drawn with its tail at. This allows ou to draw a vector equivalent to v anwhere in the plane. In the figure, v and are equivalent vectors. There are several methods for finding the resultant, or sum, of two vectors. Eample Draw the sum of and. p q p q a. Head-to-tail method Since ou can draw a vector with its tail at an point in the plane, draw q with its head at the tail of p. The resultant, or sum, p q is the vector with tail at the tail of p and head at the head of q. b. arallelogram method Draw p and q with their tails at the same point. Draw a parallelogram with sides p and q. The resultant vector is represented b a diagonal of the parallelogram. Its tail is at the common point of p and q and its head at the opposite verte of the parallelogram. Draw the resultant of the vectors using the head-to-tail method p q Geometr Reteaching

216 NME LSS DTE Draw the resultant of the vectors using the parallelogram method Skill Using trigonometric ratios to find the resultant of two vectors Trigonometric ratios can be used to find the resultant of two vectors. Eample boat is traveling at a speed of 30 mi/h at a direction of 28. The wind is blowing from the west at a speed of 25 mi/h. Draw b, the velocit of the boat, and w, the wind velocit, and find b w. Draw b w using the parallelogram method. m D 28, so m Let be the magnitude and the direction of. b w 25 D 30 the Law of osines, (25)(30) cos , so 53.4 mi/h. Suppose and in the eample are changed as indicated. Find the magnitude to the nearest tenth and the direction to the nearest degree of the resultant vector. b sin sin 152 the Law of Sines,, so sin and w 7. The direction of is 20 ; and the magnitude of are unchanged. 8. The directions of both vectors are unchanged, but b 35 mi/h and w 15 mi/h. b b w 134 Reteaching 10.6 Geometr

217 NSWERS 11. g 18.4, m F 41, m H m V 143, m W 19, m X 18 Lesson (0.71, 2.12), (4.95, 0.71), ( 2.12, 4.95) 11. (4.19, 2.73), (1.71, 4.70), (4.95, 3.93) 12. (7.31, 6.82), ( 7.80, 0.44), ( 7.90, 3.26) 13. ( 4.69, 9.33), ( 1.22, 10.37), ( 7.88, 1.39) 14. (0.62, 7.04), ( 3.31, 5.39), ( 8.95, 4.90) (1.22, 0.71), (2.26, 4.57), (6.57, 1.34), (7.61, 5.21) Reteaching hapter 11 Lesson ft d m mm ft in mi/h; mi/h; 20 Lesson (0, 2) 2. ( 2.60, 1.5) 3. ( 0.64, 1.26) 4. (3.57, 0.52) 5. ( 2.66, 3.59) 6. ( 2.94, 1.16) 7. (1, 2) 8. (2, 1) 9. ( 2, 2) TU QT heck drawings. Lesson heck drawings ; ; ; ; ; ; O 178 nswers Geometr

218 NME LSS DTE Reteaching 10.7 Rotations in the oordinate lane Skill Using transformations equations to identif the image of a point under a rotation about the origin In Lesson 10.3, ou saw that the -coordinate of the image of the point (1, 0) under a rotation of about the origin is cos. The -coordinate of the image is sin. In general, if R is the point (, ), the -coordinate of the image under the same rotation is cos sin, and the -coordinate of the image is sin cos. Two transformation equations describe the image R'(', ') of R(, ) under a rotation of degrees about the origin. ' cos sin and ' sin cos Eample The point (4, 2) is rotated 72 about the origin. Find the coordinates of the image ' rounded to the nearest hundredth. The angle of rotation is 72 ; cos and sin ' cos sin 4 cos 72 ( 2) sin ' sin cos 4 sin 72 2 cos The image of is ' (3.14, 3.19). point and an angle of rotation are given. Find the coordinates of the image of to the nearest hundredth. 1. (2, 0); (0, 3); (1, 1); (3, 2); (2, 4); (3, 1); 220 The triangle shown in the figure is rotated 90 about the origin. Find the coordinates of the images of the vertices. 7. ' (2, 2) 8. ' O 9. ' ( 2, 1) (1, 2) Geometr Reteaching

219 NME LSS DTE Skill Using rotation matrices to identif the image of a point under a rotation about the origin rotation of about the origin can be represented b the rotation matri cos sin. The coordinates of the image '(',') of a point (,) can sin cos be found b multipling the rotation matri b the matri representing the point,. cos sin sin cos The polgon with the given vertices is rotated about the origin through the given angle. Find the coordinates of the vertices of the image. Round each coordinate to the nearest hundredth. 10. (2, 1), (4, 3), (2, 5); (3, 4), (0, 5), (6, 2); (0, 10), (5, 6), (3, 8); (10, 3), ( 10, 3), (0, 8); 100 Rotation matrices can be used to rotate more than one point at a time. Simpl add a column to the matri for each point. For eample, use the matri to rotate the points ( 1, 1 ), ( 2, 2 ), and ( 3, ). If ou use a graphics calculator to multipl the matrices, ou do not need to find the values of sin and cos first. You can simpl enter the indicated values in the matrices. For eample, cos 77 sin 77 the rotation matri for a 77 rotation would be entered. sin 77 cos 77 Eample triangle with vertices (4, 6), (2, 3), and (7, 5) is rotated 82 about the origin. Find the coordinates of the images of the vertices. cos 82 sin 82 The rotation matri is. sin 82 cos 82 cos 82 sin sin 82 cos To the nearest hundredth, the images of the vertices are '( 5.38, 4.80), '(3.25, 1.56), and '(5.93, 6.24). 14. (5, 5), (2, 6), ( 2, 10); 50 [ cos sin sin cos ] 15. (1, 1), (5, 1), (3, 6), (7, 6); Reteaching 10.7 Geometr

220 NSWERS 11. g 18.4, m F 41, m H m V 143, m W 19, m X 18 Lesson (0.71, 2.12), (4.95, 0.71), ( 2.12, 4.95) 11. (4.19, 2.73), (1.71, 4.70), (4.95, 3.93) 12. (7.31, 6.82), ( 7.80, 0.44), ( 7.90, 3.26) 13. ( 4.69, 9.33), ( 1.22, 10.37), ( 7.88, 1.39) 14. (0.62, 7.04), ( 3.31, 5.39), ( 8.95, 4.90) (1.22, 0.71), (2.26, 4.57), (6.57, 1.34), (7.61, 5.21) Reteaching hapter 11 Lesson ft d m mm ft in mi/h; mi/h; 20 Lesson (0, 2) 2. ( 2.60, 1.5) 3. ( 0.64, 1.26) 4. (3.57, 0.52) 5. ( 2.66, 3.59) 6. ( 2.94, 1.16) 7. (1, 2) 8. (2, 1) 9. ( 2, 2) TU QT heck drawings. Lesson heck drawings ; ; ; ; ; ; O 178 nswers Geometr

221 NME LSS DTE Reteaching 11.1 Golden onnections Skill Using the golden ratio to find side lengths of golden rectangles rectangle is a golden rectangle if, when a square is cut off one end of the rectangle, the remaining rectangle and the original rectangle are similar. Let D be a rectangle with longer side and shorter side s. D is a golden rectangle if. s s s In the figure, D is a golden rectangle. D s s s The ratio of the length of the longer side to the length of the shorter side of a golden rectangle, called the golden ratio, is the same for all golden rectangles. Let QRS be a golden rectangle with longer side and shorter side 1. The ratio for this golden rectangle is. The numerical value of the golden ratio can be determined b cross-multipling and using the quadratic formula to solve the resulting quadratic equation ( 1) ( 1) a 1, b 1, and c 1 2 4(1)( 1) 2(1) 1 5 (Since is a length, must be positive.) Eample The shorter side of a golden rectangle is 18 meters long. Find the length of the longer side to the nearest hundredth of a meter , so s 18 The longer side is approimatel m long. Given the length of one side of a golden rectangle, find the length of the other side. Round our answers to the nearest hundredth. 1. longer side 10 ft 2. shorter side 4 d shorter side longer side Geometr Reteaching

222 NME LSS DTE 3. shorter side 7.5 m 4. shorter side 22 mm longer side longer side 5. longer side 200 ft 6. longer side 9.5 in. shorter side shorter side Skill onstructing a golden rectangle You can use a square to construct a golden rectangle. Eample QRS is a square. onstruct a golden rectangle with shorter side Q. Draw Q and SR. onstruct the midpoint, M, of Q. Draw an arc with center M and radius MR, intersecting at T. onstruct a perpendicular to Q Q at T, intersecting SR at U. Draw TU.To show that TUS is a golden rectangle, let Q 2. Then MQ and MT MR (2) So, T 5 TU , the golden ratio. S R Q S R U M Q T Refer to QTUR in the figure above. omplete. 7. TU 8. QT 9. Show that QTUR is a golden rectangle. D is a square. onstruct a golden rectangle with shorter side. 10. D 11. D 138 Reteaching 11.1 Geometr

223 NSWERS 11. g 18.4, m F 41, m H m V 143, m W 19, m X 18 Lesson (0.71, 2.12), (4.95, 0.71), ( 2.12, 4.95) 11. (4.19, 2.73), (1.71, 4.70), (4.95, 3.93) 12. (7.31, 6.82), ( 7.80, 0.44), ( 7.90, 3.26) 13. ( 4.69, 9.33), ( 1.22, 10.37), ( 7.88, 1.39) 14. (0.62, 7.04), ( 3.31, 5.39), ( 8.95, 4.90) (1.22, 0.71), (2.26, 4.57), (6.57, 1.34), (7.61, 5.21) Reteaching hapter 11 Lesson ft d m mm ft in mi/h; mi/h; 20 Lesson (0, 2) 2. ( 2.60, 1.5) 3. ( 0.64, 1.26) 4. (3.57, 0.52) 5. ( 2.66, 3.59) 6. ( 2.94, 1.16) 7. (1, 2) 8. (2, 1) 9. ( 2, 2) TU QT heck drawings. Lesson heck drawings ; ; ; ; ; ; O 178 nswers Geometr

224 NME LSS DTE Reteaching 11.2 Taicab Geometr Skill Finding the taidistance between two points In the taicab grid at the right, has coordinates ( 2, 1) and has coordinates (3, 2). The dashed line shows one path from to. The taidistance between two points is the smallest number of blocks a tai must drive to get from one point to the other. (Notice that the tai could not drive on a O diagonal, but onl along the horizontal and vertical streets.) The taidistance between two points ( 1, 1 ) and ( 2, 2 ) is If and lie on the same line, there is onl one minimum distance path between them. If not, there is more than one such path. Eample Find the taidistance between points and in the figure above. has coordinates ( 2, 1) and has coordinates (3, 2) ( 2) The taidistance from point to point is 8 blocks. Find the taidistance, d, between points and and draw at least two paths from to that are d blocks long O d d d O O Find the taidistance, d, between the points. 4. ( 7, 5) and (12, 4) 5. (14, 8) and ( 6, 5) 6. ( 7, 7) and (4, 6) d d d Geometr Reteaching

225 NME LSS DTE Skill lotting taicab circles In taicab geometr, a circle is the set of all points in the taicab plane that are a given distance r from a point, called the center. Unlike a Euclidean circle, a taicab circle is square and contains a finite number of points. In fact, if the radius of the circle is n, then the circle contains 4n points. The circumference of a taicab circle with radius r is 8r. Eample 1 taicab circle has radius 18. Determine the number of points on the circle and find its circumference. Since r 18, the circle has 4(18) 72 points and its circumference is 8(18) 144. Eample 2 lot the taicab circle with center (2, 3) and radius 4. lot the point. Then plot each point in the plane that is 4 blocks from. O Find the number, n, of points on the taicab circle with the given radius. Then find the circumference,, of the circle. 7. r r 6 9. r 10 n n n 10. r r r 16 n n n lot the taicab circle with center and radius r. 13. ( 3, 4); r ( 3, 0); r ( 2, 2); r Reteaching 11.2 Geometr

226 NSWERS 11. g 18.4, m F 41, m H m V 143, m W 19, m X 18 Lesson (0.71, 2.12), (4.95, 0.71), ( 2.12, 4.95) 11. (4.19, 2.73), (1.71, 4.70), (4.95, 3.93) 12. (7.31, 6.82), ( 7.80, 0.44), ( 7.90, 3.26) 13. ( 4.69, 9.33), ( 1.22, 10.37), ( 7.88, 1.39) 14. (0.62, 7.04), ( 3.31, 5.39), ( 8.95, 4.90) (1.22, 0.71), (2.26, 4.57), (6.57, 1.34), (7.61, 5.21) Reteaching hapter 11 Lesson ft d m mm ft in mi/h; mi/h; 20 Lesson (0, 2) 2. ( 2.60, 1.5) 3. ( 0.64, 1.26) 4. (3.57, 0.52) 5. ( 2.66, 3.59) 6. ( 2.94, 1.16) 7. (1, 2) 8. (2, 1) 9. ( 2, 2) TU QT heck drawings. Lesson heck drawings ; ; ; ; ; ; O 178 nswers Geometr

227 NSWERS 14. Lesson 11.5 O 1. segment consists of two points on a line and all the points of the line between them (an arc of a great circle); sample: EF 2. D and the small circle through and ; no; the small circle is not a line F and HD 4. G 5. not a line 6. not a line 7. not a line O Lesson ; ; ; ; ; ; n 2. 1; 3; 7; 15; 31; 63; 2 n 1 1 Lesson Yes; the graph would have eactl two odd vertices, and. 2. Yes; the graph would have eactl two odd vertices, and E. 3. 1; 2; 3; 4; 5; 6; n 1 4. The total length of the branches increases without limit Yes. 4. Yes. 5. No. 6. Yes. 7. Yes. 8. Yes. 9. No; es. 10. No; es No Lesson Yes. 2. No. 3. Yes. 4. No The torus and sphere are not topologicall equivalent. 6. The sphere is topologicall equivalent to a cube, which has Euler characteristic Geometr nswers 179

228 NME LSS DTE Reteaching 11.3 Graph Theor Skill Determining whether a graph contains an Euler path The diagram at the right is called a graph (or a network). Each point is a verte. Each segment or curve that links two vertices is called an edge. (n edge ma also connect a single verte to itself.) The degree of a verte is the number of edges at the verte. (n edge that connects a verte to itself is E counted twice.) verte is even if its degree is even, and odd if its degree is odd. n Euler path is a path that travels continuousl from one verte of a graph to each other verte, traveling along each edge eactl once. graph contains an Euler path if and onl if it has at most two odd vertices. Eample Determine whether the graph above contains an Euler path. There are three edges at verte, so has degree 3 and is odd. There are three edges at verte, so has degree 3 and is odd. There are five edges at verte, so has degree 5 and is odd. There are two edges at verte D, so D has degree 2 and D is even. There are three edges at verte E, so E has degree 3 and E is odd. Since there are more than two odd vertices in the graph, the graph does not contain an Euler path. D Suppose the indicated change were made to the graph in the figure at the top of the page. Would the resulting graph contain an Euler path? Eplain. 1. n edge is added from verte to verte E. 2. One edge from verte to is removed. Does the graph contain an Euler path? Write Yes or No Geometr Reteaching

229 NME LSS DTE Each verte in the graph at the right represents one room of an apartment. n edge connecting two vertices indicates a doorwa between the rooms. 6. Does the graph contain an Euler path? 7. an ou walk through the apartment on a continuous path, passing through each doorwa eactl once? Skill Determining whether a graph contains an Euler circuit n Euler path has a starting point and an ending point. n Euler circuit is an Euler path for which an verte is both a starting point and an ending point. For ever route into an even verte, there is another route out, so that an even verte can be anwhere on an Euler path. However, the edges at an odd verte do not occur in pairs, so if one route is taken in and another out, there is a verte left over. Then an odd verte can onl be at the beginning or end of an Euler path. It is clear that if a graph contains an odd vertices, it cannot contain an Euler circuit. n graph that contains onl even vertices contains an Euler circuit. Eample Does the graph contain an Euler circuit? If not, does it contain an Euler path? Yes. The graph has five vertices, each of which is even. Therefore, the graph contains an Euler circuit. Does the graph contain an Euler circuit? If not, does it contain an Euler path? Refer to the figure at the right. 11. Label each verte with its degree. 12. Is it possible to trace the figure in one stroke without lifting the pencil or retracing an part of the figure? 142 Reteaching 11.3 Geometr

230 NSWERS 14. Lesson 11.5 O 1. segment consists of two points on a line and all the points of the line between them (an arc of a great circle); sample: EF 2. D and the small circle through and ; no; the small circle is not a line F and HD 4. G 5. not a line 6. not a line 7. not a line O Lesson ; ; ; ; ; ; n 2. 1; 3; 7; 15; 31; 63; 2 n 1 1 Lesson Yes; the graph would have eactl two odd vertices, and. 2. Yes; the graph would have eactl two odd vertices, and E. 3. 1; 2; 3; 4; 5; 6; n 1 4. The total length of the branches increases without limit Yes. 4. Yes. 5. No. 6. Yes. 7. Yes. 8. Yes. 9. No; es. 10. No; es No Lesson Yes. 2. No. 3. Yes. 4. No The torus and sphere are not topologicall equivalent. 6. The sphere is topologicall equivalent to a cube, which has Euler characteristic Geometr nswers 179

231 NME LSS DTE Reteaching 11.4 Topolog: Twisted Geometr Skill Eploring topologicall equivalent plane figures Two figures are topologicall equivalent if one can be stretched, shrunk, or otherwise distorted into the other without cutting tearing, or intersecting itself, or compressing a segment or a curve into a point. You can imagine that the figures are made of etremel elastic material and picture whether ou can manipulate one (without tearing it or puncturing an holes in it) until its appearance matches the other. simple closed curve is a curve that is topologicall equivalent to a circle. ccording to the Jordan urve Theorem, ever simple closed curve in a plane divides the plane into two regions, the inside and the outside. Ever curve that connects a point inside the curve to a point outside the curve must intersect the curve. Eample Determine whether Figure and Figure are topologicall equivalent. Figure and Figure are not topologicall equivalent. Figure is actuall a simple closed curve. Figure intersects itself. re the figures topologicall equivalent? TORUS Geometr Reteaching

232 NME LSS DTE Skill Determining the Euler characteristic of a topological figure In hapter 6, ou learned Euler s formula. For an polhedron with V vertices, E edges, and F faces, V E F 2. For an space figure, the number V E F is called the Euler characteristic of the figure. (Notice, then, that the Euler characteristic of an polhedron is 2.) When a figure is deformed into a topologicall equivalent figure, the Euler characteristic remains unchanged. That is, the Euler characteristic is invariant. roperties that are changed in the process of deforming are not invariant. For eample, distance, area, and volume are not invariant. Eample The figures are topologicall equivalent. The figure on the left has 16 vertices, 32 edges, and 16 faces. The donut-shaped figure on the right is called a torus. Find the Euler characteristic of a torus. The Euler characteristic of the figure on the left is Since the torus is topologicall equivalent to the figure, the Euler characteristic of the torus is 0. Refer to the tet and the eample above. 5. Eplain wh the Euler characteristic of a sphere is not Eplain wh the Euler characteristic of a sphere is 2. The figure in Eercise 7 has 28 vertices, 56 faces, and 26 edges and is topologicall equivalent to the figure in Eercise 8. Find the Euler characteristic of each figure Reteaching 11.4 Geometr

233 NSWERS 14. Lesson 11.5 O 1. segment consists of two points on a line and all the points of the line between them (an arc of a great circle); sample: EF 2. D and the small circle through and ; no; the small circle is not a line F and HD 4. G 5. not a line 6. not a line 7. not a line O Lesson ; ; ; ; ; ; n 2. 1; 3; 7; 15; 31; 63; 2 n 1 1 Lesson Yes; the graph would have eactl two odd vertices, and. 2. Yes; the graph would have eactl two odd vertices, and E. 3. 1; 2; 3; 4; 5; 6; n 1 4. The total length of the branches increases without limit Yes. 4. Yes. 5. No. 6. Yes. 7. Yes. 8. Yes. 9. No; es. 10. No; es No Lesson Yes. 2. No. 3. Yes. 4. No The torus and sphere are not topologicall equivalent. 6. The sphere is topologicall equivalent to a cube, which has Euler characteristic Geometr nswers 179

234 NME LSS DTE Reteaching 11.5 Euclid Unparalleled Skill Identifing lines and relationships among lines in spherical geometr In Euclidean geometr, the arallel ostulate (which ou studied in hapter 5) guarantees that through a point not on a line there is eactl one line through the given point parallel to the given line. If the arallel ostulate is not true, what are the alternatives? There are two. 1. There is no line through the given point parallel to the given line. 2. There is more than one line through the given point parallel to the given line. The first statement is true for spherical geometr. The second is true for hperbolic geometr. that the terms point, line, and plane are not defined in Euclidean geometr. Suppose ou thought of a plane not as a flat surface etending infinitel in all directions, but as a sphere. The Euclidean interpretation of a straight line would make no sense; there would be no such lines in the plane. However, ou can use our understanding of Euclidean geometr to understand the concepts of lines on a sphere. In the tet that follows, the terms straight line and line will be used to distinguish between a line in a plane and a line on a sphere. If a plane intersects a sphere, the intersection is a circle. If the plane intersects the center of the sphere, the circle is a great circle that divides the sphere into two halves. If ou consider that a great circle etends infinitel in either direction, it makes sense that in spherical geometr, a great circle is a line. On a sphere, an two great circles intersect in two points. The great circle through points and ma be referred to as. Since the arallel ostulate of Euclidean Geometr is not true for spherical geometr, it is clear that man of the theorems of Euclidean geometr that rel on that postulate for proof are not true either. For eample, the sum of the measures of a triangle on a sphere is alwas between 180 and 540. Eample Tell whether each figure is a line. a. the circle that passes through and b. the circle that passes through and D c. the arc from E to F a. No; the circle through and is not a great circle. b. Yes; the great circle is D. c. No; the arc from E to F is not a great circle. Refer to the figure in the eample. E F D great circles 1. onsider the definition of segment in Euclidean geometr and write a definition of segment for spherical geometr. Then identif a segment in the figure. Geometr Reteaching

235 NME LSS DTE 2. Name two figures on the sphere in the eample on the preceding page that appear to be parallel. Is this a contradiction? Eplain. 3. Refer to the figure at right. Identif all lines. great circles D H G F E Skill Identifing lines and their relationships in hperbolic geometr There are several models of hperbolic geometr in which through an given line and a point not on the line, there are infinitel man lines parallel to the given line. One such model is the oincaré model. In this model, the surface is neither a plane nor a sphere, but a circle. line in oincaré s model is an arc that has endpoints on the circle, is inside the circle, and is orthogonal to it. (Orthogonal means perpendicular.) Eample In the figure, H is not on K. Name two different lines through H that are parallel to K. D and E Refer to the figure in the eample above. If the curve through the given points is a line, write its name. If not, write not a line. 4. and G 5. J and F 6. I and L 7. M and N D I H J K L M E F G N 146 Reteaching 11.5 Geometr

236 NSWERS 14. Lesson 11.5 O 1. segment consists of two points on a line and all the points of the line between them (an arc of a great circle); sample: EF 2. D and the small circle through and ; no; the small circle is not a line F and HD 4. G 5. not a line 6. not a line 7. not a line O Lesson ; ; ; ; ; ; n 2. 1; 3; 7; 15; 31; 63; 2 n 1 1 Lesson Yes; the graph would have eactl two odd vertices, and. 2. Yes; the graph would have eactl two odd vertices, and E. 3. 1; 2; 3; 4; 5; 6; n 1 4. The total length of the branches increases without limit Yes. 4. Yes. 5. No. 6. Yes. 7. Yes. 8. Yes. 9. No; es. 10. No; es No Lesson Yes. 2. No. 3. Yes. 4. No The torus and sphere are not topologicall equivalent. 6. The sphere is topologicall equivalent to a cube, which has Euler characteristic Geometr nswers 179

237 NME LSS DTE Reteaching 11.6 Fractal Geometr Skill Eamining the properties of fractals fractal is a geometric figure that is self-similar. The figure appears the same, whether at a distance or up close. Each part of the figure is similar to the entire figure. fractal can be produced b starting with a simple figure, changing it in some prescribed wa, and then repeating the process over and over. This repetition is called iteration. Man objects in nature can be modeled b fractals. Eample The growth of a tree can be modeled b a fractal. The fractal begins with a single branch (which will become the trunk) with a length of 1. In the first iteration, two new branches grow. Each is half the length of the original branch. For each subsequent iteration, two branches grow out of each new branch from the previous iteration. gain, each is half the length of the branches from the previous iteration. Make a table showing the number of new branches after each iteration and write an epression for the number of new branches after n iterations. iterations n number o f new branches n Refer to the eample. omplete each table. 1. iterations n length of new branches 2. iterations n total number of branches Geometr Reteaching

238 NME LSS DTE 3. iterations n total length of branches 4. Describe what happens to the total length of the branches as the number of iterations increases. Skill Using iteration to draw fractals To draw a fractal, begin with a simple figure, change it in some prescribed wa, and then repeat the process over and over. Eample n isosceles right triangle is shown. Sketch the first iteration of a fractal b drawing a right triangle on each leg of the given triangle. When the two triangles are drawn, the resulting figure will be a rectangle. To sketch the triangles, sketch a line parallel to the hpotenuse of the original triangle and then two perpendicular segments to form the rectangle. 5. Draw the second, third, and fourth iterations of the fractal in the eample. 148 Reteaching 11.6 Geometr

239 NSWERS 14. Lesson 11.5 O 1. segment consists of two points on a line and all the points of the line between them (an arc of a great circle); sample: EF 2. D and the small circle through and ; no; the small circle is not a line F and HD 4. G 5. not a line 6. not a line 7. not a line O Lesson ; ; ; ; ; ; n 2. 1; 3; 7; 15; 31; 63; 2 n 1 1 Lesson Yes; the graph would have eactl two odd vertices, and. 2. Yes; the graph would have eactl two odd vertices, and E. 3. 1; 2; 3; 4; 5; 6; n 1 4. The total length of the branches increases without limit Yes. 4. Yes. 5. No. 6. Yes. 7. Yes. 8. Yes. 9. No; es. 10. No; es No Lesson Yes. 2. No. 3. Yes. 4. No The torus and sphere are not topologicall equivalent. 6. The sphere is topologicall equivalent to a cube, which has Euler characteristic Geometr nswers 179

240 NME LSS DTE Reteaching 11.7 Other Transformations: rojective Geometr Skill Drawing images of figures under affine transformations The transformations we have studied so far (reflections, rotations, translations, and dilations) are all special tpes of affine transformations. n affine transformation transforms a plane figure in such a wa that the images of collinear points are collinear points, the images of straight lines are straight lines, the images of intersecting lines are intersecting lines, and the images of parallel lines are parallel lines. ffine transformations can be described using coordinates. Eample ( Let T be the affine transformation T(, ), 1 and let D be the 2 2 ) quadrilateral with vertices (0, 0), (6, 0), (6, 4), and D(2, 4). Draw D and its image, '''D', under T. ( ' (0), 1 (0, 0) 2 2 (0) ) ( ' (6), 1 (9, 0) 2 2 (0) ) D ( ' (6), 1 (9, 2) 2 2 (4) ) ' ( D' (2), 1 (3, 2) 2 2 (4), ' ) D' ' Given an affine transformation and the vertices of a figure, draw the figure and its image under the transformation. ( 1 2, 3 2 ) 1. T(, ) ( 2, 2) 2. T(, ) (0, 0), ( 3, 2), (2, 5) ( 3, 3), Q( 3, 3), R(4, 3), S(6, 3) Geometr Reteaching

241 NME LSS DTE Skill Drawing projection images of figures When a movie is shown in a theater, the image of a piece of film is projected onto a screen. The process involves a central projection. Given a point O, called the center of projection, the image of an point is a point ' on O. The ras from O are called projective ras. In the figure, the points and on line are projected onto line k. The center of projection is point. k ' ' rojective geometr is the stud of the properties of figures that do not change under a projection. rojective geometr includes no concepts of size, measurement, or congruence. Its theorems deal with the concepts of position of points and intersections of lines. In drawing projection images, onl an unmarked straightedge ma be used. Eample Let point O be a center of projection and draw a projection image of QR. Q O R First note that there are infinitel man possible images. egin b drawing O, OQ, and OR. hoose a random point ' on O, a random point Q' on, and a random point R on OR OQ ' Q'. Draw 'Q'R', a Q projection image of QR. O R R' Given, the center of projection, draw a projection image of the given triangle Reteaching 11.7 Geometr

242 NSWERS Lesson ' 2. triangle is a polgon and a heagon has three sides; true. 3. ine trees are evergreens or gorillas are pink; false. ' 4. The moon is a planet or Neptune is a star; false. 5. Ice hocke is a sport or braham Lincoln s portrait is on the nickel; true., ' 6. not truth functionall equivalent 2. Q S' R' 3. heck drawings. 4. heck drawings. Reteaching hapter 12 Lesson modus tollens; valid 2. dening the antecedent; invalid 3. affirming the consequent; invalid 4. modus ponens; valid 5. no conclusion 6. You will bu a pizza. 7. M age is not divisible b 9. Lesson 12.2 Q' ' 1. ll squares are rectangles and a ard is three feet long; true. R S 7. truth functionall equivalent Lesson If not, then not ; if not is true and not is false (that is, is false and is true) 2. If I finish m report, then the librar is open; if I finish m report and the librar is not open. 3. If there is not an orange in m lunch, then it is not Tuesda; if there is not an orange in m lunch and it is Tuesda. 4. If n + 1 is odd, then n is even; if n + 1 is odd and n is odd. 5. If a number is even, then it is a multiple of If a triangle is isosceles, then it has at least two congruent sides. 7. If a number is negative, then it does not have a real square root. Lesson Not a contradiction 2. Not a contradiction 3. contradiction 4. contradiction 5. Let 2 2 and assume temporaril that. 180 nswers Geometr

243 NME LSS DTE Reteaching 12.1 Truth and Validit in Logical rguments Skill Identifing valid and invalid arguments In logic, a statement is a sentence that is either true or false. n argument is a sequence of statements. The initial statements, called premises, lead to the final statement, called the conclusion. valid argument is one in which whenever the premises are true, the conclusion is true. In this case, the conclusion is said to be a logical consequence of the premises. The validit of an argument is determined b its form, not b the truth of its premises. Two valid argument forms are the argument form, called modus ponens (proposing mode), and the law of indirect reasoning, called modus tollens (removing mode). Valid argument true premises modus ponens If p then q p Therefore, q modus tollens If p then q Not q true conclusion Therefore, not p n argument is invalid if the conclusion does not follow logicall from the premises. Two invalid forms are called affirming the consequent, and dening the antecedent. In affirming the consequent, ou assume that because the conclusion of a conditional statement is true, the hpothesis is true. In dening the antecedent, ou assume that if the hpothesis is not true, then the conclusion must also not be true. oth are logical errors or fallacies. Invalid argument true premises conclusion (not necessaril true) affirming the consequent If p then q q Therefore, p deninng the antecedent If p then q Not p Therefore, not q Eample Determine whether the argument is valid. a. If urelia lives in New York it, then she lives in rgentina. urelia lives in New York it. Therefore, she lives in rgentina. b. If Louis wears warm mittens, then it is winter. Louis does not wear warm mittens. Therefore, it is not winter. a. Let p urelia lives in New York it and q she lives in rgentina. The argument can be written: If p then q; p. Therefore, q. The argument has the modus ponens form, so it is a valid argument even though the premise is not true. b. Let p Louis wears warm mittens and q it is winter. The argument can be written: If p then q; not p. Therefore, not q. The argument has the dening the antecedent form, so it is invalid. Geometr Reteaching

244 NME LSS DTE Identif the form of the argument and tell whether it is valid or invalid. 1. If a biccle has wheels, then pigeons can skate. igeons cannot skate. Therefore, a biccle does not have wheels. 2. If it raining, the parade will be canceled. It is not raining. Therefore, the parade will not be canceled. 3. If a quadrilateral is a parallelogram, then it has a pair of opposite sides that are parallel. JKLM has a pair of opposite sides that are parallel. Therefore, JKLM is a parallelogram. 4. If I pass the test, then I will pass the course. I pass the test. Therefore, I pass the course. Skill Reaching conclusions If a valid argument can be written using a sequence of true statements, then a true conclusion can be reached. Eample Determine what, if an, conclusion can be reached if both premises are true. Eplain. If a rectangle is a square, then it is a rhombus. Rectangle D is not a rhombus. Let p a rectangle is a square and q it is a rhombus. The premises can be written: If p, then q; not q. This is the form of the premises of the modus tollens argument. The conclusion that follows is not p. Rectangle D is not a square. Tell what, if an, conclusion can be reached if both premises are true. 5. If mone grows on trees, then the streets are paved with gold. Mone does not grow on trees. 6. If I rent a video, then ou will bu a pizza. I rent a video. 7. If a number is divisible b 9, then it is divisible b 3. M age is not divisible b Reteaching 12.1 Geometr

245 NSWERS Lesson ' 2. triangle is a polgon and a heagon has three sides; true. 3. ine trees are evergreens or gorillas are pink; false. ' 4. The moon is a planet or Neptune is a star; false. 5. Ice hocke is a sport or braham Lincoln s portrait is on the nickel; true., ' 6. not truth functionall equivalent 2. Q S' R' 3. heck drawings. 4. heck drawings. Reteaching hapter 12 Lesson modus tollens; valid 2. dening the antecedent; invalid 3. affirming the consequent; invalid 4. modus ponens; valid 5. no conclusion 6. You will bu a pizza. 7. M age is not divisible b 9. Lesson 12.2 Q' ' 1. ll squares are rectangles and a ard is three feet long; true. R S 7. truth functionall equivalent Lesson If not, then not ; if not is true and not is false (that is, is false and is true) 2. If I finish m report, then the librar is open; if I finish m report and the librar is not open. 3. If there is not an orange in m lunch, then it is not Tuesda; if there is not an orange in m lunch and it is Tuesda. 4. If n + 1 is odd, then n is even; if n + 1 is odd and n is odd. 5. If a number is even, then it is a multiple of If a triangle is isosceles, then it has at least two congruent sides. 7. If a number is negative, then it does not have a real square root. Lesson Not a contradiction 2. Not a contradiction 3. contradiction 4. contradiction 5. Let 2 2 and assume temporaril that. 180 nswers Geometr

246 NME LSS DTE Reteaching 12.2 nd, Or, and Not in Logical rguments Skill Writing conjunctions and determining their truth values conjunction is a compound statement in which two statements are joined b the word and. The truth table below shows when the conjunction p ND q is true and when it is false. p q p ND q T T T T F F F T F F F F The conjunction p ND q is true if and onl if both p and q are true. Eample Write a conjunction using the two statements and determine whether the conjunction is true or false: cube has 8 vertices. The merican flag is green. cube has 8 vertices and the merican flag is not green. The conjunction is false because onl one of the statements is true. Write a conjunction using the given statements and determine whether the conjunction is true or false. 1. ll squares are rectangles. ard is three feet long. 2. triangle is a polgon. heagon has three sides. Skill Writing disjunctions and determining their truth values disjunction is a compound statement in which two statements are joined b the word or. The truth table for the disjunction p OR q is shown. p q p OR q Tp Tq p ND T q T TF T The disjunction p OR q is true if either p is true, q or true, or both p and q are true. The disjunction is false if onl if both p and q are false. TF TF TF F TF F F Eample F F Write a conjunction using the two statements and determine whether the conjunction is true or false: ats have wings. aseballs are not cubes. ats have wings or baseballs are not cubes. The disjunction is true because one statements is true. Geometr Reteaching

247 NME LSS DTE Write a disjunction using the given statements and determine whether the disjunction is true or false. 3. ine trees are evergreens. Gorillas are pink. 4. The moon is a planet. Neptune is a star. 5. Ice hocke is a sport. braham Lincoln s portrait is on the nickel. Skill Determining whether statements are truth functionall equivalent Given an statement p, ou can write the statement NOT p (or p), called the negation of p. For eample, if p the car is green, p the car is not green. For an statement p, p and p have opposite truth values. Two statements are truth functionall equivalent if the have the same truth value. Eample The truth table for p q is given. omplete the table to determine whether p q and p OR q are truth functionall equivalent. p T T q T F p q T F p OR q The disjunction p OR q is true if p is true, if q is true, or if both p and q are true. Since p q and p OR q have the same truth value, the are truth functionall equivalent. Determine whether the statements are truth functionall equivalent. 6. p q p q p ND q p OR q 7. T T F F T F T F F T T F F T p q p q p OR q T T T T T F F F F T T T F F T T p q p q (p ND q) p OR q T T F F T F T F 154 Reteaching 12.2 Geometr

248 NSWERS Lesson ' 2. triangle is a polgon and a heagon has three sides; true. 3. ine trees are evergreens or gorillas are pink; false. ' 4. The moon is a planet or Neptune is a star; false. 5. Ice hocke is a sport or braham Lincoln s portrait is on the nickel; true., ' 6. not truth functionall equivalent 2. Q S' R' 3. heck drawings. 4. heck drawings. Reteaching hapter 12 Lesson modus tollens; valid 2. dening the antecedent; invalid 3. affirming the consequent; invalid 4. modus ponens; valid 5. no conclusion 6. You will bu a pizza. 7. M age is not divisible b 9. Lesson 12.2 Q' ' 1. ll squares are rectangles and a ard is three feet long; true. R S 7. truth functionall equivalent Lesson If not, then not ; if not is true and not is false (that is, is false and is true) 2. If I finish m report, then the librar is open; if I finish m report and the librar is not open. 3. If there is not an orange in m lunch, then it is not Tuesda; if there is not an orange in m lunch and it is Tuesda. 4. If n + 1 is odd, then n is even; if n + 1 is odd and n is odd. 5. If a number is even, then it is a multiple of If a triangle is isosceles, then it has at least two congruent sides. 7. If a number is negative, then it does not have a real square root. Lesson Not a contradiction 2. Not a contradiction 3. contradiction 4. contradiction 5. Let 2 2 and assume temporaril that. 180 nswers Geometr

249 NME LSS DTE Reteaching 12.3 loser Look at If-Then Statements Skill Writing the converse, the inverse, and the contrapositive of a given conditional and determining their truth values The conditional statement If p then q or p implies q can be written in smbols as p q. The hpothesis of the conditional is p, and the conclusion is q. Given the conditional p q, ou can write three related statements. Statement Smbols Description converse q p The hpothesis and conclusion are interchanged. (If q then p.) inverse p q The hpothesis and the conclusion are negated. (If not p then not q.) contrapositive q p The hpothesis and the conclusion are negated and interchanged. (If not q then not p.) s ou will see in the truth table below, a conditional is false onl if the hpothesis is true and the conclusion is false. Since the converse, inverse, and contrapositive of a conditional are all conditionals as well, each is also onl false when the hpothesis is true and the conclusion is false. p q p q p q q p p q q p T T F F T T T T T F F T F T T F F T T F T F F T F F T T T T T T conditional and its contrapositive are truth functionall equivalent. The converse and the inverse are also truth functionall equivalent. Eample Given the statement, If Rick has brown ees, then Luc has red hair write the converse, inverse, and contrapositive. Determine the conditions under which each of the four statements is false. converse: If Luc has red hair, then Rick has brown ees. The converse is false if Luc has red hair but Rick does not have brown ees. inverse: If Rick does not have brown ees, then Luc does not have red hair. The inverse is false if Rick does not have brown ees but Luc has red hair. contrapositive: If Luc does not have red hair, then Rick does not have brown ees. The contrapositive is false if Luc does not have red hair Rick does have brown ees. Geometr Reteaching

250 NME LSS DTE conditional statement is given. Write the indicated related conditional, and describe the circumstances under which the related conditional is false. 1. If, then. (contrapositive) 2. If the librar is open, then I will finish m report. (converse) 3. If there is an orange in m lunch, then it is Tuesda. (inverse) 4. If n is even, then n 1 is odd. (converse) Skill Writing conditional statements in if-then form Not ever conditional is written in if-then form. It ma be helpful to rewrite such sentences when using them in logic. Eample Write each statement in if-then form. a. ll triangles are polgons. b. No rectangles are trapezoids. a. If a figure is a triangle, then it is a polgon. b. If a figure is a rectangle, then it is not a trapezoid. Rewrite each statement in if-then form. 5. Ever even number is a multiple of ll isosceles triangles have at least two congruent sides. 7. No negative number has a real square root. 156 Reteaching 12.3 Geometr

251 NSWERS Lesson ' 2. triangle is a polgon and a heagon has three sides; true. 3. ine trees are evergreens or gorillas are pink; false. ' 4. The moon is a planet or Neptune is a star; false. 5. Ice hocke is a sport or braham Lincoln s portrait is on the nickel; true., ' 6. not truth functionall equivalent 2. Q S' R' 3. heck drawings. 4. heck drawings. Reteaching hapter 12 Lesson modus tollens; valid 2. dening the antecedent; invalid 3. affirming the consequent; invalid 4. modus ponens; valid 5. no conclusion 6. You will bu a pizza. 7. M age is not divisible b 9. Lesson 12.2 Q' ' 1. ll squares are rectangles and a ard is three feet long; true. R S 7. truth functionall equivalent Lesson If not, then not ; if not is true and not is false (that is, is false and is true) 2. If I finish m report, then the librar is open; if I finish m report and the librar is not open. 3. If there is not an orange in m lunch, then it is not Tuesda; if there is not an orange in m lunch and it is Tuesda. 4. If n + 1 is odd, then n is even; if n + 1 is odd and n is odd. 5. If a number is even, then it is a multiple of If a triangle is isosceles, then it has at least two congruent sides. 7. If a number is negative, then it does not have a real square root. Lesson Not a contradiction 2. Not a contradiction 3. contradiction 4. contradiction 5. Let 2 2 and assume temporaril that. 180 nswers Geometr

252 NME LSS DTE Reteaching 12.4 Indirect roof Skill Identifing contradictions You know that if a logical statement p is true, then its negation, p, is false. onsider the conjunction p ND p. Since a conjunction is true if and onl if both statements are true, the conjunction p ND p is never true. In fact, this statement is called a contradiction. If a logical argument results in a contradiction, the premise of the argument must be rejected. This tpe of argument is called reductio ad absurdum, which is Latin for reduction to absurdit. Eample Determine whether the given conjunction is a contradiction. a. triangle is a quadrilateral and a triangle is not a quadrilateral. b. square is a rhombus and a square is a parallelogram. a. Let p triangle is a quadrilateral. Then p a triangle is not a quadrilateral, so the given statement has the form p ND p and is a contradiction. b. Let p a square is a rectangle. The second part of the conjunction (a square is a parallelogram) is not equivalent to p. Then the statement is not a contradiction. Determine whether the given conjunction is a contradiction. If it is not, write a contradiction using one of the two statements. 1. cow is a mammal and a cow is not a biped. 2. The number is negative and the number is positive. 3. The llards live in Houston and the llards do not live in Teas. 4. regular heagon is equilateral and a regular heagon does not have congruent sides. Geometr Reteaching

253 NME LSS DTE Skill Using contradictions in indirect proof To prove a statement is true b contradiction (or indirect proof ), assume that it is false and show that this assumption leads to a contradiction. That is, to prove p q, assume that both p and q are true and show that this leads to a contradiction. DeMorgan s Laws (one of which ou learned in Lesson 12.2) are often useful in indirect proof. (p ND q) is truth functionall equivalent to p OR q. (p OR q) is truth functionall equivalent to p ND q. Eample Write an indirect proof that if ab 0, either a 0 or b 0. egin b assuming the hpothesis and the negation of the conclusion. The conclusion is in the form p OR q, the negation of which is truth functionall equivalent to p ND q. Then let ab 0, and assume temporaril that a 0 1 and b 0. Since a 0, a has a multiplicative inverse,. a ab 0 1 a (ab) 1 a 0 ( 1 a a) b 0 1 b 0 This contradicts the assumption that b 0. Therefore the assumption that a 0 and b 0 is false. That is, if ab 0, then a 0 or b 0. Write the first line of an indirect proof of the given statement. 5. If 2 2, then. 6. If the product of integers a and b is even, then a is even or b is even. Write an indirect proof. 7. If a is rational and b is irrational, then a b is irrational. (Hint: If a and a b are rational, there are integers m, n,, and such that a m and a b.) n 158 Reteaching 12.4 Geometr

254 NSWERS Lesson ' 2. triangle is a polgon and a heagon has three sides; true. 3. ine trees are evergreens or gorillas are pink; false. ' 4. The moon is a planet or Neptune is a star; false. 5. Ice hocke is a sport or braham Lincoln s portrait is on the nickel; true., ' 6. not truth functionall equivalent 2. Q S' R' 3. heck drawings. 4. heck drawings. Reteaching hapter 12 Lesson modus tollens; valid 2. dening the antecedent; invalid 3. affirming the consequent; invalid 4. modus ponens; valid 5. no conclusion 6. You will bu a pizza. 7. M age is not divisible b 9. Lesson 12.2 Q' ' 1. ll squares are rectangles and a ard is three feet long; true. R S 7. truth functionall equivalent Lesson If not, then not ; if not is true and not is false (that is, is false and is true) 2. If I finish m report, then the librar is open; if I finish m report and the librar is not open. 3. If there is not an orange in m lunch, then it is not Tuesda; if there is not an orange in m lunch and it is Tuesda. 4. If n + 1 is odd, then n is even; if n + 1 is odd and n is odd. 5. If a number is even, then it is a multiple of If a triangle is isosceles, then it has at least two congruent sides. 7. If a number is negative, then it does not have a real square root. Lesson Not a contradiction 2. Not a contradiction 3. contradiction 4. contradiction 5. Let 2 2 and assume temporaril that. 180 nswers Geometr

255 NSWERS 6. Let a and b be integers with ab even and assume temporaril that a is not even and b is not even. 7. Let a be rational and b irrational and assume temporaril that a b is rational. Then there are integers m, n,, and m such that a and a b. Then n m m b (a b) a. Since b n n is a ratio of two integers, b is rational.this contradicts the fact that b is irrational, which was given. Then the assumption that a b is rational is false and a b is irrational. Lesson p q q pnd q NOT (p ND q) T T F F T T F T T F F T F F T F F T F T 2. p q p q p OR q ( p OR q) ND p T T F F F F T F F T T T F T T F T F F F T T T F 3. (p OR q) ND (r OR s) 4. ( (p OR q)) ND ( r) 5. NOT ( p ND q) OR ( r) 6. ( p ND q) OR (p OR r) Geometr nswers 181

256 NME LSS DTE Reteaching 12.5 omputer Logic Skill onstructing input-output tables for networks of logic gates In performing an operation, a computer goes through a series of operations which can be thought of as on (or 1) and off (or 0). logic gate is an electronic circuit that responds to an electrical impulse. Three such logic gates are the NOT, ND, and OR gates, which act like their counterparts, the logical operators NOT, ND, and OR. n input-output table, similar to a truth table, ma be constructed for each logical gate with 1 corresponding to true and 0 corresponding to false. p NOT ~ p p q ND p ND q p q OR p OR q NOT Input Output p NOT p ND Input Output p q p ND q OR Input Output p q p OR q Eample onstruct an input-output table for the network. Read from left to right. The first gate is ND, so the output is p ND q. The second is NOT, and the output is NOT (p and q). onsider all combinations of possible input values. omplete the input-output table for the network. 1. p q NOT ND NOT Input p q ND Output NOT p q p ND q NOT (p and q) p q q p ND q NOT (p ND q) Geometr Reteaching