ANSWERS. LESSON 1.1 Building Blocks of Geometry. LESSON 1.3 What s a Widget? LESSON 1.2 Poolroom Math. LESSON 1.4 Polygons 17.

Transcription

1 NSWES LESSON 1.1 uilding locks of Geometry S. 9 cm. SN 4. endpoint 5. NS 6. Q 7. S 8. KN KL, NM LM, NO LO 9. E(14, 15) cm M 1.5 cm.0 cm cm 1.5 cm cm E Q Z LESSON 1. What s a Widget? 1. d. c. e 4. i 5. f 6. b 7. h 8. a 9. g 10. hey have the same measure, 1. ecause mq 77, its complement has measure 1. So m 1, which is the same as m ,,, E, F,,, E, F,, E, F, E, F, EF (15 lines) 1. ossible coplanar set: {,, H, G}; 1 different sets LESSON 1. oolroom Math 1. vertex. bisector. side E N I G M O 160 z F E 1. LESSON 1.4 olygons olygon Number of Number of name sides diagonals 1. riangle 0. Quadrilateral 4. entagon Hexagon Heptagon Octagon ecagon odecagon E 10. N Y X Q Q U iscovering Geometry ractice Your Skills NSWES Key urriculum ress

2 11.,, E 1. ossible answer: and 1. ossible answer: and For Exercises 6 10, 1, and 1, answers may vary. ossible answers are shown. 6. a b 14. ossible answer: and F b b cm G a b N LESSON 1.5 riangles For Exercises 1 7, answers will vary. ossible answers are shown. 1. GH. EF I. G FH 4. EG and GEF 5. EG and GEF 7. F 8. EFHG 9. FJ 10. FH 11. (0, ) 1. E(0, 5) 1. G(16, ) LESSON 1.7 ircles I L M S N (8, ); (, 7); (, ) G H O For Exercises 10 1, answers may vary. ossible answers are shown. 10. F(8, ) 11. (4, ) 1. G(10, ) 9. he chord goes through the center,. (It is a diameter.) LESSON 1.6 Special Quadrilaterals 1. G. S Q E M H 90 O E 5. E K I 9 NSWES iscovering Geometry ractice Your Skills 008 Key urriculum ress

3 11. Kite 4. ossible answers: a. b. c. Q LESSON 1.8 Space Geometry 1.. LESSON.1 Inductive easoning 1. 0, , , 4. 1, , , , 140. ectangular prism 4. entagonal prism y y cubes 8. x, y 1 x (, 1) x LESSON 1.9 icture Is Worth a housand Words (1, ) 1. ossible locations Gas 11. False; m m ower 1. False; , rue. ora, Ellen, harles, nica, Fred, ruce LESSON. Finding the nth erm 1. Linear. Linear. Not linear 4. Linear 5. n E F f (n) riangles cute triangles 6. n g(n) Isosceles triangles Scalene triangles 7. f(n) 4n 5; f(50) f(n) 5n 11; f(50) 9 9. f(n) 1 n 6; f(50) 1 iscovering Geometry ractice Your Skills NSWES Key urriculum ress

4 n n rea of figure n n Number of triangles n n nswers will vary. ossible answers: a b c e 5 teams, 10 games d a d b 1 4 e 6 5 c f 7 6 teams, 7 games LESSON. Mathematical Modeling 1. b a f e c d a. 40 b. 150 c. f(n) n(n ), or f(n) n 4n. H H H H a. 8 sequences b. sequences have 1 tail. c. 8 H H H HHH HH HH H HH H H. 66 different pairs. Use a dodecagon showing sides and diagonals. 6 teams, 6 games LESSON.4 eductive easoning 1. No. Explanations will vary. Sample explanation: ecause is equilateral,. ecause lies between and,, so is not equal to. hus is not equilateral, by deductive reasoning.. nswers will vary. me m (me m 90 ); deductive. a, e, f; inductive 4. eductive a. 4x ( x) 8 x he original equation. 4x 6 x 8 x x 6 8 x x 6 8 x x istributive property. ombining like terms. ddition property of equality. Subtraction property of equality. ivision property of equality. 94 NSWES iscovering Geometry ractice Your Skills 008 Key urriculum ress

5 b. 19 ( x 1) 5 x he original equation. 19 (x 1) 5(x ) Multiplication property of equality. 19 6x 5x 10 istributive property. 1 6x 5x 10 ombining like terms. 1 11x 10 ddition property of equality x Subtraction property of equality. 1 x ivision property of equality. 5. a. 16, 1; inductive b. f(n) 5n 9; 41; deductive LESSON.1 uplicating Segments and ngles 1.. XY Q S X Q. ossible answer: Y S LESSON.5 ngle elationships 1. a 68, b 11, c 68. a 17. a 5, b 40, c 5, d a 90, b 90, c 4, d 48, e 1 5. a 0, b 70, c 0, d 70, e a 70, b 55, c 5 7. Sometimes 8. lways 9. Never 10. Sometimes 11. acute obtuse 15. converse LESSON.6 Special ngles on arallel Lines LESSON. onstructing erpendicular isectors 1.. Square 1. a 54, b 54, c 54. a 115, b 65, c 115, d 65. a 7, b cannot be determined 7. a 10, b 78, c 58, d 1, e 6, f x x 0, y 5 iscovering Geometry ractice Your Skills NSWES Key urriculum ress

6 . XY 5 4 LESSON. onstructing erpendiculars to a Line 1. False. he altitude from coincides with the side so it is not shorter. M. False. In an isosceles triangle, an altitude and median coincide so they are of equal length. X W Y 4. is not unique. M 5. is not unique. Q. rue 4. False. In an acute triangle, all altitudes are inside. In a right triangle, one altitude is inside and two are sides. In an obtuse triangle, one altitude is inside and two are outside. here is no other possibility so exactly one altitude is never outside. 5. False. In an obtuse triangle, the intersection of the perpendicular bisectors is outside the triangle S Q WX YZ 7. a. and b.,, and c. and and from and (but not from and ) d. and and from and E Q W X Y Z 96 NSWES iscovering Geometry ractice Your Skills 008 Key urriculum ress

7 LESSON.4 onstructing ngle isectors a. 1 and b. 1,, and M O c.,, and 4 d. 1 and and from and 4. is the bisector of H 6. ossible answer:. x 0, me O I Z 5. hey are concurrent. LESSON.6 onstruction roblems 1. ossible answer: E 6. I. ossible answer: K 7. N GN and O HO H erimeter O M N G U LESSON.5 onstructing arallel Lines 1.. S Q. 4. S m Q iscovering Geometry ractice Your Skills NSWES Key urriculum ress

8 . ossible answers: LESSON.7 onstructing oints of oncurrency E E 1. ircumcenter. Locate the power-generation plant at the incenter. Locate each transformer at the foot of the perpendicular from the incenter to each side.. 4. ossible answer: E U O S 5. ossible answer: 6. ossible answer: 4. ossible answer: In the equilateral triangle, the centers of the inscribed and circumscribed circles are the same. In the obtuse triangle, one center is outside the triangle. S 1 S 1 S S 7. ossible answer: S 1 S 1 5. ossible answer: In an acute triangle, the circumcenter is inside the triangle. In a right triangle, it is on the hypotenuse. In an obtuse triangle, the circumcenter is outside the triangle. (onstructions not shown.) S 1 S S 98 NSWES iscovering Geometry ractice Your Skills 008 Key urriculum ress

9 LESSON.8 he entroid cm, Q 5.7 cm, 4.8 cm G LESSON 4. roperties of Isosceles riangles 1. m 64. mg 45. x m 9, perimeter of 46 cm 5. LM 16 m, mm mq 44, Q a. b. c. by the onverse of the I onjecture. 8. x 1, y mq mq 55 by V, which makes m 55 by the riangle Sum onjecture. So, Q is isosceles by the onverse of the Isosceles riangle onjecture. LESSON 4. riangle Inequalities 6 cm 10 cm 1. Yes. No 8 cm 4. (, 4) 5. 16, L 8, QM 15, a. Incenter b. entroid c. ircumcenter d. ircumcenter e. Orthocenter f. Incenter g. entroid LESSON 4.1 riangle Sum onjecture 1. p 67, q 15. x 8, y 81. a 78, b 9 4. r 40, s 40, t x 1, y y s 8 8. m m a 10. mq he sum of the measures of and is 90 because m is 90 and all three angles must be 180. So, and are complementary. 1. me me because they are vertical angles. ecause the measures of all three angles in each triangle add to 180, if equal measures are subtracted from each, what remains will be equal. Q. 19 x 5 4. b a c 5. b c a 6. a c d b 7. x x he interior angle at is 60. he interior angle at is 0. ut now the sum of the measures of the triangle is not y the Exterior ngles onjecture, x x mqs. So,mQS x. So, by the onverse of the Isosceles riangle onjecture, QS is isosceles. 11. Not possible. 1. Q 17 km 9 km 8 km LESSON 4.4 re here ongruence Shortcuts? 1. S or S. SSS. SSS 4. QM (SS) 5. IE (SSS) iscovering Geometry ractice Your Skills NSWES Key urriculum ress

10 6. annot be determined, as shown by the figure. Y 9. ll triangles will be congruent by S. ossible triangle: Z X 7. NO (SS) 8. annot be determined, as shown by the figure. Q U S 9. OG (SS) 10. Only one triangle because of SSS. U S 11. wo possible triangles. 1. Only one triangle because of SS. Z X Y LESSON 4.5 re here Other ongruence Shortcuts? 1. annot be determined 10. ll triangles will be congruent by S. ossible procedure: Use and to construct and then copy and at the ends of. LESSON 4.6 orresponding arts of ongruent riangles 1. SSS, SS, S, S. YZ WX, I onjecture. WZ XY, I onjecture 4. S YWM ZXM by SS. YW ZX by. 7. by SS. by. 8. ossible answer: E and F are both the distance between and. ecause the lines are parallel, the distances are equal. So, E F. 9. ossible answer: E F (see Exercise 8). EF FE because both are right angles, EF FE because they are the same segment. So, EF FE by SS. E F by. 10. ossible answer: It is given that and, and because they are the same segment. So by SS and by. LESSON 4.7 Flowchart hinking Q. XZY (S). (S or S) 4. S (S) 5. N (S) 6. GQK (S or S) 7. Yes, QS MOL by SSS. 8. No, corresponding sides V and WV are not congruent. 1. (See flowchart proof at bottom of page 101.). (See flowchart proof at bottom of page 101.). (See flowchart proof at bottom of page 101.) LESSON 4.8 roving Special riangle onjectures m 6. m 5, 1 4. me N erimeter NSWES iscovering Geometry ractice Your Skills 008 Key urriculum ress

11 7. (See flowchart proof at bottom of page 10.) 8. Flowchart roof is a median efinition of median SSS onjecture bisects efinition of bisect Same segment. 170 ; 6 sides sides 5. x x me E LESSON 5. Exterior ngles of a olygon 1. 1 sides. 4 sides. 4 sides 4. 6 sides 5. a 64, b a 10, b 9 7. a 156, b 1, c a 15, b 40, c 105, d LESSON 5.1 olygon Sum onjecture 1. a 10, b 10, c 97, d 8, e 154. a 9, b 44, c 51, d 85, e 44, f 16 Lesson 4.7, Exercises 1,, 1. Q S Q S QS SQ QS SQ S Q I onjecture SS onjecture QS QS Same segment. KE KI EK IK KIE is a kite E I efinition of kite KE KI SSS onjecture EK IK K bisects EKI and EI efinition of bisect K K Same segment. I onjecture is a efinition of Same segment S onjecture efinition of I onjecture iscovering Geometry ractice Your Skills NSWES Key urriculum ress

12 LESSON 5. Kite and rapezoid roperties 1. x 0. x 14, y 56. x 64, y 4 4. x 1, y S 6. a M N 7. MNO is a. y the riangle Midsegment onjecture, ON M and MN O. Flowchart roof 8. O 1_ efinition of midpoint MN 1_ Midsegment onjecture 9. ossible answer: aragraph proof: raw E with E on. E is a. E because they are corresponding angles of parallel lines. because it is given, so E, because both are congruent to. herefore, E is isosceles by the onverse of the Isosceles riangle onjecture. E because they are opposite sides of a and E because E is isosceles. herefore, because both are congruent to E. LESSON 5.4 roperties of Midsegments 1. a 89, b 54, c 91. x 1, y 7, z. x 17, y 11, z erimeter XYZ 66, Q 7, ZX M(1, 6), N(14.5, ); slope 1.6, slope MN ick a point from which and can be viewed over land. Measure and and find the midpoints M and N. MN. M 1_ efinition of midpoint ON onjecture O MN oth congruent to 1_ ON NM oth congruent to ON MN SS onjecture ON 1_ Midsegment onjecture ON M oth congruent to 1_ NM onjecture 8. aragraph proof: Looking at FG, HI FG by the riangle Midsegment onjecture. Looking at Q, FG Q for the same reason. ecause FG Q, quadrilateral FGQ is a trapezoid and E is the midsegment, so it is parallel to FG and Q. herefore, HI FG E Q. Lesson 4.8, Exercise 7 Same segment is an altitude and are right angles efinition of altitude oth are right angles S onjecture is a median efinition of median onverse of I 10 NSWES iscovering Geometry ractice Your Skills 008 Key urriculum ress

13 LESSON 5.5 roperties of arallelograms 1. erimeter 8 cm., , 7 4. a 51, b 48, c a 41, b 86, c F F 1 esultant vector F 1 F 9. a 8, b 14, c 14, d 8, e 14, f 8, g 5, h 1, i 61, j 81, k No 4. c, d, f, g 5. d, e, g 6. f, g 7. h 8. c, d, f, g 9. c, d, f, g 10. d, g 11. d, g 1. None 1. arallelogram 14. LESSON 5.7 roving Quadrilateral roperties 1. (See flowchart proof at bottom of page.). Flowchart roof SSS onjecture Same segment. Flowchart roof LESSON 5.6 roperties of Special arallelograms 1. OQ 16, mqs 90,. mokl 45, mmol 90, perimeter KLMN. O 6, 11, mo 90 efinition of rhombus O O iagonals of bisect each other O O O O O O SSS onjecture iagonals of a bisect each other Lesson 5.7, Exercise 1 Q I onjecture Q Q S onjecture Q and Q bisect each other efinition of bisect Q I onjecture iscovering Geometry ractice Your Skills NSWES Key urriculum ress

14 4. Flowchart roof Opposite sides of LESSON 6.1 angent roperties 1. w 16. mqx 65. a. mnq 90, mmq 90 b. rapezoid. ossible explanation: M and NQ are both perpendicular to Q, so they are parallel to each other. he distance from M to Q is M, and the distance from N to Q is NQ. ut the two circles are not congruent, so M NQ. herefore, MN is not a constant distance from Q and they are not parallel. Exactly one pair of sides is parallel, so MNQ is a trapezoid. 4. y 1 x ossible answer: angent segments from a point to a circle are congruent. So,,, and. herefore,. 6. a cm b cm 7. X Y I onjecture X Y S onjecture X Y LESSON 6. hord roperties efinition of 1. a 95, b 85, c v cannot be determined, w 90 X Y oth are 90. z w 100, x 50, y w 49, x 1.5, y x 16 cm, y cannot be determined 7. Kite. ossible explanation: OM ON because congruent chords and are the same distance from the center. M N because they are halves of congruent chords. So, MON has two pairs of adjacent congruent sides and is a kite. 8. he perpendicular segment from the center of the circle bisects the chord, so the chord has length 1 units. ut the diameter of the circle is 1 units, and the chord cannot be as long as the diameter because it doesn t pass through the center of the circle. 9. (0,1), M(4, ) 10. m 49, m 5, m 156, m ossible answer: Fold and crease to match the endpoints of the arc. he crease is the perpendicular bisector of the chord connecting the endpoints. Fold and crease so that one endpoint falls on any other point on the arc. he crease is the perpendicular bisector of the chord between the two matching points. he center is the intersection of the two creases. enter LESSON 6. rcs and ngles 1. mxnm 40, mxn 180, mmn 100. x 10, y 60, z 10. a 90, b 55, c 5 4. a 50, b 60, c x m 90, m 7, m 6, m m 140, m 0, m 60, m p 18, q 87, r 58, s a 50, b 50, c 80, d 50, e 10, f 90, g 50, h 50, j 90, k 40, m 80, n NSWES iscovering Geometry ractice Your Skills 008 Key urriculum ress

15 LESSON 6.4 roving ircle onjectures 1. Flowchart roof X X angent Segments onjecture X X ransitivity X X angent Segments onjecture 4. Flowchart roof onstruct radii O, O, O, and O. O O efinition of arc measure bisects at X efinition of segment bisector O O adii of same circle O O adii of same circle. ngles are numbered for reference. 1 O O SS onjecture 4 O aragraph roof It is given that OE, so 1 by the onjecture. ecause O and O are radii, they are congruent, so O is isosceles. herefore 4 1 by the I onjecture. oth and 4 are congruent to 1, so 4. y the I onjecture, 4, so. he measure of an arc equals the measure of its central angle, so because their central angles are congruent, E E.. Flowchart roof X X angent Segments onjecture X XQ Q Segment addition X XQ X XS ddition roperty of Equality Q S ransitivity XQ XS E angent Segments onjecture X XS S Segment addition LESSON 6.5 he ircumference/iameter atio 1. 1 cm. r 1.5 cm. 60 cm 4. d 4 cm cm 6. d 4.0 cm, r 1.0 cm in. 8. Yes; about.0 in. 5 in. in cm 10. ress the square against the tree as shown. Measure the tangent segment on the square. he tangent segment is the same length as the radius. Use r to find the circumference. Q S efinition of congruent segments ree cm iscovering Geometry ractice Your Skills NSWES Key urriculum ress

16 LESSON 6.6 round the World 1. t least 7 olive pieces. bout.5 rotations ( ) 085 m/s (about km/s or just ( ) under mi/s) cm or 9. cm ( Sitting speed ) (64.5 4) 107,500 km/h LESSON 6.7 rc Length or EXLOION Intersection Secants, angents, and hords 1. x 1. m 70, me 150. m 114, me me 75, m x 80, y 110, z x 4, y 150, z 1 7. x 11, y 68, z 5 8. x 8, y ossible answers: he two points where the figure and the image intersect determine. Or connect any two corresponding points and construct the perpendicular bisector, which is. E E N E 5. -fold rotational symmetry, lines of reflection 6. -fold rotational symmetry 7. 1 line of reflection 8. 1 line of reflection 9. -fold rotational symmetry, lines of reflection 10. -fold rotational symmetry line of reflection N E LESSON 7.1 ransformations and Symmetry 1. I I 1. 4-fold rotational symmetry, 4 lines of reflection. L Q L 106 NSWES iscovering Geometry ractice Your Skills 008 Key urriculum ress

17 LESSON 7. roperties of Isometries 1. otation y (, ) (5, 1) (, 1) x (, 1) ( 5, 1) (, ). ranslation (, 6) ( 5, 4) (, 4) ( 1, ) y (, 0) x (, ) 6. (x, y) (x 1, y 6); translation; (8, 8), (8, 4) 7. (x, y) (x, y); reflection across the y-axis; (7, ), (4, 5) 8. (x, y) (y, x); reflection across the line y x; (7, 0), (0, ) LESSON 7. ompositions of ransformations 1. ranslation by, 5. otation 45 counterclockwise. ranslation by 16, 0 4. otation 180 about the intersection of the two lines 5. ranslation by 16, 0 6. otation 180 about the intersection of the two lines 7. eflection across the line x 8. eflection across the line x 9. mo 50 ; rotation 100 clockwise about O. eflection (0, 5) y (4, 5) O (, ) (0, ) (4, ) (6, ) x 10. ranslation cm along the line perpendicular to k and in the direction from k to. 4. S Fence 1 cm k S F 5. Fence S S F Fence S iscovering Geometry ractice Your Skills NSWES Key urriculum ress

18 LESSON 7.4 essellations with egular olygons 4. Sample answer: 1. n 15. n 0. ossible answer: regular tessellation is a tessellation in which the tiles are congruent regular polygons whose edges exactly match. 4. ossible answer: 1-uniform tiling is a tessellation in which all vertices are identical. LESSON 8.1 reas of ectangles and arallelograms cm. 7.5 cm. 110 cm cm m 6. No. ossible answer: cm 4 cm 5 cm 17 cm units 8. 7 units 9. No. arpet area is 0 yd 180 ft. oom area is (1.5 ft)(16.5 ft) 06.5 ft. ana will be ft short. LESSONS essellations 1.. Sample answer: LESSON 8. reas of riangles, rapezoids, and Kites ft. 0 cm. b 1 in cm cm cm units cm LESSON 8. rea roblems 1. a ft b. 40 bundles; $ L. ossible answer:. Sample answer: NSWES iscovering Geometry ractice Your Skills 008 Key urriculum ress

19 4. It is too late to change the area. he length of the diagonals determines the area. LESSON 8.4 reas of egular olygons cm. a 7.8 cm. p 4.6 cm 4. n s 4 cm, a.8 cm, 8 cm cm 1.4 cm 7. r 10 cm 8. x r 7 cm LESSON 8.7 Surface rea cm. 40 cm cm cm cm cm cm cm 9. 1 sheet: front rectangle: ; back rectangle: 1 71 ; bottom rectangle: 6; side trapezoids: 8; total 6 ft. rea of 1 sheet 4 8 ft. ossible pattern: 6. ossible answer (s will vary): s.1 cm, a.7 cm, 45.9 cm 1 1_ ft Side 1_ ft Front 1 1_ ft ft Left over ft 1_ ft Side 1 1_ ft ack 1_ ft ottom ft ft ft ft LESSON 9.1 he heorem of ythagoras 7. pproximately 1.5 cm : area of square 6; area of square within angle ; area of octagon 10; area of octagon within angle ; shaded area cm LESSON 8.5 reas of ircles cm cm. cm 4. 4 cm cm cm cm cm cm or.6 cm 11. (64 18) square units 1. 5 cm LESSON 8.6 ny Way You Slice It cm 6.54 cm. cm.51 cm. 1 cm 7.70 cm 4. (16 ) cm 18.7 cm cm 4.41 cm 1. a 1 cm. p.9 cm. x 8 ft 4. h 14. in. 5. rea 19.0 ft 6. (11, 1); r 5 7. rea 49.7 cm 8. V 15.4 cm 9. If the base area is 16 cm, then the radius is 4 cm. he radius is a leg of the right triangle; the slant height is the hypotenuse. he leg cannot be longer than the hypotenuse. 10. rea 150 in ; hypotenuse Q 5 in.; altitude to the hypotenuse 1 in. LESSON 9. he onverse of the ythagorean heorem 1. No. Yes. Yes 4. Yes 5. rea 1. cm 6. he top triangle is equilateral, so half its side length is.5. triangle with sides.5, 6, and 6.5 is a right triangle because So, the angle marked 95 should be x cm. y the onverse of the ythagorean heorem, is a right triangle, and is a right angle. and are supplementary, so is also a right triangle. Use the ythagorean heorem to find x. iscovering Geometry ractice Your Skills NSWES Key urriculum ress

20 cm 9. No. ecause, of is not a right angle. 10. annot be determined. he length of is unknown. One possible quadrilateral is shown. 11. Yes. Using SSS,. hat means that the four angles of the quadrilateral are all congruent by. ecause the four angles must sum to 60 and they are all congruent, they must be right angles. So, is a rectangle. LESSON 9. wo Special ight riangles 1. a 14 cm. a 1 cm, b 4 cm. a 1 cm, b 6 cm cm 5. erimeter 6 6 cm; area cm 6. 0 cm; 0 0 cm; area cm cm 8. 1, 9. (6, 6) 10. ossible answer: LESSON 9.4 Story roblems he foot is about 8.7 ft away from the base of the building. o lower it by ft, move the foot an additional. ft away from the base of the building.. bout 6.4 km 0.4 km linear feet of trim must be painted, or 4. feet. wo coats means ft of coverage. Just over 1 quarts of paint is needed. If Hans buys quarts, he would have almost 1 quart left. It is slightly cheaper to buy 1 gallon and have about 1 1 quarts left. he choice is one of money versus conserving. Students may notice that the eaves extend beyond the exterior walls of the house and adjust their answer accordingly in., in in., in in km 1.7 km.1 km 1. km 0.9 km. km LESSON 9.5 istance in oordinate Geometry units. 0 units. 17 units 4. is a rhombus: ll sides 4, slope 5, slope, 5 so is not a right angle, and is not a square. 5. UVW is an isosceles trapezoid: U and VW have slope 1, so they are parallel. UV and W have length 0 and are not parallel (slope UV 1, slope W ). 6. Isosceles; perimeter units 7. M(7, 10); N(10, 14); slope MN 4 ; slope 4 ; MN 5; 10; the slopes are equal; MN (x 1) (y 5) 4 9. enter (0, ), r he distances from the center to the three points on the circle are not all the same: 61, 61, NSWES iscovering Geometry ractice Your Skills 008 Key urriculum ress

21 LESSON 9.6 ircles and the ythagorean heorem 1. (5 4) cm, or about 54.5 cm. (7 4) cm, or about 49. cm. (58 7) cm 6.1 cm 4. rea cm cm cm 10.7 cm 6. S LESSON 10.1 he Geometry of Solids 1. oblique. the axis. the altitude 4. bases 5. a radius 6. right 7. ircle or 10. or 11. ight pentagonal prism 1. E and FGHIJ 1. F, G, H, I, EJ 14. ny of F, G, H, I, EJ or their lengths 15. False. he axis is not perpendicular to the base in an oblique cylinder. 16. False. rectangular prism has six faces. Four are called lateral faces and two are called bases. 17. rue LESSON 10. Volume of yramids and ones cm cm cm 4. V 840x 5. V 8 a b 6. V 4xy 7. : 18 cubic units, : 144 cubic units. is larger. 8. : 5 cubic units, : 5 cubic units. hey have equal volumes. 9. : 9x cubic units, : 7x cubic units. is larger. LESSON 10.4 Volume roblems cm. 178 in. 4 cans; 58 in.07 ft ; 4.6% lb (about 1 ton) 5. Note that E and E. V 8 cm ; S (8 4) cm 1.7 cm 6. bout 110,447 gallons truckloads LESSON 10.5 isplacement and ensity ll answers are approximate cm. 7.8 g/cm g/cm in. 5. No, it s not gold (or at least not pure gold). he mass of the nugget is 165 g, and the volume is cm, so the density is 9.4 g/cm. ure gold has density 19. g/cm. LESSON 10.6 Volume of a Sphere cm, or about cm. 18 cm, or about 56.5 cm LESSON 10. Volume of risms and ylinders cm. 144 cm cm 4. V 4xy(x ), or 8x y 1xy 5. V 1 4 p h 6. V 6 1 x y 7. 6 ft. 7 cm, or about 6. cm 4. 8 cm, or about 9. cm 5. 4 cm, or about 157. cm cm, or about 18. cm cm in in in ; 47.6% LESSON 10.7 Surface rea of a Sphere 1. V cm ; S cm. V 184. cm ; S 16.4 cm iscovering Geometry ractice Your Skills NSWES Key urriculum ress

22 . V cm ; S cm 4. V 4.1 cm ; S 61.1 cm 5. bout.9 cm 6. bout 57. cm 7. 9 quarts LESSON 11.1 Similar olygons 1. 8 cm; EI 7 cm; SN 15 cm; Y 1 cm. SL 5. cm; MI 10 cm; m 10 ; mu 85 ; m 80. Yes. ll corresponding angles are congruent. oth figures are s, so opposite sides within each are equal. he corresponding sides are proportional Yes. orresponding angles are congruent by the onjecture. orresponding sides are proportional 4 = 6 = No Yes. ll angles are right angles, so corresponding angles are congruent. he corresponding side lengths have the ratio 4, 7 so corresponding side lengths are proportional (0, 1) 8. 4 to 1 y y (, ) 5 (, 4) (1.5, 1.5) (, 0.5) x cm 7. E. ossible explanation: E and by I, so by the Similarity onjecture, the triangles are similar. 8. Q S. ossible explanation: S and Q because each pair is inscribed in the same arc, so by the Similarity onjecture, the triangles are similar. 9. MLK NOK. ossible explanation: MLK NOK by and K K because they are the same angle, so by the Similarity onjecture, the two triangles are similar. LESSON 11. Indirect Measurement with Similar riangles 1. 7 ft ft mi 4. bout 18.5 ft m, 1. m, 1.8 m,.4 m, and.0 m LESSON 11.4 orresponding arts of Similar riangles 1. h 0.9 cm; j 4.0 cm..75 cm, 4.50 cm, 5.60 cm. WX cm; 1 cm; 1 cm; YZ 8 cm; XZ cm 4. x cm; y cm 5. a 8 cm; b. cm; c.8 cm 6. 4 cm; 5.5 cm; 8.75 cm LESSON 11.5 roportions with rea cm 9. 4 cm :5 7. : cm tiles F(4, ) 5 E(8, ) x LESSON 11.6 roportions with Volume 1. Yes. No. 16 cm 4. 0 cm 5. 8: ft LESSON 11. Similar riangles 1. M 10.5 cm. Q X; Q 4.8 cm; QS 11. cm. E; 1.5 cm; 10 cm 4. S 15 cm; Q 51 cm 5. Similarity onjecture LESSON 11.7 roportional Segments etween arallel Lines 1. x 1 cm. Yes. No 4. NE 1.5 cm 5. 6 cm; Q 4 cm; I 1 cm 6. a 9 cm; b 18 cm 11 NSWES iscovering Geometry ractice Your Skills 008 Key urriculum ress

23 7. S.5 cm, E 0 cm 8. x 0 cm; y 7. cm 9. p cm; q 8.6 cm LESSON 1.1 rigonometric atios 1. sin p r. cos q r. tan p q 4. sin Q q r 5. sin cos tan sin x x x m m m 0 w 15. sin 40 ; 8 w 18.0 cm x 16. sin 8 ; 1 4 x 7.4 cm 17. cos 17 7 y ; y 76. cm 18. a t z 76 LESSON 1. roblem Solving with ight riangles 1. rea cm. rea 5 ft. rea 109 in 4. x y a 7.6 in. 7. iameter 0.5 cm bout.0 m 11. bout 445. ft 1. bout.6 ft LESSON 1. he Law of Sines 1. rea 46 cm. rea 4 m. rea 45 ft 4. m 14 cm 5. p 17 cm 6. q 1 cm 7. m 66, m 8. m 7, mq mk 81, mm Second line: about 15 ft, between tethers: about 15 ft LESSON 1.4 he Law of osines 1. t 1 cm. b 67 cm. w 4 cm 4. m 76, m 45, m m 77, m 66, ms 7 6. ms 46, mu 85, mv bout 4 8. bout 4.0 cm 9. bout 4.7 in. LESSON 1.5 roblem Solving with rigonometry 1. bout.85 mi/h; about 15. m 50.64, m 59.70, m bout 8.0 km from ower 1, 5.1 km from ower 4. bout 85 miles 5. bout 50 ft of fencing; about 11,656 ft LESSON 1.1 he remises of Geometry 1. a. b. istributive property c. Subtraction property d. ddition property e. ivision property. False. False 4. rue; transitive property of congruence and definition of congruence 5. Q LESSON 1. lanning a Geometry roof roofs may vary. 1. Flowchart roof Q M Q ostulate Q Q ostulate Q ostulate Q Q ostulate Q S ostulate Q hird ngle heorem iscovering Geometry ractice Your Skills NSWES Key urriculum ress

24 . Flowchart roof Q S. Flowchart roof a b c 180 riangle Sum heorem Q SU I heorem Q US hird ngle heorem U onverse of E heorem a x V heorem x b c 180 Substitution x y c 180 Substitution x y z 180 Substitution Q SU b y V heorem c z V heorem. roof: Statement eason is bisector 4. onverse of ngle of isector heorem efinition of angle bisector 6. is a right 6. efinition of angle perpendicular 7. is a right 7. efinition of angle perpendicular ight ngles re ongruent heorem S heorem. Flowchart roof MN QM MN NO ransitivity NO QM LESSON 1. riangle roofs roofs may vary. 1. Flowchart roof XY ZY XZ WY WY WY eflexive property WXY WZY SS heorem XZY is isosceles efinition of isosceles triangle YM is the altitude from vertex Y efinition of altitude and vertex angle YM is angle bisector of XYZ Isosceles riangle Vertex ngle heorem XYM ZYM efinition of angle bisector QMN and NMO are supplementary Linear air ostulate MNO is isosceles efinition of isosceles triangle NMO NO I heorem QMN ON Supplements of ongruent ngles heorem ON and NO are supplementary Linear air ostulate 4. roof: Statement eason is isosceles. efinition of isosceles triangle.. I heorem 4. E NSWES iscovering Geometry ractice Your Skills 008 Key urriculum ress

25 5. E 5. ransitivity 6. E 6. onverse of ostulate 7. E 7. ostulate is a right 9. efinition of angle perpendicular 10. E is a right 10. efinition of right angle angle, transitivity 11. E 11. efinition of perpendicular LESSON 1.4 Quadrilateral roofs roofs may vary. 1. : is a Show: and bisect each other at M Flowchart roof I heorem M M. : M M, M M Show: is a roof: Statement eason 1. M M 1.. M M is a efinition of I heorem M M S ostulate and bisect each other at M efinition of bisect, definition of congruence M M M M. efinition of congruence Opposite Sides heorem. M M. 4. M M 4. efinition of congruence 5. M M 5. V heorem 6. M M 6. SS ostulate onverse of I heorem 9. M M 9. V heorem 10. M M 10. SS ostulate onverse of I heorem 1. is a 1. efinition of. : is a rhombus Show: and bisect M each other at M and Flowchart roof is a efinition of rhombus and bisect each other arallelogram iagonals heorem M and M are supplementary Linear air ostulate is a rhombus M M hombus ngles heorem M M SS ostulate M M M is a right angle ongruent and Supplementary heorem efinition of perpendicular efinition of rhombus M M eflexive property iscovering Geometry ractice Your Skills NSWES Key urriculum ress

26 4. : and bisect each other at M and Show: is a rhombus Flowchart roof (See flowchart at bottom of page.) 5. : is a trapezoid with and Show: is isosceles roof: Statement eason E 1. is a trapezoid 1. with. onstruct E. arallel ostulate. E is a. efinition of 4. E 4. Opposite Sides ongruent heorem 5. E 5. ostulate E 7. ransitivity 8. E is isosceles 8. onverse of I heorem 9. E 9. efinition of isosceles triangle ransitivity 11. is isosceles 11. efinition of isosceles trapezoid M 6. : is a trapezoid with and Show: is isosceles roof: Statement eason 1. is a trapezoid 1. with. onstruct E. arallel ostulate. and E intersect. Line Intersection at F ostulate 4. F is a 4. efinition of 5. F 5. Opposite Sides ongruent heorem F 7. ransitivity 8. F is isosceles 8. efinition of isosceles triangle 9. F F 9. I heorem 10. F 10. Opposite ngles heorem 11. F 11. I heorem ransitivity eflexive property SS ostulate is isosceles 16. efinition of isosceles trapezoid F E Lesson 1.4, Exercise 4 is a onverse of the arallelogram iagonals heorem Opposite Sides heorem ll 4 sides are congruent ransitivity is a rhombus efinition of rhombus and bisect each other at M M M efinition of bisect, definition of congruence Opposite Sides heorem M M M M SS ostulate eflexive property M and M are right angles efinition of perpendicular M M ight ngles ongruent heorem 116 NSWES iscovering Geometry ractice Your Skills 008 Key urriculum ress

27 7. False 8. False herefore the assumption,, is false, so.. aragraph roof: ssume 9. rue : with and Show: is a Flowchart roof and are supplementary Interior Supplements heorem Supplements of ongruent ngles heorem is a and are supplementary Interior Supplements heorem It is given that. y the reflexive property. So by SS,. hen by. ut this contradicts the given that. So.. : with Show: aragraph roof: ssume If, then by the onverse of the I heorem, is isosceles and. ut this contradicts the given that. herefore,. 4. : oplanar lines k,, and m, k, and m k intersecting k m Show: m intersects aragraph roof: ssume m does not intersect If m does not intersect, then by the definition of parallel, m. ut because k, by the arallel ransitivity heorem, k m. his contradicts the given that m intersects k. herefore, m intersects. onverse of Opposite ngles heorem LESSON 1.6 ircle roofs LESSON 1.5 Indirect roof roofs may vary. 1. ssume ase 1: If, then is isosceles, by the definition of isosceles. y the I heorem,, which contradicts the given that m m. So,. ase : is isosceles. 1. : ircle O with Show: Flowchart roof onstruct O, O, O, O Line ostulate O 4 1 y the Exterior ngle heorem, m1 m m4, so m1 m4. y the ngle Sum ostulate, m m m, so m m. ut is isosceles, so m4 m by the I heorem. So, by transitivity, m1 m4 m m, or m1 m, which contradicts the given that m m. So,. O O efinition of circle, definition of radii O O SSS ostulate O O efinition of congruence, definition of arc measure, transitivity O O efinition of circle, definition of radii iscovering Geometry ractice Your Skills NSWES Key urriculum ress

28 . aragraph roof: hords,, and E are congruent because the pentagon is regular. y the proof in Exercise 1, the arcs,, and E are congruent and therefore have the same measure. me 1 me by the Inscribed ngles Intercepting rcs heorem. Similarly, m 1 m and m 1 m. y transitivity and algebra, the three angles have the same measure. So, by the definition of trisect, the diagonals trisect E.. aragraph roof: onstruct the common internal tangent U (Line ostulate, definition of tangent). Label the intersection of the tangent and S as U. U S LESSON 1.7 Similarity roofs 1. Flowchart roof Similarity ostulate efinition of similar triangle eflexive property Multiplication property U U SU by the angent Segments heorem. U is isosceles by definition because U U. So, by the I heorem, U. all this angle measure x. SU is isosceles because U SU, and by the I heorem, S US. all this angle measure y. he angle measures of S are then x, y, and (x y). y the riangle Sum heorem, x y (x y) 180. y algebra (combining like terms and dividing by ), x y 90. ut ms x y, so by transitivity and the definition of right angle, S is a right angle.. : rapezoid with, and and intersecting at E Show: E E E E Flowchart roof I heorem I heorem E 4. aragraph roof: onstruct tangent (Line ostulate, definition of tangent). and both have the same intercepted arc,. Similarly, and have the same intercepted arc,. So, by transitivity, the Inscribed ngles Intercepting rcs heorem, and algebra, and are congruent. herefore, by the onverse of the ostulate,. E E Similarity ostulate E E E E efinition of similarity 118 NSWES iscovering Geometry ractice Your Skills 008 Key urriculum ress

29 . : with right, Show: Flowchart roof is right efinition of perpendicular ight ngles re ongruent heorem 4. : with right angles and, Show: is a rectangle Similarity ostulate efinition of similarity Multiplication property is right eflexive property roof: Statement eason 1. onstruct 1. Line ostulate. and are. right angles.. ight ngles re ongruent heorem eflexive property HL ongruence heorem m m 8. efinition of congruence 9. m m 9. riangle Sum m 180 heorem 10. m efinition of right angle 11. m m 11. Subtraction property m m 1. Substitution m m 1. ngle ddition m ostulate 14. m ransitivity 15. m efinition of right angle 16. m m 16. Quadrilateral Sum m m heorem m Substitution property and subtraction property efinition of congruence 19. is a rectangle 19. Four ongruent ngles ectangle heorem iscovering Geometry ractice Your Skills NSWES Key urriculum ress