Reteaching Inequalities in Two Triangles

Transcription

1 Name ate lass Inequalities in Two Triangles INV You have worked with segments and angles in triangles. Now ou will eplore inequalities with triangles. Hinge Theorem If two sides of one triangle are congruent to those of another triangle, and if the included angle of the first triangle is greater than that of the second triangle, then the third side of the first triangle is longer than that of the second triangle. ample: ompare the measures of and. Step 1: ompare the angle measures of the sides opposite and. The opposite angle of. The opposite angle of 9. Step : Make a comparison. The side opposite the greater angle is longer. omplete the statements to solve each problem. 1. ompare the measures of and.. ompare the lengths of KL and NP J K 15 M 86 1 N opposite angle opposite angle 9 ompare the measures. L KL opposite angle 8 NP opposite angle 86 KL NP P. ompare the measures of VW and YZ.. ompare the measures of JL and MP. V Y J 1 K M 1 N T 8 W X 8 YZ VW Z JL MP L P Saon. ll rights reserved. 87 Saon Geometr

2 continued INV onverse of the Hinge Theorem If two sides of one triangle are congruent to those of another triangle, and if the third side of the first triangle is longer than that of the second triangle, then the measure of the angle opposite the third side of the first triangle is greater than that of the second triangle. ample: ompare the measures of and. Step 1: ompare the measures of the lengths of the sides opposite the angles. The side opposite. The side opposite.5. Step : Make a comparison. The angle opposite the greater side is larger. m m omplete the statements to solve each problem. 5. ompare the measures of and. 6. ompare the measures of and side opposite side opposite.5 side opposite side opposite The angle opposite the greater side is larger. The angle opposite the greater side is larger. m m m m ompare the measures. 7. ompare the measures of and. 8. ompare the measures of and m m m m 10 Saon. ll rights reserved. 88 Saon Geometr

3 Name ate lass Ratios, Proportions, and Similarit 1 You have worked with corresponding sides and angles. Now ou will work with ratios and proportions. proportion is an equation stating that two ratios are equal. In ever proportion, the product of the etremes equals the product of the means. ross Products Propert In a proportion, if a b c and b and d 0, d then ad bc. a and d are the etremes. a b = c d b and c are the means. You can solve a proportion like 5 b finding the cross products (56) 8(5) ross Products Propert Simplif. 5 ivide both sides b 56. omplete the statements to solve each proportion w w(6) 9(8) ross Products Propert (16) 15(5) ross Products Propert 6 w 5 Simplif Simplif. w 7 ivide both sides b 6. 0 ivide both sides b 16. Solve each proportion Saon. ll rights reserved. 89 Saon Geometr

4 continued 1 In similar polgons, the corresponding angles are congruent, and the corresponding sides are proportional. similarit ratio is the ratio of two corresponding linear measurements in a pair of similar figures. ind the unknown side length in the two similar triangles. The triangles are similar so corresponding sides are in proportion. Write a proportion and solve for (8) 17(6) ross Products Propert 8 10 Simplif ivide both sides b 8. The length of the missing side is = omplete the statement to solve for ().8 (8.) ross Products Propert 7.5 Simplif..6 ivide Solve for the unknown side lengths in the similar figures ; 15 w w ; 10 Saon. ll rights reserved. 90 Saon Geometr

5 Name ate lass inding istance from a Point to a Line You have worked with finding distance on a coordinate plane. Now ou will find the shortest distance between a point and a line. inding istance The shortest distance between a point and a line is alwas a perpendicular segment. ind the distance from the point (1, ) to the ind the distance from the point (, ) to the line line O units - - O units Step 1: raw a point on the line 1, perpendicular from the given point. Step : ount the number of units between the two points. Step : The distance between the points is units. Step 1: raw a point on the line 1, perpendicular from the given point. Step : ount the number of units between the two points. Step : The distance between the points is units. omplete the statements to find the distance between the point and the line. 1. ind the distance between the point (, ) and the line O The point on the line perpendicular to the given point is (, 1 ). - ount the number of units between the two points: units. - The distance is units.. ind the distance between the point (, ) and the line 5.. ind the distance between the line and the point (, 1). - - O - - O units 7 units Saon. ll rights reserved. 91 Saon Geometr

6 inding the losest Point Given the equation 1 and the point W (, ), find the point on the line that is closest to W. ind the distance from W to the line. Step 1: raw the line and point on the coordinate grid. Step : etermine the slope of the line. The slope of the line is. Step : etermine the slope of the line perpendicular to the given line. The negative reciprocal of is 1. Step : Use the slope to determine more points on the perpendicular line. raw the line. The lines intersect at ( 1, ). Step 5: ind the distance between (, ) and ( 1, ). The distance from the point to the line is about.16 units. d ( 1 ) ( 1 ) ( ( 1)) ( ( )) () ( 1) continued - - The distance from the point to the line is about.16 units. - - O omplete the statements to find the distance.. Given the equation 1 and the point (, ), find the point on the line that is closest to the point. The slope of the line is. The negative reciprocal is 1. nother point on the line is (0, 1). - - O - Use the distance formula. - d ( 1 ) ( 1 ) ( 0) ( ( 1)) () ( 1) 5. The distance is about. units. Saon. ll rights reserved. 9 Saon Geometr

7 Name ate lass You have working with radii and diameters of a circle. Now, ou will work with other tpes of lines and segments in circles. Lines that Intersect ircles chord is a segment whose endpoints lie on a circle. diameter is a chord. secant is a line that intersects a circle at two points. tangent is a line in the same plane as a that of a circle that intersects the circle at eactl one point, called the point of tangenc. Radii and diameters also intersect circles. hords, Secants, and Tangents omplete the statements to identif each line or segment that intersects the circle. 1. hord: G and are chords. is a secant. is a point of tangenc. l l is a point of tangent. Secant: line l G Tangent: line m iameter: G Radii: H and HG Identif each line or segment that intersects each circle. H m l.. K N J M L I H M hord: LM Secant: Tangent: LM MN iameter: none Radii: JK Point of tangenc: M hord: Secant: Tangent: iameter: Radii:, I, HG I G MI, M, M Point of tangenc: H Saon. ll rights reserved. 9 Saon Geometr

8 continued hord iameter Relationships If a diameter is perpendicular to a chord, then it bisects the chord and the chord s arcs. If a diameter bisects a chord other than another diameter, then it is perpendicular to the chord. The perpendicular bisector of a chord contains the center of the circle. In a circle or congruent circle, chords equidistant from the center are congruent, and congruent chords are equidistant from the center of the circle. ample: ind the length of GH. The radius of circle K is 10 meters, and KJ is 6 meters. irst, draw a radius KH. Then, use the Pthagorean theorem to find JH. KJ JH KH 6 b 10 6 b 100 JH 8m b 6 b 8 Since KJ is perpendicular to GH, JH JG. Therefore, the length of GH , or 16 meters. N G 6 J M K P H L omplete the statements to find the missing distance.. circle has a diameter of 50 centimeters. hord JK is centimeters. How far is JK from the center of the circle to the nearest hundredth? raw a perpendicular line from point P to chord JK. L P M Label the point of intersection. J K P K PK PK 5 cm; K 1 cm P 1 5 P 1 65 P 81 P 1.9 Saon. ll rights reserved. 9 Saon Geometr

9 Name ate lass You have worked with proportions to find unknown sides in similar triangles. Now ou will work with other tpes of similar polgons. ppling Similiarit Using Similarit to ind Unknown Measures You can use properties of similar polgons to solve proportions. The figures in the diagram are similar. ind the values of and. Step 1: Write a similarit 7 ratio Step : Write and solve a proportion to solve for and The figures in the diagram are similar. omplete the statements to find the missing values (10) 15 (8) The figures in the diagram are similar. Solve for the missing lengths... The similarit ratio is : w w 7; 7.5 ; 6 Saon. ll rights reserved. 95 Saon Geometr

10 continued ppling Similarit to Perimeter igures and GHIJ are similar polgons. Their corresponding sides have a ratio of :5. If the perimeter of figure is 6 inches, what is the perimeter of GHIJ? Step 1: The figures are similar polgons. The ratio of the perimeters is :5. Step : Set up a proportion using the ratio :5. ratio ratio GHIJ perimeter of perimeter GHIJ J G I H The perimeter of GHIJ is 60 inches. omplete the statements to find the perimeter of the similar figure.. igures and GHIJ are similar polgons. Their corresponding sides have a ratio of :7. If the perimeter of figure is inches, what is the perimeter of GHIJ? ratio ratio GHIJ perimeter of perimeter GHIJ The perimeter is 17 inches. 5. Heagons and GHIJKL are similar figures. The perimeter of is 8 centimeters. The similarit ratio of to GHIJKL is :9. What is the perimeter of GHIJKL? 6 centimeters 6. Rectangle is similar to GHIJ. The perimeter of is 56 centimeters. The similarit ratio of to GHIJ is 1:16. What is the perimeter of GHIJ? 896 centimeters 7. igures LMNP and QRST are similar figures. The ratio of their corresponding sides is :8. If the perimeter of LMNP is 18 centimeters what is the perimeter of QRST? 8 centimeters Saon. ll rights reserved. 96 Saon Geometr

11 Name ate lass You have worked with a coordinate plane and proofs. Now ou will work with proofs on a coordinate plane. Introduction to oordinate Proofs 5 Positioning a igure in the oordinate Plane coordinate proof is a proof that uses coordinate geometr and algebra. In a coordinate proof, the first step is to position a figure in a plane. There are several was ou can do this to make our proof easier. Keep the figure in Quadrant 1 b using the origin as a verte. enter a side of the figure at the origin. Use one or both aes as sides of the figure. O - O O Position each figure in the coordinate plane and give the coordinates of each verte. 1. a square with side length. a right triangle with leg lengths and units units (0, ) (, ) 6 (0, 0) O - - (, 0) (0, ) (0, 0) (, 0) - O a triangle with base 8 units and height. a rectangle with length 6 units and units width units (-, ) (, ) (-, 0) - O (0, ) (, 0) - (-, 0) O (, 0) Saon. ll rights reserved. 97 Saon Geometr

12 continued 5 You can prove that a statement about a figure is true without knowing the side lengths. To do this, assign variables as the coordinates of the vertices. Position a right triangle with leg lengths c and d in the coordinate plane and give the coordinates of each verte. (0, d) a right triangle with leg lengths c and d Step 1: Place one verte at the origin and label (0, 0). Step : The verte on the -ais is c units awa. Label (c, 0). (0, 0) (c, 0) Step : The verte on the -ais is d units up. Label (0, d ). Position each figure in the coordinate plane and give the coordinates of each verte. 5. a right triangle with leg lengths s and t 6. a square with side lengths k (0, t) (0, k) (k, k) (0, 0) (s, 0) (0, 0) (0, k) 7. a rectangle with leg lengths l and w 8. a triangle with base b and height h (0, w) (l, w) (0, h) (0, 0) (l, 0) (0, 0) (b, 0) Saon. ll rights reserved. 98 Saon Geometr

13 Name ate lass Triangle Similarit:, SS, SSS 6 You have worked with coordinate proofs. Now ou will work with triangle similarit. Triangle Similarit:, SSS ngle-ngle () Similarit If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar Side-Side-Side (SSS) Similarit If the three sides of one triangle are proportional to the three corresponding sides of another triangle, then the triangles are similar. TU XY UV YZ TUV XYZ VT ZX Given the values of the two triangles as shown, show that the triangles are similar triangles. Step 1: Write similarit ratios for pairs of sides. Let the lengths of be the top number of the proportions. : MP : MN MP MN : NP NP T X V U Z Y 1 Step : Since the ratio of each pair of sides reduces to the same fraction, all of the side pairs of the two triangles are proportional. Therefore, MNP b SSS Similarit P 10 N M omplete the steps to show that the two triangles are similar. 1. Q T b the definition of congruent angles. m S 9 b the Triangle ngle Sum Theorem. S V b the definition of congruent angles. Q R 9 9 S U 9 T 9 V Therefore, QRS TUV b Similarit. plain how the triangles are similar and write a similarit statement.. RS VW TR ZV TS ZW 1 ; Therefore; RST VWZ b SSS Similarit R 6 10 S 8 T W 18 V 0 Z Saon. ll rights reserved. 99 Saon Geometr

14 continued 6 Triangle Similarit: SS Side-ngle-Side (SS) Similarit 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 Show that Then, find. Step 1: b the Refleive Propert. 1 6 ; 1 b SS Similarit. Step : ind..5 6 () 6(.5) Proportional orresponding Sides Substitute. ross Products Propert.5 7 Solve. omplete the steps to show that the triangles are similar. Then, find GK. J K. JK H, so J H and K b the lternate Interior ngles Theorem. 8 G JGK GH b Similarit To find GK : 8 1 GK 11 H 8( 11 ) 1 ( GK ) 88 1(GK) 88 1 GK 7 1 GK T plain wh the triangles are similar and then find US.. It is given that TSU WVU. U U b the Refleive Propert. TSU WVU. b Similarit US 9. S V 8 6 W U Saon. ll rights reserved. 100 Saon Geometr

15 Name ate lass ircles and Inscribed ngles 7 Inscribed ngles The measure of an inscribed angle is half the measure of its intercepted arc. In the picture, is an inscribed angle and is its intercepted arc on the circle. m 1 m If an inscribed angle intercepts a semicircle, then it is a right angle. intersects the semicircle, so m 90. If inscribed angles intercept the same arc, then the are congruent. and intercept, so. ind m GJ and m H. Step 1: Since m GJ 110, m GJ 1 m GJ 1 (110 ) 55. Step : Since m JH 6, then 1 m H m JH 6. Solve for m H : 1 m H 6 m H (6 ) H 6 J G 110 m H 7 omplete the steps to solve for. 1. Since RQS intercepts a semicircle, m RQS 90. (5 8) R Q (5 + 8) S ind each value.. z. m JH 9 5 H 5z J (z + 9) G Saon. ll rights reserved. 101 Saon Geometr

16 continued 7 Inscribed Quadrilaterals If a quadrilateral is inscribed in a circle, then it has supplementar opposite angles. is inscribed in circle ind m G. Step 1: ind the value of z. m m G 180 GH is inscribed in a circle. 8z 1z z 198 z 9 Substitute. Simplif. 8z (1z - 18) G Step : ind m G. m G 1z 18 1(9) H omplete the steps to find the measure of the given angles.. R 5. T S (8 + 8) m R m T 180 m T T m T 11(10) R m T V m R 7 7(10) 70 ind the measures of the given angles in RSTV. 6. S V 9 ind the angle measures of each quadrilateral JKLM (6 + ) K (z + 5) (z - 17) L 10 ( + 1) J (z + ) M 10, 75, 60, 105 1, 90, 8, 90 Saon. ll rights reserved. 10 Saon Geometr

17 Name ate lass Indirect Proofs 8 You have worked with inscribed angles in circles. Now ou will work with indirect proofs. Indirect Proofs In direct reasoning, ou begin with a true hpothesis and prove that a conclusion is true. In an indirect proof, ou begin b assuming that the conclusion is false. You then show that this assumption leads to a contradiction. To write an indirect proof, first identif the given information and the conclusion to be proved. Net, assume either that the conclusion is false or that the opposite of the conclusion is true. Identif the conclusion to be proved and the opposite of the conclusion to be assumed for the following statement. If two angles are not congruent, then the are not both right angles. Step 1: Identif the conclusion to be proved. Prove: 1 and are not both right angles. Step : When writing an indirect proof, assume that the opposite of the conclusion is true. ssume: 1 and are both right angles. omplete the steps to write the conclusion and the opposite of the conclusion for each statement. 1. If a b, then a b, for an real numbers, a and b. onclusion: a b ssume: a b. If the sum of the measures of two angles is not equal to 180, then the two angles are not supplementar. onclusion: 1 and are not supplementar. ssume: 1 and are supplementar. Write the conjecture and the opposite of the conclusion for each statement.. If 1 is not congruent to 1, then line a intersects line b. onclusion: Line a intersects line b. ssume: Line a is parallel to line b.. If RST is acute, then it does not have a right angle. onclusion: RST does not have a right angle. ssume: RST has a right angle. 1 a b Saon. ll rights reserved. 10 Saon Geometr

18 continued 8 Writing an Indirect Proof onsider the statement: Two acute angles with a common side do not form a linear pair. Here are the steps to write an indirect proof of the statement. Steps ample 1. Identif the conjecture to be proved. Given: 1 and are acute angles with a common side. Prove: 1 and do not form a linear pair.. ssume the opposite of the conclusion ssume: 1 and form a linear pair. is true.. Use direct reasoning to show that the assumption leads to a contradiction.. onclude that the assumption is false. Therefore, the original conjecture must be true. m 1 m 180 b the definition of a linear pair. Since m 1 90 and m 90, m 1 m 180. This is a contradiction. The assumption that 1 and form a linear pair is false. Therefore, 1 and do not form a linear pair. omplete the indirect proof b following the directions in each step. 5. n obtuse triangle cannot have a right angle. is an obtuse angle. Steps 1. Identif the conjecture to be proved.. ssume the opposite of the conclusion is true. Write the assumption.. Use direct reasoning to show that the assumption leads to a contradiction.. onclude that the assumption is false. Therefore, the original conjecture must be true. Proof Given: is an obtuse triangle. is an obtuse angle. Prove: cannot have a right angle. ssume: has a right angle. Let be a right angle. If is a right angle, then m m 90. ut m 90 since is an obtuse angle. This is a contradiction. The assumption that has a right angle is false. Therefore, cannot have a right angle. Saon. ll rights reserved. 10 Saon Geometr

19 Name ate lass Introduction to Solids 9 You have worked with indirect proofs. Now ou will work with solid geometr. Three-dimensional figures, or solids, can have flat or curved surfaces. Prisms and pramids are named b the shape of their bases. ach flat surface is called a face. n edge is the segment where two faces intersect verte is the point where three or more faces intersect. In a cone, it is where the curved surface comes to a point. Prisms Pramids linders ones verte bases bases bases base triangular rectangular triangular rectangular prism prism pramid pramid Neither clinders nor cones have edges. Identif the figure. Name the vertices, edges, and bases. Step 1: The figure is a rectangular pramid. Step : The vertices are,,,, and. The edges are,,,,,,, and. The base is quadrilateral. omplete the steps to identif the figure and name the vertices, edges, and bases. R 1. This is a triangular pramid. S The vertices are Q, R, S, and T. The edges are RQ, SR, QS, QT, ST, and RT. Q T The base is QST. Identif each figure. Name the vertices, edges, and bases.. clinder; vertices: none; edges: none; bases: circle, circle. triangular prism: vertices:,,,, G, H; edges:,,, G, GH, H,, G, H ; bases:, GH G H Saon. ll rights reserved. 105 Saon Geometr

20 uler s ormula closed three-dimensional figure formed b four or more polgons that intersect onl at their edges is called a polhedron; for eample, prisms and pramids. There is a unique relationship between the number of faces, vertices, and edges of an polhedron. uler s formula: or an polhedron with V vertices, edges, and faces, V. ample: Verif uler s formula for a rectangular prism. Step 1: ind the number of vertices, edges, and faces. V 8, 1, 6 Step : Substitute the values into uler s formula and simplif. V continued 9 omplete the steps to find the missing number.. triangular prism has 5 faces and 5. rectangular pramid has 5 faces and 9 edges. 5 vertices. How man vertices does it have? How man edges does it have? V V V V 10 V 6 It has 6 vertices. Use uler s formula to solve each eercise. 8 8 It has 8 edges 6. polhedron has 6 vertices and 7. polhedron has 1 faces and faces. 18 vertices. How man edges does it have? 8 8 How man edges does it have? 8. polhedron has 16 edges and 9. polhedron has 9 vertices and 10 vertices. 1 edges. How man faces does it have? 8 5 How man faces does it have? Saon. ll rights reserved. 106 Saon Geometr

21 Name ate lass Geometric Mean 50 You have worked with solids. Now ou will work with geometric mean. Geometric Mean The geometric mean for two positive numbers, a and b, is the positive number such that a b. ind the geometric mean of and 1. Let represent the geometric mean of and 1. Write the algebraic epression for finding the geometric mean and solve for : a b Substitude and 1. 1 () (1) ind the cross product ind the square root. Round to the nearest tenth. omplete the steps to find the geometric mean for each pair of numbers. Round our answer to the nearest tenth and 9 /. 8 and and ( ) 9 ( ) 8 (15) ( ) 10(6) ind the geometric mean for each pair of numbers. Round our answer to the nearest tenth.. 7 and and and and and 6 9. and 16 8 Saon. ll rights reserved. 107 Saon Geometr

22 When an altitude is drawn from the verte of the 90 angle of a right triangle to the hpotenuse of the right triangle, some useful relationships are formed. If the altitude is drawn to the hpotenuse of a right triangle, then the two triangles formed are similar both to each other and to the original triangle. JLM MLK JMK continued 50 J K M L The length of the altitude is the geometric mean between the segments of the hpotenuse. The length of a leg is the geometric mean between the hpotenuse and the segment of the hpotenuse that is closer to that leg. ample: Given XYZ, find the missing value, t. Step 1: Since XW is an altitude of a right triangle, t is the geometric mean of the segments of the hpotenuse. The segments measure 7 and. Since t is the geometric mean of 7 and, write 7 t Step : Solve for t : 7 t t. t 81 t. X t Y W 7 t 9 Z 10. ind the value of a and b. Since a and b are the legs of the triangle, each one is a geometric mean of the hpotenuse and the segment closer to that leg a b 1 b 1 a 5 1 (a) (5) 5 1 (b) (1) 1 a 5 1 b 1 a 1.9 b 11.1 Saon. ll rights reserved. 108 Saon Geometr