H P T E R 7 TRINGLE ONGRUENE LESSON 7-1 pp. 364 369 1. Triangulation is the use of triangles to determine the location of an object. 2. a. Sample: V ngle measures: 22, 50, 108 b. ny opinion is allowed. The answer is yes. 3. Sample: Z 3 c m X 4. a. Sample: X 5 c m 4 5 8 0 b. ny opinion is allowed. The answer is no. 5. 180 (32 102) 180 134 46 The third angle has measure 46. 6. The measure of the third angle is 180 (x y), or 180 x y. 7. nswers may vary. Mathematically, they should agree. 8. Sample: Yes; given the lengths of three segments, a unique triangle is determined. 9. Sample: No; given the measures of three angles, a unique triangle is not determined. 10. Sample: Yes; given the lengths of two segments and the measure of the angle they form, a unique triangle is determined. Z W Y Y 4 c m E t 11. a. Sample: b. ny opinion is allowed. The answer will be found to be no. 12. a. Sample: = 4 m = 6 0 b. ny opinion is allowed. The answer will be found to be no. 13. a. Sample: b. ny opinion is allowed. The answer will be found to be yes. 14. a. Sample: = 2 m = 4 0 m = 6 0 b. ny opinion is allowed. The answer will be found to be yes. 15. a. No unique triangle can be drawn. b. The answer is no. 16. a. Sample: E 1 8 2 ' m = 7 0 m = 3 8 6 2 = 3 m = 4 0 m = 6 0 10 0 USMP Geometry Scott, Foresman and ompany
16. b. 1.1 cm 182 E 2.3 cm n n (2.3 182) 1.1 381 E 381 c. S; 2 angles and the side between them 17. sum of angle measures of octagon: (n 2) 180 (8 2) 180 6 180 1080 measure of each angle: 1080 8 135 Each angle in a stop sign has measure 135. 18. a. IR TH b. GHT RI 19. If, G, H, and R are not collinear, then these points are vertices of a quadrilateral called an isosceles trapezoid. 20. Sample: LMN is the image of FGH under an isometry. 21. If m P 40, then m m P 40; m m m m m P m P 90; m m m P 50 22. a. Sample: n b. r m r n is a rotation. 23. N M P m 24. S Q 1 1 (, 6 2 2) 10 b o o k 9 te le p h o ne re ce iv e r (2, 1 ) 2 p a rro t ca r (4, 6) 8 (7, 10) 7 ro lle r s k a te b e ll(7 1, 6) (6, 21 ) 2 2 6 a irp la ne 1 (7, 4 2) 5 b u g le ( 1 2, 4) 4 mo u s e (1, 5) ro o s te r s h e a d 1 (4 2, 8) 3 2 w o o l ca p (8, 9 ) b a th ro b e 1 (2, 7) y o -y o Pu rita n s h a t ( 1 2, 9 ) 0 (8, 0) 1 2 3 4 5 6 7 8 USMP Geometry Scott, Foresman and ompany 101
LESSON 7-2 pp. 370 377 1. Four conditions that lead to triangle congruence are SSS, SS, S, and S. 2. SSS ongruence 3. Symmetry of a kite was used in the proof of the SSS ongruence. 4. Symmetry of an isosceles triangle was used in the proof of the SS ongruence. 5. S ongruence : If two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent. 6. In the S ongruence, the corresponding sides are not included by the pairs of congruent angles as they are in the S ongruence. 7. is included by and. 8. F E by SS 9. not enough information to know 10. not enough information to know 11. OYN YOX by S 12. GO IPN by S 13. STE TE by S 14. SSS explains why all triangles are rigid. 15. a. PO JKL b. If O 5 yards, then LJ is 5 yards. c. O L d. If m P 73, then m K 73. 16. a. E b. F c. E d. FE and E 17. by the S ongruence. 18. Yes, the triangles will be congruent by the S ongruence. 19. Given: m FE m E; G,, and E are collinear. To prove: EG bisects F. 0. m FE m E; Given 0. G,, and E are collinear. 1. m FG Linear Pair 0. 180 m FE; (and algebra) 0. m G 0. 180 m E 2. m FG Transitive Property 0. m G of Equality (and substitution) 3. EG bisects F. 20. a. n 2 b. 21. a. n 5 b. 22. a. n 3 b. definition of angle bisector 23. Given: EF. If m 32 and m 64, then m 180 (32 64) 180 96 84. Since m m F, then m F 84. 24. Given: OL NEW Sample conclusions: O N, L E, W, OL NE, L EW, O NW Justification: PF 102 USMP Geometry Scott, Foresman and ompany
25. x 26. Sample: LESSON 7-3 pp. 378 383 1. a. G y b. S ongruence c. Given: is the midpoint of E; E; E To prove: E 0. is the midpoint of Given 0. E; E; 0. E 1. E definition of midpoint 2. E S ongruence 2. Given: Y is the midpoint of WV; m Z m X To prove: WXY VZY 0. Y is the midpoint Given 0. of WV; m Z m X E F E H z 1. WY VY definition of midpoint 2. XYW ZYV Vertical ngles 3. WXY VZY S ongruence 3. Given: ; To prove: 0. ; Given 1. Reflexive Property of ongruence 2. SSS ongruence 3. PF 4. a. and are congruent. b. Isosceles Triangle ase ngles onverse 5. a. Reflexive Property of ongruence b. S ongruence c. PF 6. Given: X Y, and X Y To prove: X Y 0. X Y, and Given 0. X Y 1. Reflexive Property of ongruence 2. X Y SSS ongruence 3. X Y PF 7. Given: NK bisects JNM; NM NK; NL NJ To prove: JNK LNM 0. NK bisects JNM; Given 0. NM NK; NL NJ 1. JNK LNM definition of angle bisector 2. JNK LNM SS ongruence USMP Geometry Scott, Foresman and ompany 103
8. Given:, and To prove: 0., and Given 0. 1. Reflexive Property of Equality 2. Lines I 3. SS ongruence 9. Given: m m ; Given: m m To prove: 0. m m ; Given 0. m m 1. Reflexive Property of Equality 2. S ongruence 3. PF 10. a. G,, J, JG, GJ, G, JG, J b. G J, GJ JG, G J 11. a. SS ongruence b. PF 12. not enough information to know 13. EFG IHG; S ongruence 14. a. True b. False 15. a. Yes, assuming the units for the length measurements are the same b. SS ongruence 16. If m 50, then m m m 50, m m 80, and m m 100. 17. = ' ' = ' ' 18. The sum of the measures of the acute angles in a right triangle is 180 90 90. 19. Sample: ' = ' 20. The Triangle Inequality Postulate states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side. ecause 3 cm 6 cm 9 cm 11 cm, the Triangle Inequality Postulate is violated. 21. Yes. Since, both, and (PF ). s a consequence, and (I Lines ). Therefore, is a parallelogram. LESSON 7-4 pp. 384 388 1. a. There are 8 triangles in the figure: QUS, US, S, SQ, QU, U, Q, QU. b. U appears congruent to QU. c. QU appears congruent to Q. 2. a. Given b. Reflexive Property of ongruence c. SS ongruence d. PF 3. Given: Regular pentagon LIVE To prove: EV VIE 0. Regular pentagon Given 0. LIVE 1. E IV definition of regular pentagon 2. EV EV Reflexive Property of ongruence 3. EV IVE definition of regular pentagon 4. EV IVE SS ongruence 5. EV VIE PF = ' 104 USMP Geometry Scott, Foresman and ompany
4. Given: QU, and Q U To prove: 1 2 0. QU, and Given 0. Q U 1. U U Reflexive Property of ongruence 2. QU U SSS ongruence 3. 1 2 PF 5. Given: E, and m m E To prove: E 0. E, and Given 0. m m E 1. Reflexive Property of ongruence 2. E S ongruence 3. E PF 6. Given: PR QS, and PS QR To prove: m P m Q 0. PR QS, and Given 0. PS QR 1. RS RS Reflexive Property of Equality 2. PRS QSR SSS ongruence 3. m P m Q PF 7. Since l, the perpendicular bisector of and E, is a symmetry line of EF by the Regular Polygon Symmetry, r l () and r l () F. Therefore, F because reflections preserve distance. 8. Given:,, E, and are collinear; E; To prove: E 0.,, E, and Given 0. are collinear; 0. E; 1. Isosceles Triangle ase ngles 2. E SS ongruence 3. E PF 9. Given: QR TU, and S is the midpoint of QU. To prove: S is the midpoint of RT. 0. QR TU, and S is Given 0. the midpoint of QU. 1. QS US definition of midpoint 2. m QSR m UST Vertical ngles 3. m Q m U lines I 4. QSR UST S ongruence 5. ST SR PF 6. S is the midpoint definition of 0. of RT. midpoint 10. P P by the SS ongruence. So, P P by the PF. 11. a. Vertical ngles b. SS ongruence c. PF d. Vertical ngles e. Transitive Property of ongruence 12. a. Sample: 32 110 b. No, because the condition does not determine a unique triangle. USMP Geometry Scott, Foresman and ompany 105
13. sum of angle measures of a pentagon: (n 2) 180 (5 2) 180 3 180 540 measure of each angle: 540 5 108 14. True 15. True 16. 0 180 2y 2y 180 y 90 The angle is acute. 17. E,, and point F. 18. WX XY YZ WZ 13 XY 17 45 XY 15 19. Samples are given. a. b. c. heptagon octagon LESSON 7-5 pp. 389 396 1. Sample: 2. a. right triangle has sides represented by H and L. b. H represents hypotenuse; L represents leg. 3. The proof of the HL ongruence makes use of the symmetry of the isosceles triangle. 4. Given: ; To prove: 0. ; Given 1. Reflexive Property of Equality 2. HL ongruence 5. The SS ondition leads to congruence when the sides opposite the congruent angles in each triangle are longer than the other congruent sides. 6. a. HL ongruence b. FE 7. a. SS ongruence b. GHI KJL 8. a. HL ongruence b. MNP ONP 9. a. c. reduced size: W WY X 50 9 cm 9 cm 11 cm nonagon decagon Z d. No. There are two possible triangles XZW, as shown in the drawing. 10. Given: VX WY, WZ VY, and XZ XY To prove: W V 0. VX WY, WZ VY, Given 0. and XZ XY 106 USMP Geometry Scott, Foresman and ompany
10. (continued) 1. WXZ and VXY definition of 0. are right angles. perpendicular lines 2. WXZ VXY HL ongruence 3. W V PF 11. Given: P R; P and R are right angles. To prove: PR is a kite. 0. P R; P and Given 0. R are right angles. 1. Reflexive Property of Equality 2. P R HL ongruence 3. P R PF 4. PR is a kite. definition of kite 12. a. Given b. definition of circle c. definition of circle d. Ss ongruence (Steps 1, 2, and Given) 13. Let T be the top of the maypole, M the point on the maypole that is the same height as June and pril s hands, J the position of June s hand, and the position of pril s hand. T J M ground J is parallel to the ground, since J and are equal in height (given). So, J TM by the Perpendicular to Parallels. Since TM TM and TJ T, then TJM TM by the HL ongruence. Thus, JM M by the PF. So, since JM is the distance from June s hand to the maypole, and M is the distance from pril s hand to the maypole, and JM M, their hands are the same distance from the maypole. 14. a. definition of isosceles trapezoid b. Isosceles Trapezoid USMP Geometry Scott, Foresman and ompany c. Reflexive Property of ongruence d. SS ongruence (steps 1 3) e. PF 15. f. Isosceles Trapezoid Symmetry g. Reflections preserve distance. 16. nswers may vary. Sample: I prefer the argument in Question 15 because it is shorter. 17. a. reduced size: R Q 45 S b. Yes, by the S ongruence. 18. Given: ; E a. To prove: E E 0. ; Given 0. E 1. Isosceles Triangle ase ngles 2. E SS ongruence 3. E E PF b. To prove: E is isosceles. (continuing from the proof above) 1. E Isosceles Triangle ase ngles onverse 2. E is definition of isosceles 0. isosceles. triangle 19. GY, EX, EY 20. a. EF b. r m ( EF) c. Sample: r m () E 21. The conjecture is false. ounterexample: E H G F 80 2" 107
LESSON 7-6 pp. 397 403 13. 1. tessellation is a covering of a plane with nonoverlapping congruent regions and no holes. 2. Samples: brickwork, floor or wall tiles, wallpaper 3. Steps 1 4: heck students drawings. Step 5: is a parallelogram. Rotations preserve angle measure, so and. y the I Lines, and. Therefore, is a parallelogram by the definition of parallelogram. 4. heck students drawings. 5. Sample (reduced size): 6. Sample (reduced size): 7. Key question: an a given region tessellate the plane? 8. One museum where many tessellations can be found is the lhambra. 9. Yes. Since the sum of the angle measures in is 360, it is possible to have a different angle from each of the four congruent quadrilaterals meeting at a single point. 10. (c); the measure of each angle of a regular pentagon is 108, and 360 is not evenly divisible by 108. 11. No; the measure of each angle of a regular heptagon is about 129, and 360 is not evenly divisible by 129. 12. new type of tessellating pentagon was discovered as recently as 1985. 14. a. Sample (reduced size): Y ' Z X O Y' Z' ' Z" " * " b. They are the four angles of O, so the sum of their measures is 360. c. Translate the figure repeatedly in all directions. d. nswers may vary. 15., so and (PF ). So and (I Lines), and is a parallelogram (definition of parallelogram). 16., so, (PF ). So is a kite (definition of kite). 17. Given: ; ; To prove: 0. ; Given 0. ; 1. Reflexive Property of Equality 108 USMP Geometry Scott, Foresman and ompany
17. (continued) 2. HL ongruence 3. PF 4. I Lines 18. Given: ; To prove: 0. ; Given 1. Reflexive Property of ongruence 2. SSS ongruence 3. PF 19. Given: m m ; Given: m m To prove: 0. m m ; Given 0. m m 1. Reflexive Property of ongruence 2. S ongruence 3. PF 20. (1) X; S ongruence (2) YZ; SS ongruence (3) Z; S ongruence 21. a. No b. There is no such segment, unless PQR is isosceles with PQ PR. 22. Playfair s Parallel Postulate: Through a point not on a line, there is exactly one line parallel to the given line. 23. Sample: The Moors were a North frican people of erber and rab descent who invaded and conquered Spain in the eighth century. They were driven back to frica during centuries of wars and inquisitions, the final Moors leaving in 1492. 24. Sample (reduced size): IN-LSS TIVITY p. 404 S tep 1. ; ; ; S tep 2. 0. Parallelogram Given 0. 1. ; definition of 0. parallelogram 2. m Lines I 0. m ; 0. m 0. m 3. Reflexive Property of Equality 4. S ongruence S tep 3. ; ; USMP Geometry Scott, Foresman and ompany 109
Step 4. E E E; E E; ; Step 5. ; ; E E; E E Step 6. The parallelogram possesses 2-fold rotation symmetry. LESSON 7-7 pp. 405 410 1. a. is congruent to. b. 5 is congruent to 1. c. The midpoints of and are the same point (E). 2. a.,, and are congruent to. b. 2, 5, and 6 are congruent to 1. c. The midpoints of and are the same point (E). 3. a. definition of parallelogram b. lines I c. definition of parallelogram d. lines I e. S ongruence f. PF 4. Properties of a Parallelogram, part (a) 5. This image of is. 6. Given: Parallelogram To prove: rawing: raw auxiliary diagonal. 0. Parallelogram Given 1. ; Properties of a Parallelogram (a) 2. Reflexive Property of ongruence 3. SSS ongruence 4. PF 7. Given: Parallelogram ; E To prove: E is the midpoint of and. rawing: E 0. Parallelogram Given 1. definition of parallelogram 2. lines I 3. E E Vertical ngles 4. Properties of a Parallelogram (a) 5. E E S ongruence 6. E E; E E PF 7. E is the midpoint definition of 0. of and. midpoint 8. a. E E 7 b. 12 9. a. E E x b. 2y c. 2 E 2x 10. a. R and OP are congruent. b. The distance between two given parallel lines is constant. 11. (1) ' ' ' ' 110 USMP Geometry Scott, Foresman and ompany
11. (continued) (2) ' (3) ' ' ' ' ' 12. (b) is false, is true. 13. : The distance between two given parallel lines is constant. 14. a., b. parallelogram (R) kite (rd) rectangle (rpb) (R) ' sq uare (rd) (rpb) (R) rhombus (rd) (R) 15. From the Properties of a Parallelogram,, EF, EF GH. From the Transitive Property of Equality, GH. 16. Given: is a parallelogram; M is the midpoint of. To prove: If M M, then is a rectangle. 0. is a Given 0. parallelogram; 0. M is the midpoint 0. of ; M M 1. M M definition of midpoint 2. Properties of a Parallelogram (a) 3. M M SSS ongruence 4. PF 5. ; Properties of a 0. Parallelogram (b) 6. Transitive Property of 0. ongruence ' USMP Geometry Scott, Foresman and ompany 7. m m Quadrilateral-Sum 0. m 0. m 360 8. 4 m 360 Substitution 9. m m Multiplication 0. m Property of Equality 0. m 90 and Substitution 10. is a definition of rectangle 0. rectangle. 17. Sample: 18. Given: Trapezoid F; E F; F; F ; F To prove: F 0. Trapezoid F; Given 0. E F; F; 0. F ; F 1. EF and definition of 0. are right angles. perpendicular 2. E Two Perpendiculars 3. E is a definition of 0. parallelogram. parallelogram 4. E Properties of a Parallelogram (a) 5. EF HL ongruence 6. F PF 19. No; the congruent sides are not corresponding sides. corresponds to XZ, not YZ. 20. a. 8 triangles are formed. b. { RIT, GHT }, { RTH, GTI } { }, RIH, IRG, HGI, GHR 21. It cannot be justified. 22. 23. Transitive Property of ongruence 24. Two Perpendiculars 111
25. K L 3. a. nswers will vary. Sample: M I J P H square is formed. 26. LESSON 7-8 pp. 411 415 1. ccording to the definition of parallelogram, two pairs of parallel sides makes a quadrilateral a parallelogram. 2. a. nswers will vary. Sample: 6.5 cm O 6.5 cm b. The quadrilateral formed is a parallelogram. c. Yes. You can form a rectangle and a rhombus, but they are also parallelograms. G N F E b. parallelogram is formed. 4. Sufficient conditions for parallelograms other than its definitions: (1) oth pairs of opposite sides are congruent. (2) oth pairs of opposite angles are congruent. (3) The diagonals bisect each other. (4) One pair of sides is parallel and congruent. 5. Given: Quadrilateral PQRS with PQ RS and PS QR. To prove: PQRS is a parallelogram. 0. PQ RS; Given 0. PS QR 1. raw QS. Unique Line ssumption 2. QS QS Reflexive Property of ongruence 3. PQS RSQ SSS ongruence 4. PQS RSQ PF 5. PQ SR I Lines 6. PSQ RQS PF 7. PS QR I Lines 8. PQRS is a definition of 0. parallelogram. parallelogram 6. No, since the congruent angles are not opposite each other. 7. Yes, since both pairs of opposite sides are congruent. 8. Not necessarily. For example, could be an isosceles trapezoid with bases and. 9. parallelogram is formed. 10. parallelogram, a kite, or a rectangle can be formed. 112 USMP Geometry Scott, Foresman and ompany
11. Given: Quadrilateral ; diagonals and intersect at their midpoint, O. To prove: is a parallelogram. rawing: O 0. O is the midpoint Given 6. of and. 1. O O; O O definition of midpoint 2. O O Vertical ngles 3. O O SS ongruence 4. PF 5. I Lines 6. O O Vertical ngles 7. O O SS ongruence 8. PF 9. I Lines 10. is a definition of 6. parallelogram. parallelogram 12. (5x 30) (7x 90) (3x 10) (4x 50) 360 19x 20 360 19x 380 x 20 m N 5x 30 5(20) 30 130 m S 4x 50 4(20) 50 130 m E 7x 90 7(20) 90 50 m T 3x 10 3(20) 10 50 Yes, NEST is a parallelogram because both pairs of opposite angles are congruent. 13. The first scout was right. Since the diagonals intersect at their midpoints, is a parallelogram. So opposite angles are congruent:, and. ut also, by SSS. So. Then all 4 angles of the parallelogram are congruent, so each must be a right angle. Therefore, is a rectangle. 14. No; WX is not parallel to ZY. 15. m m 70 m 180 m 180 70 110 m m 110 16. E E 11 20 2 E 2 11 22 17. PQRS is a parallelogram, so PQ RS (Properties of a Parallelogram ). 18. No; each angle of a regular pentagon measures 108, but 360 is not evenly divisible by 108. So a regular pentagon will not tessellate. 19. Given: PTS is isosceles with vertex angle T, and m PTQ m STR. a. To prove: TPQ TSR 0. PTS is Given 0. isosceles with 0. vertex angle T, 0. and m PTQ 0. m STR. 1. PT ST definition of isosceles triangle 2. m P m S Isosceles Triangle ase ngles 3. TPQ TSR S ongruence b. To prove: TQR is isosceles. 0. TPQ TSR from part a 1. TQ TR PF 2. TQR is isosceles. definition of isosceles triangle 20. Yes; sample path: Q to T to P to Q to R to S to T to R USMP Geometry Scott, Foresman and ompany 113
21. The conjecture is false. Sample counterexample: LESSON 7-9 pp. 416 422 1. a. x 90 50 180 x 140 180 x 40 b. x y 180 40 y 180 y 140 c. z x 40 2. m m m, by the Exterior ngle 3. 2 and 1 4. 1 and 3 5. 2 and 3 6. a. Exterior ngle : In a triangle, the measure of an exterior angle is equal to the sum of the measures of the interior angles at the other two vertices of the triangle. b. Exterior ngle Inequality: In a triangle, the measure of an exterior angle is greater than the measure of the interior angle at each of the other two vertices. 7. a. is the largest angle, since the side opposite,, is the longest. b. is the smallest angle, since the side opposite,, is the shortest. 8. a. m m E m F 180 61 60 m F 180 121 m F 180 m F 59 EF is the longest side, since the angle opposite EF,, is the largest. b. E is the shortest side, since the angle opposite E, F, is the smallest. 9. a. 77 135 148 360 b. Exterior angle at : 80 at : 180 87 93 at : 180 90 90 at : 180 83 97 Sum: 80 93 90 97 360 10. Sample: U X 70 W V a. m V 70 b. m U 180 70, so m U 110 c. m UWX 180 70, so m UWX 110 11. True 12. L shortest longest V U middle If m U 60, then m L 60 and m V 60, and the sum of the measures of the interior angles of LUV 180. If m V 60, then m L 60 and m U 60, and the sum of the measures of the interior angles of LUV 180. If m L 60, then m U 60 and m V 60, and it is possible for the sum of the interior angles of LUV to equal 180. So L has measure 60. 13. m GHJ 18 15 33 m G 180 (40 33) 107 In GHJ, GHJ is the smallest angle, so GJ is the shortest side. 14. a. 3 is an exterior angle of PRT. Therefore, by the Exterior ngle Inequality, m 3 m 1. b. m 2 m Q 90. Therefore, by the Unequal ngles, PQ PS. 15. a. m m x b. m m m y x c. m m m 180 y 2x m 180 m 180 y 2x 114 USMP Geometry Scott, Foresman and ompany
16. m and m E each equal m m. and E also form vertical angles, so they have the same measure. 17. E is a parallelogram; Sufficient onditions for a Parallelogram, part (c) 18. Yes; Sufficient onditions for a Parallelogram, part (d) 19. No; congruent sides are not opposite each other. 20. VS QV 3x TS QR y QS 2 QV 2 3x 6x 21. Sample: a. E b. HL ongruence 22. Given: M is the midpoint of ; N M To prove: M is the midpoint of N. 0. M is the midpoint Given 0. of ; N M 1. M M definition of midpoint 2. NM M Vertical ngles 3. NM M S ongruence 4. NM M PF 5. M is the midpoint definition of 0. of N. midpoint 23. a. Sample: sum 72 72 72 72 72 5(72) sum 360 b. Sample: sum 60 60 60 60 60 60 sum 6(60) 360 c. The sum of the measures of the exterior angles of a convex polygon, one at each vertex, is 360. H P T E R 7 PROGRESS SELF-TEST pp. 426 427 1. a. m 180 m 180 60 120 b. m 60, by the Exterior ngle Inequality c. m 60, by the Exterior ngle Inequality 2. a. Sample (reduced size): M b. Yes c. by the S ongruence 3. a. Sample: 1" 10 cm 100 45 30 70 b. Yes c. by the S ongruence 4. a. No b. The given information is the SS condition but not Ss ongruence, so it does not yield a unique triangle. 5. a. Yes b. SSS ongruence 6. S Triangle ongruence : If two angles and a non-included side of one triangle are congruent to two angles and the corresponding non-included side of a second triangle, then the triangles are congruent. 7. a. by the S ongruence. b. WVU ZXY by the HL ongruence. L N USMP Geometry Scott, Foresman and ompany 115
8. a. b. S ongruence 9. Given: M is the midpoint of ; To prove: M M 0. M is the midpoint Given 0. of ; 1. M M definition of midpoint 2. M M Vertical ngles 3. M M Lines I 4. M M S ongruence 10. Given: WX WY; WUY WVX To prove: WUV is isosceles. 0. WX WY; Given 0. WUY WVX 1. W W Reflexive Property of ongruence 2. UWY VWX S ongruence 3. WU WV PF 4. WUV is isosceles. definition of isosceles triangle 11. Since,, and and are both right angles, by the HL ongruence. Thus, by the PF. 12. is a parallelogram because one pair of opposite sides is parallel and congruent. 13. OR PO x LR P y PR 2 PO 2x 14. Yes, is a parallelogram because the diagonals bisect each other. 15. Sample (reduced size): 16. (f) YZ; m XWY 180 (34 97) 180 131 49, so the shortest segment in WXY is WY. m WYZ 180 97 83, and m ZWY 180 (83 55) 180 138 42, so the shortest segment in WZY is YZ. So YZ must be the shortest segment in the figure. H P T E R 7 REVIEW pp. 428 432 1. a. No b. There are many noncongruent triangles that fit the given information. 2. a. Yes b. SSS ongruence 3. a. Yes b. S ongruence 4. a. Yes b. S ongruence 5. a. Sample: O O N 5 cm b. No. There are many noncongruent triangles that fit the given information. 6. a. Sample: Q 1.5" 1" 1" P 30 R P b. No. There are two noncongruent triangles that fit the given information. 7. a. Sample: T 2 cm S 60 4 cm U b. Yes, by the Ss ongruence. M 116 USMP Geometry Scott, Foresman and ompany
8. a. Sample: V 2" W 55 1" b. Yes, by the SS ongruence. 9. a. m QSR 180 m QST 180 132 48 b. m Q 132, by the Exterior ngle Inequality c. m Q m R m QST 132, by the Exterior ngle 10. a. q 2q 138 3q 138 q 46 m X q 46 b. m Y 2q 2(46) 92 11. 1 is the largest. 1 is an exterior angle of TWX, TWY, and TWZ. Therefore, m 1 m 2, m 1 m 3, and m 1 m 4. 12. m 2 360 (m m m ) 360 (110 100 90) 360 300 60 m 1 180 m 2 180 60 120 13. MOP MNP by the S ongruence 14. not enough information to tell 15. KLM KJM by HL ongruence 16. not enough information to tell 17. FG by S ongruence 18. not enough information to tell 19. Sample: a. F b. SSS ongruence 20. Sample: a. HI KL b. S ongruence X 21. Given: ; To prove: 0. ; Given 0. 1. Reflexive Property of ongruence 2. S ongruence 22. Given: WZ ZY; WX XY; WZ WX To prove: WZY WXY 0. WZ ZY; WX XY; Given 0. WZ WX 1. Z and X are definition of 0. right angles. perpendicular 2. WY WY Reflexive Property of ongruence 3. WZY WXY HL ongruence 23. Given: P and Q intersect at and. To prove: PQ PQ 0. P and Q Given 0. intersect at and. 1. P P; Q Q definition of circle 2. PQ PQ Reflexive Property of Equality 3. PQ PQ SSS ongruence 24. Given: UW bisects YUV; UW UY; V UXY To prove: UVW UXY 0. UW bisects YUV; Given 0. UW UY; 0. V UXY 1. YUX WUV definition of angle bisector 2. UVW UXY S ongruence USMP Geometry Scott, Foresman and ompany 117
25. annot be proved. This is the SS condition, but the longer congruent sides are not opposite the congruent angles. So the Ss ongruence does not apply. 26. Given: ; To prove: 0. ; Given 0. 1. Reflexive Property of ongruence 2. SS ongruence 27. Given: ; To prove: 0. ; Given 1. Reflexive Property of ongruence 2. SSS ongruence 3. PF 28. Given: EFGH is a regular octagon. To prove: 0. EFGH is a Given 0. regular octagon. 1. definition of regular polygon 2. definition of regular polygon 3. SS ongruence 4. PF 29. Given: JK bisects MJL; MJ LJ To prove: M L 0. JK bisects MJL; Given 0. MJ LJ 1. KJM KJL definition of angle bisector 2. JK JK Reflexive Property of ongruence 3. KJM KJL SS ongruence 4. M L PF 30. Given: ; ; To prove: 0. ; ; Given 0. 1. and are definition of 0. right angles. perpendicular lines 2. Reflexive Property of ongruence 3. HL ongruence 4. PF 31. Given: N is the midpoint of OE; l m To prove: E UO 0. N is the midpoint Given 0. of OE; l m 1. ON NE definition of midpoint 2. NUO NE; lines I 0. NOU NE 3. NUO NE S ongruence 4. E UO PF 32. Given: SPQ RQP; S R To prove: QS PR 0. SPQ RQP; Given 0. S R 1. PQ PQ Reflexive Property of ongruence 2. SPQ RQP S ongruence 3. QS PR PF 33. E E 12 19 2 E 2 12 24 34. E E 3x y 2 2 E 2 3x 6x 118 USMP Geometry Scott, Foresman and ompany
35. m m 130 m 180 m 180 130 50 m m 50 36. Z W 49. Sample: Y X 37. Yes, since both pairs of opposite sides are congruent. 38. Not necessarily, since the diagonals may or may not bisect each other. 39. Not necessarily. For example, could be an isosceles trapezoid with bases and, and. 40. Yes, since both pairs of opposite angles are congruent. 41. is the largest angle, since is the longest side. 42. is the smallest angle, since is the shortest side. 43. m F 180 (m m E) 180 (47 68) 180 115 65 sides from shortest to longest: EF, E, F 44. In GHJ, HJ is the shortest side. ut HJ is also in HJI, where HI is the shortest side. So the shortest segment in the figure is HI. 45. RP P by the HL ongruence. So, by the PF, RP P. 46. The SSS ongruence states that triangles with all three sides congruent are congruent. 47. X X X by the S ongruence, so by the PF. 48. a. QR and ZP have lengths equal to the width of the river. b. ZXP RXQ by the S ongruence, so QR ZP by the PF. 50. Sample: 51. Sample: 52. No; each angle in a regular octagon measures 135, but 360 is not evenly divisible by 135. So a regular octagon will not tessellate. 53. is approximately in the shape of a parallelogram. 54. Let their intersection be the midpoint of both PQ and MN. USMP Geometry Scott, Foresman and ompany 119
55., and EF by the Properties of a Parallelogram (opposite sides are congruent). Therefore, EF by the Transitive Property of Equality. 56. Yes; by the Properties of a Parallelogram (opposite angles are congruent). H P T E R 7 REFRESHER p. 433 1. 6(2 x) 12 6x 2. x(4 y) 4x xy 3. (h l)a ah al 4. 1 2 b(w t) 1 2 bw 1 2 bt 5. (a 2b)b ab 2b 2 6. 2xy(y x) 2xy 2 2x 2 y 7. ax 2x x(a 2) 8. 9c 2 12c 3c(3c 4) 9. 4b 20h 4(b 5h) 10. r 2 h 2 (r 2 h 2 ) 11. 1 2 bh 1 2 lh 1 h(b l) 2 12. 12ar 2 6ar 6ar(2r 1) 13. (y 2)(y 4) y 2 2y 4y 8 y 2 6y 8 14. (a 7)(3b 2) 3ab 21b 2a 14 15. (2r t)(2r t) 4r 2 2tr 2rt t 2 4r 2 t 2 16. (s 4) 2 (s 4)(s 4) s 2 4s 4s 16 s 2 8s 16 17. (m n) 2 (m n)(m n) m 2 mn mn n 2 m 2 2mn n 2 18. 1 2 (e f )(y h) 1 (ey eh f y f h) 2 1 2 ey 1 2 eh 1 2 f y 1 2 f h 19. 2 1.41 20. 4 3 6.93 21. 4 2 5 2 16 25 41 6.40 17 22. 2.06 4 23. 100 10 24. 40 4 10 2 10 25. 18 9 2 3 2 26. 2 45 2 9 5 2 3 5 6 5 27. x 2 25 x 5 28. y 2 9 25 y 2 16 y 4 29. 40 2z 2 2z 2 40 z 2 20 z 20 2 5 4.47 30. 3 r 2 r 2 3 r 2 3 3 r 0.98 120 USMP Geometry Scott, Foresman and ompany