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CHAPTER 7 11. a . Sample: T RIANGLE C ONGRUENCE LESSON 7-1 C m/CAB = 7 0 m/CBA = 3 8 pp. 364–369 1. Triangulation is the use of triangles to determine the location of an object. 2 . a . Sample: Z X Y A B b . Any opinion is allowed. The answer will be found to be no. 12 . a . Sample: AB = 4 m/CAB = 6 0 C t V W A Angle measures: < 22°, 50°, 108° b . Any opinion is allowed. The answer is yes. 3 . Sample: Z 3 cm X 5 cm Y B C D 45° B 6 7 8 9 10 E 80° A b . Any opinion is allowed. The answer is no. . 180 2 (32 1 102) 5 180 2 134 5 46 The third angle has measure 46. . The measure of the third angle is 180 2 (x 1 y), or 180 2 x 2 y. . Answers may vary. Mathematically, they should agree. . Sample: Yes; given the lengths of three segments, a unique triangle is determined. . Sample: No; given the measures of three angles, a unique triangle is not determined. . Sample: Yes; given the lengths of two segments and the measure of the angle they form, a unique triangle is determined. 10 0 b . Any opinion is allowed. The answer will be found to be no. 13 . a . Sample: BC = 3 C m/CAB = 4 0 m/CBA = 6 0 4 cm 4 . a . Sample: 5 B A b . Any opinion is allowed. The answer will be found to be yes. 14 . a . Sample: AB = 2 m/CAB = 4 0 m/CBA = 6 0 C A B b . Any 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 D C 182' A 62° B UCSMP Geometry © Scott, Foresman and Company 16. b. CD 1.1 cm 1829 < 5 DE 2.3 cm n 21. If m/PAB 5 40, then m/ACD 5 m/PBC 5 40; m/D 5 m/DCB 5 m/CBA 5 m/BAD 5 m/APB 5 m/CPB 5 90; m/DAC 5 m/BCA 5 m/PBA 5 50 22. a. Sample: n 5 (2.3 ? 182) ÷ 1.1 < 381 DE < 3819 c . ASA; 2 angles and the side between them 17. sum of angle measures of octagon: (n 2 2) ? 180 5 (8 2 2) ? 180 5 6 ? 180 5 1080 measure of each angle: 1080 ÷ 8 5 135 Each angle in a stop sign has measure 135. 18. a. IR > TH b. /GHT > /BRI 19. If B, G, H, and R are not collinear, then these points are vertices of a quadrilateral called an isosceles trapezoid. 20. Sample: 4LMN is the image of 4FGH under an isometry. n A C B D m b. rm ◦ rn is a rotation. 23. N M P Q S , 24. 1 ( 21], 6 2] ) book 10 9 te le p1h o ne r e ce iv e r (2, 2] ) 8 p a rro t (7, 10) ca r (4, 6) r o lle r s k a te (6, 21] ) b e ll (721], 6) 7 2 6 a ir p la ne (7, 421] ) 5 b u g le (1] , 4) 2 4 mo u s e (1, 5) r o o s te r ’s h e a d (4 12] , 8) 3 2 w o o l ca p (8, 9 ) b a th r o b e 1 (2, 7) y o -y o Pu r ita n’s h a t 0 ( 12] , 9 ) (8, 0) 1 UCSMP Geometry © Scott, Foresman and Company 2 3 4 5 6 7 8 101 LESSON 7-2 pp. 370–377 1. Four conditions that lead to triangle congruence are SSS, SAS, ASA, and AAS. 2. SSS Congruence Theorem 3. Symmetry of a kite was used in the proof of the SSS Congruence Theorem. 4. Symmetry of an isosceles triangle was used in the proof of the SAS Congruence Theorem. 5. ASA Congruence Theorem: 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 AAS Congruence Theorem, the corresponding sides are not included by the pairs of congruent angles as they are in the ASA Congruence Theorem. 7. AD is included by /CAD and /ADC. 8. 4ABF > 4CED by SAS 9. not enough information to know 10. not enough information to know 11. 4OY N > 4YOX by ASA 12. 4CGO > 4IPN by ASA 13. 4STE > 4BTE by AAS 14. SSS explains why all triangles are rigid. 15. a. 4DPO > 4JKL b. If OD 5 5 yards, then LJ is 5 yards. c. /O > /L d . If m/P 5 73, then m/K 5 73. 16. a. /C > /E b. /A > /F c. CB > ED d . AC > FE and CB > ED 17. 4ABD > 4CBD by the AAS Congruence Theorem. 18. Yes, the triangles will be congruent by the ASA Congruence Theorem. 19. Given: m/FDE 5 m/C9DE; G, D, and E are collinear. @##$ bisects /FDC9. To prove: EG Argument: Conclusions Justifications 0. m/FDE 5 m/C9DE; Given 0. G, D, and E are collinear. 102 1. m/FDG 5 0. 180 2 m/FDE; 0. m/C9DG 5 0. 180 2 m/C9DE 2. m/FDG 5 0. m/C9DG @##$ bisects /FDC9. 3. EG Linear Pair Theorem (and algebra) Transitive Property of Equality (and substitution) definition of angle bisector 20. a. n 5 2 b. 21. a. n 5 5 b. 22. a. n 5 3 b. 23. Given: 4ABC > 4DEF. If m/A 5 32 and m/B 5 64, then m/C 5 180 2 (32 1 64) 5 180 2 96 5 84. Since m/C 5 m/F, then m/F 5 84. 24. Given: 4OLD > 4NEW Sample conclusions: /O > /N, /L > /E, /D > /W, OL > NE, LD > EW, OD > NW Justification: CPCF Theorem UCSMP Geometry © Scott, Foresman and Company y 25. 1. WY > VY 2. /XY W > /ZY V x G z H 26. Sample: C E A B LESSON 7-3 1. a. pp. 378–383 B A D F D C E b. ASA Congruence Theorem c. Given: C is the midpoint of AE; /BAC > /DEC; /BCA > /DCE To prove: 4ABC > 4EDC Argument: Conclusions Justifications 0. C is the midpoint of Given 0. AE; /BAC > /DEC; 0. /BCA > /DCE 1. AC > EC definition of midpoint 2. 4ABC > 4EDC ASA Congruence Theorem 2. Given: Y is the midpoint of WV; m/Z 5 m/X To prove: 4WXY > 4VZY Argument: Conclusions Justifications 0. Y is the midpoint Given 0. of WV; m/Z 5 m/X UCSMP Geometry © Scott, Foresman and Company 3. 4WXY > 4VZY definition of midpoint Vertical Angles Theorem AAS Congruence Theorem 3. Given: AB 5 CD; BC 5 AD To prove: /BAC > /DCA Argument: Conclusions Justifications 0. AB 5 CD; BC 5 AD Given 1. AC > AC Reflexive Property of Congruence 2. 4ABC > 4CDA SSS Congruence Theorem 3. /BAC > /DCA CPCF Theorem 4. a. AB and BC are congruent. b. Isosceles Triangle Base Angles Converse Theorem 5. a. Reflexive Property of Congruence b. AAS Congruence Theorem c. CPCF Theorem 6. Given: XA 5 Y A, and XC 5 Y C To prove: /XAC > /Y AC Argument: Conclusions Justifications 0. XA 5 Y A, and Given 0. XC 5 Y C 1. AC > AC Reflexive Property of Congruence 2. 4XAC > 4Y AC SSS Congruence Theorem 3. /XAC > /Y AC CPCF Theorem ###$ bisects /JNM; NM > NK; 7. Given: NK NL > NJ To prove: 4JNK > 4LNM Argument: Conclusions Justifications ###$ bisects /JNM; 0. NK Given 0. NM > NK; NL > NJ 1. /JNK > /LNM definition of angle bisector 2. 4JNK > 4LNM SAS Congruence Theorem 103 8. Given: BC k AD, and BC 5 AD To prove: 4ABC > 4CDA Argument: Conclusions Justifications 0. BC k AD, and Given 0. BC 5 AD 1. AC 5 AC Reflexive Property of Equality 2. /BCA > /DAC k Lines ⇒ AIA > Theorem 3. 4ABC > 4CDA SAS Congruence Theorem 9. Given: m/ADB 5 m/CDB; Given: m/ABD 5 m/CBD To prove: AB 5 CB Argument: Justifications Conclusions 0. m/ADB 5 m/CDB; Given 0. m/ABD 5 m/CBD 1. BD 5 BD Reflexive Property of Equality 2. 4ABD > 4CBD ASA Congruence Theorem 3. AB 5 CB CPCF Theorem 10. a. 4ADG, 4ADB, 4BDJ, 4DJG, 4AGJ, 4AGB, 4BJG, 4BJA b. 4ADG > 4BDJ, 4AGJ > 4BJG, 4AGB > 4BJA 11. a. SAS Congruence Theorem b. CPCF Theorem 12. not enough information to know 13. 4EFG > 4IHG; AAS Congruence Theorem 14. a. True b. False 15. a. Yes, assuming the units for the length measurements are the same b. SAS Congruence Theorem 16. If m/DAC 5 50, then m/BAC 5 m/DCA 5 m/BCA 5 50, m/D 5 m/B 5 80, and m/DAB 5 m/DCB 5 100. 17. B = A' B' A D 104 C = D' 18. The sum of the measures of the acute angles in a right triangle is 180 2 90 5 90. 19. Sample: C B' = A D B = A' C' 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. Because 3 cm 1 6 cm 5 9 cm , 11 cm, the Triangle Inequality Postulate is violated. 21. Yes. Since 4ABC > 4CDA, both /BAC > /DCA, and /BCA > /DAC (CPCF Theorem). As a consequence, AB k DC and BC k DA (AIA > ⇒ k Lines Theorem). Therefore, ABCD is a parallelogram. LESSON 7-4 pp. 384–388 1. a. There are 8 triangles in the figure: 4QUS, 4USA, 4ASD, 4DSQ, 4QUA, 4UAD, 4ADQ, 4DQU. b. 4DAU appears congruent to 4QUA. c. 4QUD appears congruent to 4DAQ. 2. a. Given b. Reflexive Property of Congruence c. SAS Congruence Theorem d. CPCF Theorem 3. Given: Regular pentagon ALIVE To prove: /EAV > /VIE Argument: Conclusions Justifications 0. Regular pentagon Given 0. ALIVE 1. AE > IV definition of regular pentagon 2. EV > EV Reflexive Property of Congruence 3. /AEV > /IVE definition of regular pentagon 4. 4AEV > 4IVE SAS Congruence Theorem 5. /EAV > /VIE CPCF Theorem C' UCSMP Geometry © Scott, Foresman and Company 4. Given: QU > AD, and QA > UD To prove: /1 > /2 Argument: Conclusions Justifications 0. QU > AD, and Given 0. QA > UD 1. UA > UA Reflexive Property of Congruence 2. 4QUA > 4DAU SSS Congruence Theorem 3. /1 > /2 CPCF Theorem 5. Given: AD 5 AE, and m/D 5 m/E To prove: EB 5 CD Argument: Justifications Conclusions 0. AD 5 AE, and Given 0. m/D 5 m/E 1. /A > /A Reflexive Property of Congruence 2. 4AEB > 4ADC ASA Congruence Theorem 3. EB 5 CD CPCF Theorem 6. Given: PR 5 QS, and PS 5 QR To prove: m/P 5 m/Q Argument: Conclusions Justifications 0. PR 5 QS, and Given 0. PS 5 QR 1. RS 5 RS Reflexive Property of Equality 2. 4PRS > 4QSR SSS Congruence Theorem 3. m/P 5 m/Q CPCF Theorem 7. Argument: Since l, the perpendicular bisector of AB and ED, is a symmetry line of ABCDEF by the Regular Polygon Symmetry Theorem, rl(A) 5 B and rl(C) 5 F. Therefore, AC > BF because reflections preserve distance. 8. Given: A, D, E, and B are collinear; AD > BE; AC > BC To prove: DC > CE UCSMP Geometry © Scott, Foresman and Company Argument: Conclusions 0. A, D, E, and B 0. are collinear; 0. AD > BE; AC > BC 1. /CAB > /CBA Justifications Given Isosceles Triangle Base Angles Theorem 2. 4CAD > 4CBE SAS Congruence Theorem 3. DC > CE CPCF Theorem 9. Given: QR k TU, and S is the midpoint of QU. To prove: S is the midpoint of RT. Argument: Conclusions Justifications 0. QR k TU, and S is Given 0. the midpoint of QU. 1. QS 5 US definition of midpoint 2. m/QSR 5 m/UST Vertical Angles Theorem 3. m/Q 5 m/U k lines ⇒ AIA > Theorem 4. 4QSR > 4UST ASA Congruence Theorem 5. ST 5 SR CPCF Theorem 6. S is the midpoint definition of 0. of RT. midpoint 10. 4PCA > 4PCB by the SAS Congruence Theorem. So, PA 5 PB by the CPCF Theorem. 11. a. Vertical Angles Theorem b. SAS Congruence Theorem c. CPCF Theorem d. Vertical Angles Theorem e . Transitive Property of Congruence 12. a. Sample: C A 32° 110° B b. No, because the AA condition does not determine a unique triangle. 105 13. sum of angle measures of a pentagon: (n 2 2) ? 180 5 (5 2 2) ? 180 5 3 ? 180 5 540 measure of each angle: 540 ÷ 5 5 108 14. True 15. True 16. 0 , 180 2 2y 2y , 180 y , 90 The angle is acute. 17. 4ADC > 4AEB 5 AB, AC, and point F. 18. WX 1 XY 1 Y Z 5 WZ 13 1 XY 1 17 5 45 XY 5 15 19. Samples are given. a. b. c. heptagon octagon LESSON 7-5 pp. 389–396 1. Sample: 2. a. A right triangle has sides represented by H and L. b. H represents hypotenuse; L represents leg. 3. The proof of the HL Congruence Theorem makes use of the symmetry of the isosceles triangle. 4. Given: BD ⊥ AC; AB 5 BC To prove: 4ABD > 4CBD Argument: Conclusions Justifications 0. BD ⊥ AC; AB 5 BC Given 1. BD 5 BD Reflexive Property of Equality 2. 4ABD > 4CBD HL Congruence Theorem 5. The SSA Condition leads to congruence when the sides opposite the congruent angles in each triangle are longer than the other congruent sides. 6. a. HL Congruence Theorem b. 4ABC > 4FED 7. a. SAS Congruence Theorem b. 4GHI > 4KJL 8. a. HL Congruence Theorem b. 4MNP > 4ONP 9. a.–c. reduced size: W WY 50° 9 cm 9 cm 11 cm X nonagon decagon Z d. No. There are two possible triangles XZW, as shown in the drawing. 10. Given: VX ⊥ WY , WZ 5 VY , and XZ 5 XY To prove: /W > /V Argument: Conclusions Justifications 0. VX⊥WY , WZ 5 VY , Given 0. and XZ 5 XY 106 UCSMP Geometry © Scott, Foresman and Company 10. (continued) 1. /WXZ and /VXY 0. are right angles. 2. 4WXZ > 4VXY definition of perpendicular lines HL Congruence Theorem 3. /W > /V CPCF Theorem 11. Given: AP 5 AR; /P and /R are right angles. To prove: PBRA is a kite. Argument: Conclusions Justifications 0. AP 5 AR; /P and Given 0. /R are right angles. 1. AB 5 AB Reflexive Property of Equality 2. 4PBA > 4RBA HL Congruence Theorem 3. PB > RB CPCF Theorem 4. PBRA is a kite. definition of kite 12. a. Given b. definition of circle c. definition of circle d. SsA Congruence Theorem (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 April’s hands, J the position of June’s hand, and A the position of April’s hand. T J A M ground JA is parallel to the ground, since J and A are equal in height (given). So, JA ⊥ TM by the Perpendicular to Parallels Theorem. Since TM > TM and TJ > TA, then 4TJM > 4TAM by the HL Congruence Theorem. Thus, JM > AM by the CPCF Theorem. So, since JM is the distance from June’s hand to the maypole, and AM is the distance from April’s hand to the maypole, and JM 5 AM, their hands are the same distance from the maypole. 14. a. definition of isosceles trapezoid b. Isosceles Trapezoid Theorem UCSMP Geometry © Scott, Foresman and Company c. Reflexive Property of Congruence d. SAS Congruence Theorem (steps 1–3) e. CPCF Theorem 15. f . Isosceles Trapezoid Symmetry Theorem g . Reflections preserve distance. 16. Answers may vary. Sample: “I prefer the argument in Question 15 because it is shorter.” 17. a. reduced size: R 80° 2" Q 45° S b. Yes, by the AAS Congruence Theorem. 18. Given: AB 5 AC; BE 5 DC a. To prove: /ADE > /AED Argument: Conclusions Justifications 0. AB 5 AC; Given 0. BE 5 DC 1. /B > /C Isosceles Triangle Base Angles Theorem 2. 4ABE > 4ACD SAS Congruence Theorem 3. /ADE > /AED CPCF Theorem b. To prove: 4ADE is isosceles. (continuing from the proof above) 1. AD > AE Isosceles Triangle Base Angles Converse Theorem 2. 4ADE is definition of isosceles 0. isosceles. triangle 19. 4AGY , 4DEX, 4DEY 20. a. 4ABC > 4DEF b. rm(4DEF) 5 4ABC c. Sample: rm(AB) 5 DE 21. The conjecture is false. Counterexample: A B D C E H F G 107 LESSON 7-6 pp. 397–403 13. 1. A 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: Check students’ drawings. Step 5: ABCD is a parallelogram. Rotations preserve angle measure, so /BAC > /DCA and /BCA > /DAC. By the AIA > ⇒ k Lines Theorem, AD k BC and AB k DC. Therefore, ABCD is a parallelogram by the definition of parallelogram. 4. Check students’ drawings. 5. Sample (reduced size): 14. a. Sample (reduced size): Y A 6. Sample (reduced size): Z C 108 O X Z" A* 7. Key question: Can a given region tessellate the plane? 8. One museum where many tessellations can be found is the Alhambra. 9. Yes. Since the sum of the angle measures in ABCD 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. A new type of tessellating pentagon was discovered as recently as 1985. B C' Y' Z' A' C" B" b. They are the four angles of ABOC, so the sum of their measures is 360. c. Translate the figure repeatedly in all directions. d. Answers may vary. 15. 4ABC > 4CDA, so /BAC > /DCA and /BCA > /DAC (CPCF Theorem). So AB k CD and BC k AD (AIA > ⇒ k Lines), and ABCD is a parallelogram (definition of parallelogram). 16. 4ABC > 4ADC, so AB 5 AD, BC 5 DC (CPCF Theorem). So ABCD is a kite (definition of kite). 17. Given: AD 5 BC; DA ⊥ AB; BC ⊥ CD To prove: AB k DC Argument: Conclusions Justifications 0. AD 5 BC; Given 0. DA ⊥ AB; BC ⊥ CD 1. BD 5 BD Reflexive Property of Equality UCSMP Geometry © Scott, Foresman and Company 17. (continued) 2. 4ABD > 4CDB 3. /ABD > /CDB 4. AB k DC 24. Sample (reduced size): HL Congruence Theorem CPCF Theorem AIA > ⇒ k Lines Theorem 18. Given: AC > BD; AD > BC To prove: /ADB > /BCA Argument: Conclusions Justifications 0. AC > BD; AD > BC Given 1. AB > AB Reflexive Property of Congruence 2. 4ABD > 4BAC SSS Congruence Theorem 3. /ADB > /BCA CPCF Theorem 19. Given: m/ABD 5 m/BDC; Given: m/ADB 5 m/DBC To prove: AB 5 CD Argument: Conclusions Justifications 0. m/ABD 5 m/BDC; Given 0. m/ADB 5 m/DBC 1. BD > BD Reflexive Property of Congruence 2. 4BAD > 4DCB ASA Congruence Theorem 3. AB 5 CD CPCF Theorem 20. (1) /B > /X; ASA Congruence Theorem (2) AC > Y Z; SAS Congruence Theorem (3) /C > /Z; AAS Congruence Theorem 21. a. No b. There is no such segment, unless 4PQR is isosceles with PQ 5 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 African people of Berber and Arab descent who invaded and conquered Spain in the eighth century. They were driven back to Africa during centuries of wars and inquisitions, the final Moors leaving in 1492. UCSMP Geometry © Scott, Foresman and Company IN-CLASS ACTIVITY S t ep 1. A D p. 404 B C /B > /D; /DAB > /BCD; /BAC > /DCA; /DAC > /BCA S t ep 2. Argument: Conclusions Justifications 0. Parallelogram Given 0. ABCD 1. AD k CB; definition of 0. AB k DC parallelogram 2. m/DAC k Lines ⇒ AIA 5 0. 5 m/BCA; Theorem 0. m/BAC 0. 5 m/DCA 3. AC 5 AC Reflexive Property of Equality 4. 4ADC > 4CBA ASA Congruence Theorem S t ep 3. AD > BC; AB > CD; /D > /B 109 Step 4. A 2. BD > BD B E D C 4ADE > 4CBE; 4CED > 4AEB; 4ABD > 4CDB; 4ABC > 4CDA Step 5. /ADB > /CBD; /ABD > /CDB; AE > CE; DE > BE Step 6. The parallelogram possesses 2-fold rotation symmetry. LESSON 7-7 pp. 405–410 1. a. BC is congruent to AD. b. /5 is congruent to /1. c. The midpoints of AC and BD are the same point (E). 2. a. AB, BC, and CD are congruent to AD. b. /2, /5, and /6 are congruent to /1. c. The midpoints of AC and BD are the same point (E). 3. a. definition of parallelogram b. k lines ⇒ AIA > Theorem c. definition of parallelogram d. k lines ⇒ AIA > Theorem e. ASA Congruence Theorem f. CPCF Theorem 4. Properties of a Parallelogram Theorem, part (a) 5. This image of ABCD is CDAB. 6. Given: Parallelogram ABCD To prove: /DAB > /BCD Drawing: B A D C Argument: Draw auxiliary diagonal BD. Conclusions Justifications 0. Parallelogram ABCD Given 1. AD > CB; AB > CD Properties of a Parallelogram Theorem (a) 110 Reflexive Property of Congruence 3. 4ABD > 4CDB SSS Congruence Theorem 4. /A > /C CPCF Theorem 7. Given: Parallelogram ABCD; AC > BD 5 E To prove: E is the midpoint of AC and BD. Drawing: B A E C D Argument: Conclusions 0. Parallelogram ABCD 1. AD k BC 2. /ADB > /CBD 3. /AED > /CEB 4. AD > CB 5. 4ADE > 4CBE Justifications Given definition of parallelogram k lines ⇒ AIA > Theorem Vertical Angles Theorem Properties of a Parallelogram Theorem (a) AAS Congruence Theorem CPCF Theorem definition of midpoint 6. AE > CE; DE > BE 7. E is the midpoint 0. of AC and BD. 8. a. CE 5 AE 5 7 b. BC 5 AD 5 12 9. a. ED 5 BE 5 x b. AD 5 BC 5 2y c. BD 5 2 ? BE 5 2x 10. a. RD and OP are congruent. b. The distance between two given parallel lines is constant. 11. (1) D' A' A B C C' D B' UCSMP Geometry © Scott, Foresman and Company 11. (continued) (2) A C' B B' D D' (3) A' C B' B A C' 7. m/A 1 m/B 1 0. m/BCD 1 0. m/CDA 5 360 8. 4 ? m/A 5 360 9. m/A 5 m/B 5 0. m/BCD 5 0. m/CDA 5 90 10. ABCD is a 0. rectangle. 17. Sample: Quadrilateral-Sum Theorem Substitution Multiplication Property of Equality and Substitution definition of rectangle A' D C D' 12. (b) A is false, B is true. 13. Theorem: The distance between two given parallel lines is constant. 14. a., b. kite (rd) parallelogram (R) rectangle (rpb) (R) rhombus (rd) (R) sq uare (rd) (rpb) (R) 15. From the Properties of a Parallelogram Theorem, AB 5 CD, CD 5 EF, EF 5 GH. From the Transitive Property of Equality, AB 5 GH. 16. Given: ABCD is a parallelogram; M is the midpoint of AB. To prove: If MD 5 MC, then ABCD is a rectangle. Argument: Conclusions Justifications 0. ABCD is a Given 0. parallelogram; 0. M is the midpoint 0. of AB; MD 5 MC 1. AM 5 BM definition of midpoint 2. AD 5 BC Properties of a Parallelogram Theorem (a) 3. 4AMD > 4BMC SSS Congruence Theorem 4. /A > /B CPCF Theorem 5. /B > /CDA; Properties of a 0. /A > /BCD Parallelogram Theorem (b) 6. /A > /B > Transitive Property of 0. /BCD > /CDA Congruence UCSMP Geometry © Scott, Foresman and Company 18. Given: Trapezoid ABCF; AE ⊥ CF; BD ⊥ CF; @#$ k FC @##$ AF 5 BC; AB To prove: /F > /C Argument: Conclusions Justifications 0. Trapezoid ABCF; Given 0. AE ⊥ CF; BD ⊥ CF; @#$ k FC @##$ 0. AF 5 BC; AB 1. /AEF and /BDC definition of 0. are right angles. perpendicular 2. AE k BD Two Perpendiculars Theorem 3. ABDE is a definition of 0. parallelogram. parallelogram 4. AE 5 BD Properties of a Parallelogram Theorem (a) 5. 4AEF > 4BDC HL Congruence Theorem 6. /F > /C CPCF Theorem 19. No; the congruent sides are not corresponding sides. AC corresponds to XZ, not Y Z. 20. a. 8 triangles are formed. © ª © ª b. ©4RIT, 4GHT , 4RTH, 4GTI ª , 4RIH, 4IRG, 4HGI, 4GHR 21. It cannot be justified. 22. A D B C 23. Transitive Property of Congruence 24. Two Perpendiculars Theorem 111 25. K 3. a. Answers will vary. Sample: L M J I B A P E N C D F O H G A square is formed. 26. LESSON 7-8 pp. 411–415 1. According to the definition of parallelogram, two pairs of parallel sides makes a quadrilateral a parallelogram. 2. a. Answers will vary. Sample: D A 6.5 cm 6.5 cm C B b. The quadrilateral formed is a parallelogram. c. Yes. You can form a rectangle and a rhombus, but they are also parallelograms. 112 b. A parallelogram is formed. 4. Sufficient conditions for parallelograms other than its definitions: (1) Both pairs of opposite sides are congruent. (2) Both 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. Argument: Conclusions Justifications 0. PQ > RS; Given 0. PS > QR 1. Draw QS. Unique Line Assumption 2. QS > QS Reflexive Property of Congruence 3. 4PQS > 4RSQ SSS Congruence Theorem 4. /PQS > /RSQ CPCF Theorem 5. PQ k SR AIA > ⇒ k Lines Theorem 6. /PSQ > /RQS CPCF Theorem 7. PS k QR AIA > ⇒ k Lines Theorem 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, ABCD could be an isosceles trapezoid with bases AB and CD. 9. A parallelogram is formed. 10. A parallelogram, a kite, or a rectangle can be formed. UCSMP Geometry © Scott, Foresman and Company 11. Given: Quadrilateral ABCD; diagonals AC and BD intersect at their midpoint, O. To prove: ABCD is a parallelogram. Drawing: A B O D C Argument: Conclusions 0. O is the midpoint 6. of AC and BD. 1. AO > CO; DO > BO 2. /AOB > /COD 3. 4AOB > 4COD 4. /ABD > /CDB 5. AB k CD 6. /AOD > /COB 7. 4AOD > 4COB 8./ADB > /CBD 9. AD k BC Justifications Given definition of midpoint Vertical Angles Theorem SAS Congruence Theorem CPCF Theorem AIA > ⇒ k Lines Theorem Vertical Angles Theorem SAS Congruence Theorem CPCF Theorem AIA > ⇒ k Lines Theorem definition of parallelogram 10. ABCD is a 6. parallelogram. 12. (5x 1 30) 1 (7x 2 90) 1 (3x 2 10) 1 (4x 1 50) 5 360 19x 2 20 5 360 19x 5 380 x 5 20 m/N 5 5x 1 30 5 5(20) 1 30 5 130 m/S 5 4x 1 50 5 4(20) 1 50 5 130 m/E 5 7x 2 90 5 7(20) 2 90 5 50 m/T 5 3x 2 10 5 3(20) 2 10 5 50 Yes, NEST is a parallelogram because both pairs of opposite angles are congruent. UCSMP Geometry © Scott, Foresman and Company 13. The first scout was right. Since the diagonals intersect at their midpoints, ABCD is a parallelogram. So opposite angles are congruent: /DAB > /BCD, and /ADC > /CBA. But also, 4ADC > 4BCD by SSS Theorem. So /ADC > /BCD. Then all 4 angles of the parallelogram are congruent, so each must be a right angle. Therefore, ABCD is a rectangle. 14. No; WX is not parallel to ZY . 15. m/BCD 5 m/BAD 5 70 m/ADC 5 180 2 m/BAD 5 180 2 70 5 110 m/ABC 5 m/ADC 5 110 16. DE 5 BE 5 11 AD 5 BC 5 20 BD 5 2 ? BE 5 2 ? 11 5 22 17. PQRS is a parallelogram, so PQ 5 RS (Properties of a Parallelogram Theorem). 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: 4PTS is isosceles with vertex angle T, and m/PTQ 5 m/STR. a. To prove: 4TPQ > 4TSR Argument: Conclusions Justifications 0. 4PTS is Given 0. isosceles with 0. vertex angle T, 0. and m/PTQ 0. 5 m/STR. 1. PT 5 ST definition of isosceles triangle 2. m/P 5 m/S Isosceles Triangle Base Angles Theorem 3. 4TPQ > 4TSR ASA Congruence Theorem b. To prove: 4TQR is isosceles. Argument: Conclusions Justifications 0. 4TPQ > 4TSR from part a 1. TQ 5 TR CPCF Theorem 2. 4TQR 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 113 21. The conjecture is false. Sample counterexample: A D B C LESSON 7-9 pp. 416–422 1. a. x 1 90 1 50 5 180 x 1 140 5 180 x 5 40 b. x 1 y 5 180 40 1 y 5 180 y 5 140 c. z 5 x 5 40 2. m/ABD 5 m/C 1 m/D, by the Exterior Angle Theorem 3. /2 and /1 4. /1 and /3 5. /2 and /3 6. a. Exterior Angle Theorem: 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 Angle 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. /B is the largest angle, since the side opposite /B, AC, is the longest. b. /C is the smallest angle, since the side opposite /C, AB, is the shortest. 8. a. m/D 1 m/E 1 m/F 5 180 61 1 60 1 m/F 5 180 121 1 m/F 5 180 m/F 5 59 EF is the longest side, since the angle opposite EF, /D, is the largest. b. DE is the shortest side, since the angle opposite DE, /F, is the smallest. 9. a. 77 1 135 1 148 5 360 114 b. Exterior angle at C: 80 at B: 180 2 87 5 93 at A: 180 2 90 5 90 at D: 180 2 83 5 97 Sum: 80 1 93 1 90 1 97 5 360 10. Sample: U X 70° W V a. m/V , 70 b. m/U , 180 2 70, so m/U , 110 c. m/UWX , 180 2 70, so m/UWX , 110 11. True 12. L shortest longest V middle U If m/U 5 60, then m/L . 60 and m/V . 60, and the sum of the measures of the interior angles of 4LUV . 180. If m/V 5 60, then m/L , 60 and m/U , 60, and the sum of the measures of the interior angles of 4LUV , 180. If m/L 5 60, then m/U , 60 and m/V . 60, and it is possible for the sum of the interior angles of 4LUV to equal 180. So /L has measure 60. 13. m/GHJ 5 18 1 15 5 33 m/G 5 180 2 (40 1 33) 5 107 In 4GHJ, /GHJ is the smallest angle, so GJ is the shortest side. 14. a. /3 is an exterior angle of 4PRT. Therefore, by the Exterior Angle Inequality, m/3 . m/1. b. m/2 , m/Q 5 90. Therefore, by the Unequal Angles Theorem, PQ , PS. 15. a. m/ACB 5 m/BCD 5 x b. m/CBD 5 m/A 1 m/ACB 5 y 1 x c. m/A 1 m/ACD 1 m/D 5 180 y 1 2x 1 m/D 5 180 m/D 5 180 2 y 2 2x UCSMP Geometry © Scott, Foresman and Company 16. m/BCD and m/ACE each equal m/A 1 m/B. /BCD and /ACE also form vertical angles, so they have the same measure. 17. ADCE is a parallelogram; Sufficient Conditions for a Parallelogram Theorem, part (c) 18. Yes; Sufficient Conditions for a Parallelogram Theorem, part (d) 19. No; congruent sides are not opposite each other. 20. VS 5 QV 5 3x TS 5 QR 5 y QS 5 2 ? QV 5 2 ? 3x 5 6x 21. Sample: a. AB 5 DE b. HL Congruence Theorem 22. Given: M is the midpoint of AB; /N > /MCB To prove: M is the midpoint of NC. Argument: Conclusions Justifications 0. M is the midpoint Given 0. of AB; /N > /MCB 1. AM 5 BM definition of midpoint 2. /NMA > /CMB Vertical Angles Theorem 3. 4ANM > 4BCM AAS Congruence Theorem 4. NM 5 CM CPCF Theorem 5. M is the midpoint definition of 0. of NC. midpoint 23. a. Sample: sum 5 72 1 72 1 72 1 72 1 72 5 5(72) sum 5 360 b. Sample: sum 5 60 1 60 1 60 1 60 1 60 1 60 sum 5 6(60) 5 360 UCSMP Geometry © Scott, Foresman and Company c. The sum of the measures of the exterior angles of a convex polygon, one at each vertex, is 360. CHAPTER 7 PROGRESS SELF-TEST pp. 426–427 1. a. m/CBD 5 180 2 m/ABD 5 180 2 60 5 120 b. m/C , 60, by the Exterior Angle Inequality c. m/D , 60, by the Exterior Angle Inequality 2. a. Sample (reduced size): M 10 cm L 30° 70° N b. Yes c. by the ASA Congruence Theorem 3. a. Sample: A 1" 100° 45° B C Yes by the AAS Congruence Theorem No The given information is the SSA condition but not SsA Congruence, so it does not yield a unique triangle. 5. a. Yes b. SSS Congruence Theorem 6. AAS Triangle Congruence Theorem: 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. 4ABC > 4CDA by the ASA Congruence Theorem. b. 4WVU > 4ZXY by the HL Congruence Theorem. b. c. 4. a. b. 115 8. a. 4ABC > 4CDA b. ASA Congruence Theorem @#$ k CD @##$ 9. Given: M is the midpoint of AC; AB To prove: 4MBA > 4MDC Argument: Justifications Conclusions 0. M is the midpoint Given @#$ k CD @##$ 0. of AC; AB 1. AM > CM definition of midpoint 2. /AMB > /CMD Vertical Angles Theorem 3. /BAM > /DCM k Lines ⇒ AIA > Theorem 4. 4MBA > 4MDC ASA Congruence Theorem 10. Given: WX 5 WY ; /WUY > /WVX To prove: 4WUV is isosceles. Argument: Conclusions Justifications 0. WX 5 WY ; Given 0. /WUY > /WVX 1. /W > /W Reflexive Property of Congruence 2. 4UWY > 4VWX AAS Congruence Theorem 3. WU > WV CPCF Theorem 4. 4WUV is isosceles. definition of isosceles triangle 11. Since AB 5 AC, AD 5 AD, and /ADB and /ADC are both right angles, 4ADB > 4ADC by the HL Congruence Theorem. Thus, BD 5 CD by the CPCF Theorem. 12. ABCD is a parallelogram because one pair of opposite sides is parallel and congruent. 13. OR 5 PO 5 x LR 5 AP 5 y PR 5 2 ? PO 5 2x 14. Yes, ABCD is a parallelogram because the diagonals bisect each other. 15. Sample (reduced size): 16. (f ) Y Z; m/XWY 5 180 2 (34 1 97) 5 180 2 131 5 49, so the shortest segment in 4WXY is WY . m/WY Z 5 180 2 97 5 83, and m/ZWY 5 180 2 (83 1 55) 5 180 2 138 5 42, so the shortest segment in 4WZY is Y Z. So Y Z must be the shortest segment in the figure. CHAPTER 7 REVIEW pp. 428–432 1. a. No b. There are many noncongruent triangles that fit the given information. 2. a. Yes b. SSS Congruence Theorem 3. a. Yes b. ASA Congruence Theorem 4. a. Yes b. AAS Congruence Theorem 5. a. Sample: O O N M 5 cm b. No. There are many noncongruent triangles that fit the given information. 6. a. Sample: Q 1" P 1" P 1.5" 30° R 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 SsA Congruence Theorem. 116 UCSMP Geometry © Scott, Foresman and Company 8. a. Sample: W 2" 55°1" V X b. Yes, by the SAS Congruence Theorem. 9. a. m/QSR 5 180 2 m/QST 5 180 2 132 5 48 b. m/Q , 132, by the Exterior Angle Inequality c. m/Q 1 m/R 5 m/QST 5 132, by the Exterior Angle Theorem 10. a. q 1 2q 5 138 3q 5 138 q 5 46 m/X 5 q 5 46 b. m/Y 5 2q 5 2(46) 5 92 11. /1 is the largest. /1 is an exterior angle of 4TWX, 4TWY, and 4TWZ. Therefore, m/1 . m/2, m/1 . m/3, and m/1 . m/4. 12. m/2 5 360 2 (m/A 1 m/B 1 m/C) 5 360 2 (110 1 100 1 90) 5 360 2 300 5 60 m/1 5 180 2 m/2 5 180 2 60 5 120 13. 4MOP > 4MNP by the AAS Congruence Theorem 14. not enough information to tell 15. 4KLM > 4KJM by HL Congruence Theorem 16. not enough information to tell 17. 4ACB > 4DFG by AAS Congruence Theorem 18. not enough information to tell 19. Sample: a. AC > DF b. SSS Congruence Theorem 20. Sample: a. HI > KL b. ASA Congruence Theorem UCSMP Geometry © Scott, Foresman and Company 21. Given: /DAC > /BAC; /DCA > /BCA To prove: 4ADC > 4ABC Argument: Conclusions Justifications 0. /DAC > /BAC; Given 0. /DCA > /BCA 1. AC > AC Reflexive Property of Congruence 2. 4ADC > 4ABC ASA Congruence Theorem 22. Given: WZ ⊥ ZY ; WX ⊥ XY ; WZ 5 WX To prove: 4WZY > 4WXY Argument: Conclusions Justifications 0. WZ ⊥ ZY ; WX ⊥ XY ; Given 0. WZ 5 WX 1. /Z and /X are definition of 0. right angles. perpendicular 2. WY > WY Reflexive Property of Congruence 3. 4WZY > 4WXY HL Congruence Theorem 23. Given: (P and (Q intersect at A and B. To prove: 4APQ > 4BPQ Argument: Conclusions Justifications 0. (P and (Q Given 0. intersect at A and B. 1. PA 5 PB; QA 5 QB definition of circle 2. PQ 5 PQ Reflexive Property of Equality 3. 4APQ > 4BPQ SSS Congruence Theorem ####$ 24. Given: UW bisects /Y UV; UW > UY ; /V > /UXY To prove: 4UVW > 4UXY Argument: Conclusions Justifications ####$ bisects /Y UV; 0. UW Given 0. UW > UY ; 0. /V > /UXY 1. /Y UX > /WUV definition of angle bisector 2. 4UVW > 4UXY AAS Congruence Theorem 117 25. Cannot be proved. This is the SSA condition, but the longer congruent sides are not opposite the congruent angles. So the SsA Congruence Theorem does not apply. 26. Given: AB > DC; /ABC > /DCB To prove: 4ACB > 4DBC Argument: Conclusions Justifications 0. AB > DC; Given 0. /ABC > /DCB 1. BC > BC Reflexive Property of Congruence 2. 4ACB > 4DBC SAS Congruence Theorem 27. Given: AB > AC; BD > DC To prove: /BAD > /CAD Argument: Conclusions Justifications 0. AB > AC; BD > DC Given 1. AD > AD Reflexive Property of Congruence 2. 4ABD > 4ACD SSS Congruence Theorem 3. /BAD > /CAD CPCF Theorem 28. Given: ABCDEFGH is a regular octagon. To prove: AC > BD Argument: Conclusions Justifications 0. ABCDEFGH is a Given 0. regular octagon. 1. AB > BC > CD definition of regular polygon 2. /ABC > /DCB definition of regular polygon 3. 4ABC > 4DCB SAS Congruence Theorem 4. AC > BD CPCF Theorem ##$ bisects /MJL; MJ > LJ 29. Given: JK To prove: /M > /L Argument: Conclusions Justifications ##$ 0. JK bisects /MJL; Given 0. MJ > LJ 1. /KJM > /KJL definition of angle bisector 2. JK > JK Reflexive Property of Congruence 118 3. 4KJM > 4KJL SAS Congruence Theorem 4. /M > /L CPCF Theorem 30. Given: AD ⊥ DC; AB ⊥ BC; AD > BC To prove: AB > CD Argument: Conclusions Justifications 0. AD ⊥ DC; AB ⊥ BC; Given 0. AD > BC 1. /D and /B are definition of 0. right angles. perpendicular lines 2. AC > AC Reflexive Property of Congruence 3. 4ADC > 4CBA HL Congruence Theorem 4. AB > CD CPCF Theorem 31. Given: N is the midpoint of OE; l k m To prove: AE > UO Argument: Conclusions Justifications 0. N is the midpoint Given 0. of OE; l k m 1. ON > NE definition of midpoint 2. /NUO > /NAE; k lines ⇒ AIA > 0. /NOU > /NEA Theorem 3. 4NUO > 4NAE AAS Congruence Theorem 4. AE > UO CPCF Theorem 32. Given: /SPQ > /RQP; /S > /R To prove: QS > PR Argument: Conclusions Justifications 0. /SPQ > /RQP; Given 0. /S > /R 1. PQ > PQ Reflexive Property of Congruence 2. 4SPQ > 4RQP AAS Congruence Theorem 3. QS > PR CPCF Theorem 33. DE 5 BE 5 12 AD 5 BC 5 19 BD 5 2 ? BE 5 2 ? 12 5 24 34. AE 5 CE 5 3x AD 5 BC 5 y 2 2 AC 5 2 ? CE 5 2 ? 3x 5 6x UCSMP Geometry © Scott, Foresman and Company 35. m/DCB 5 m/DAB 5 130 m/ADC 5 180 2 m/DAB 5 180 2 130 5 50 m/ABC 5 m/ADC 5 50 36. Z W 49. Sample: C 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, ABCD could be an isosceles trapezoid with bases AB and CD, and AB Þ CD. 40. Yes, since both pairs of opposite angles are congruent. 41. /A is the largest angle, since BC is the longest side. 42. /C is the smallest angle, since AB is the shortest side. 43. m/F 5 180 2 (m/D 1 m/E) 5 180 2 (47 1 68) 5 180 2 115 5 65 sides from shortest to longest: EF, ED, DF 44. In 4GHJ, HJ is the shortest side. But HJ is also in 4HJI, where HI is the shortest side. So the shortest segment in the figure is HI. 45. 4ARP > 4ABP by the HL Congruence Theorem. So, by the CPCF Theorem, RP 5 PB. 46. The SSS Congruence Theorem states that triangles with all three sides congruent are congruent. 47. 4XBD > 4XBA > 4XBC by the AAS Congruence Theorem, so DB 5 AB 5 CB by the CPCF Theorem. 48. a. QR and ZP have lengths equal to the width of the river. b. 4ZXP > 4RXQ by the ASA Congruence Theorem, so QR > ZP by the CPCF Theorem. UCSMP Geometry © Scott, Foresman and Company 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. ABCD is approximately in the shape of a parallelogram. 54. Let their intersection be the midpoint of both PQ and MN. 119 55. AB 5 CD, and CD 5 EF by the Properties of a Parallelogram Theorem (opposite sides are congruent). Therefore, AB 5 EF by the Transitive Property of Equality. 56. Yes; by the Properties of a Parallelogram Theorem (opposite angles are congruent). CHAPTER 7 REFRESHER p. 433 1. 6(2 1 x) 5 12 1 6x 2. x(4 2 y) 5 4x 2 xy 3. (h 1 l)a 5 ah 1 al 1 2 1 2 1 2 4. b(w 1 t) 5 bw 1 bt 5. (a 2 2b)b 5 ab 2 2b2 6. 2xy(y 1 x) 5 2xy2 1 2x2y 7. ax 1 2x 5 x(a 1 2) 8. 9c2 2 12c 5 3c(3c 2 4) 9. 4b 2 20h 5 4(b 2 5h) 10. pr 2 1 ph2 5 p(r 2 1 h2) 1 2 1 2 1 2 11. bh 1 lh 5 h(b 1 l) 12. 12ar 2 1 6ar 5 6ar(2r 1 1) 13. (y 1 2)(y 1 4) 5 y2 1 2y 1 4y 1 8 5 y2 1 6y 1 8 14. (a 2 7)(3b 1 2) 5 3ab 2 21b 1 2a 2 14 15. (2r 1 t)(2r 2 t) 5 4r 2 1 2tr 2 2rt 2 t2 5 4r 2 2 t2 16. (s 1 4)2 5 (s 1 4)(s 1 4) 5 s2 1 4s 1 4s 1 16 5 s2 1 8s 1 16 2 17. (m 1 n) 5 (m 1 n)(m 1 n) 5 m2 1 mn 1 mn 1 n2 5 m2 1 2mn 1 n2 120 1 1 2 2 1 1 1 1 ey 1 eh 1 f y 1 f h 2 2 2 2 18. (e 1 f )(y 1 h) 5 (ey 1 eh 1 f y 1 f h) 5 p 19. 2 < 1.41 p 20. 4 3 < 6.93 p p p 21. 42 1 52 5 16 1 25 5 41 < 6.40 22. s 17 < 2.06 4 p 23. 100 5 10 p p p 24. 40 5 4 ? 10 5 2 10 p p p 25. 18 5 9 ? 2 5 3 2 p p p p 26. 2 45 5 2 9 ? 5 5 2 ? 3 5 5 6 5 27. x2 5 25 x 5 65 28. y2 1 9 5 25 y2 5 16 y 5 64 29. 40 5 2z2 2z2 5 40 z2 5 20p p z 5 6 20 5 62 5 < 64.47 30. 3 5 pr 2 pr 2 5 3 r2 5 3 p r56 s 3 < 60.98 p UCSMP Geometry © Scott, Foresman and Company