LESSON 10.3 Answers for the lesson Apply Properties of Chords Copyright Houghton Mifflin Harcourt Publishing Company. All rights reserved. Skill Practice 1. Sample answer: Point Y bisects C XZ if C XY > C YZ. 2. No. Sample answer: Inscribe a square in a circle using the intersection of the square s diagonals as the center of the circle. 3. 758 4. 1168 5. 8 6. 7; use Theorem 10.5 and solve 4x 5 3x 1 7. 7. 5; use Theorem 10.5 and solve 5x 2 6 5 2x 1 9. 8. 11; since SQ and TQ are radii they have the same measure. Solve 6x 1 9 5 8x 2 13. 9. 5; use Theorem 10.6 and solve 18 5 5x 2 7. 10. 3; use Theorem 10.6 and 10.5 and solve 6x 1 4 5 22. 11. 7 ; use Theorem 10.6 and 3 solve 4x 1 1 5 x 1 8. 12. AB is a diameter of the circle; Theorem 10.4. 13. JH bisects FG and C FG ; Theorem 10.5. 14. NP > LM ; Theorem 10.6. 15. D 16. CD should equal EF using Theorem 10.6. 17. You don t know that AC DB therefore you can t show C BC > C CD. 18. Diameter; use Theorem 10.4. 19. Diameter; the two triangles are congruent by the SAS Congruence Postulate which makes AB the perpendicular bisector of CD. Use Theorem 10.4. 20. Not a diameter; if AB were a diameter CE would equal DE. 21. Using the facts that napb is equilateral which makes it equiangular and that m C AC 5 308, you can conclude that m APD 5 m BPD 5 308. You now know that m C BC 5 308 which makes AC > BC. napd > nbpd by the SAS Congruence Postulate since BP > AP and PD > PD. Because corresponding parts of congruent triangles are congruent, AD > BD. Along with DC > DC you have nadc > nbdc by the SSS Congruence Postulate. Answer Transparencies for Checking Homework 301
22. a. If the diameter of a circle bisects a chord of the circle and its arc, then the diameter is perpendicular to the chord; in the converse of Theorem 10.5 you are given the diameter, while in Theorem 10.4 you are to prove that the first chord is the diameter. b. P S R T C Q PC > RC since they are both radii of the same circle. CT > CT by the Reflexive Property of Congruence. ncpt > ncrt by the SSS Congruence Postulate. Since RTP is a straight angle and m RTC 5 m PTC, both angles must be 908,which makes SQ PR. c. Since SQ is a diameter of (C and it bisects both PR and C PR, SQ is a perpendicular bisector of PR. Therefore, any point on SQ is equidistant from the endpoints of the segment, so QP 5 QR. 23. From the diagram m C AC 5 m C CB and m C AB 5 x8, so you know that m C AC 1 m C CB 1 x8 5 3608. Replacing m C CB by m C AC and solving for m C AC, you get m C 3608 2 x8 AC 5. This along 2 with the fact that all arcs have integer measures implies that x is even. 24. about 16.38 Problem Solving 25. AB should be congruent to BC. 26. (1) Construct the perpendicular bisector of the control panel, extend this segment to the other control panel, and then find the midpoint of the segment. (2) Draw the 2 diagonals from the top of one control panel to the bottom of the other. The center will be at their intersection. Answer Transparencies for Checking Homework 302
27. Given AB > CD. Since PA, PB, PC, and PD are radii of (P, they are congruent. By the SSS Congruence Postulate, npcd > npab. Because corresponding parts of congruent triangles are congruent, CPD > APB. With m CPD 5 m APB and the fact they are both central angles, you now have m C CD 5 m C AB which leads to C CD > C AB. 28. Given: AB and CD are chords and C AB > C CD. From this m C AB 5 m C CD which implies m APB 5 m CPD which implies APB > CPD. PA, PB, PC, and PD are radii of (P, so they are congruent. Using the SAS Congruence Postulate, napb > ndpc. AB > CD. 29. a. A longer chord B C D b. The length of a chord in a circle increases as the distance from the center of the circle to the chord decreases. c. Given: Radius r and real numbers a and b such that r > a > b > 0. Let a be the distance from one chord to the center of the circle and b be the distance from a second chord to the center of the circle. Using the Pythagorean Theorem, the length of the chord a units away from the center is 2 Ï r 2 2 a 2 and the length of the chord b units away from the center is 30. a. 200 ft 2 Ï r 2 2 b 2. Using properties of real numbers, Ï r 2 2 b 2 > Ï r 2 2 a 2. b. about 45.7 mi/h Answer Transparencies for Checking Homework 303
31. Given: QS is the perpendicular bisector of RT in (L. Suppose center L is not on QS. Since LT and LR are radii of the circle they are congruent. With PL > PL, nrlp > ntlp by the SSS Congruence Postulate. RPL and TPL are congruent and they form a linear pair. This makes them right angles and leads to PL being perpendicular to RT. By the Perpendicular Postulate, L must be on QS (which is a contradiction). Thus QS must be a diameter. 32. Given: EG is a diameter of (L and EG DF. Since all radii are congruent, LD > LF. Since EG DF, ndlc and nflc are right triangles. Since LC > LC, ndcl > nflc by the HL Congruence Theorem. CD > CF and DLC > FLC. With m DLC 5 m FLC, we now have m DLC 5 m C DG and m FLC 5 m C FG. This leads to C DG > C FG. 33. Case 1: In the same circle or in congruent circles, if two chords are equidistant from the center of the circle, the chords are congruent. Given: EF 5 EG, EF AB, and EG CD, then EF > EG. Draw radii EA, EB, EC, and ED. Since all radii are congruent and neaf, nebf, necg, and nedg are right triangles, then neaf > nebf > necg > nedg by the HL Congruence Theorem. Using corresponding parts of congruent triangles are congruent, AF > FB > CG > GD. This implies that all four segments have the same length. Using the Segment Addition Postulate, AF 1 FB 5 AB and CG 1 GD 5 CD. From this you can use substitution to conclude that AB 5 CD, which implies AB > CD. Case 2: In the same circle or in congruent circles, if two chords are congruent, the distance from the center to each chord is the same. Given: Congruent chords CD and AB in (E. Draw radii EA, EB, EC, and ED. Since all radii are congruent, necd > neab by the SSS Congruence Postulate. Answer Transparencies for Checking Homework 304
33. (cont.) D > A. EF AB and EG CD, so EFA and EGD are right angles and are congruent. Then neaf > nedg using the AAS Congruence Theorem. EF > EG. This implies that EF 5 EG. 34. Given: C AB with the bottom panel of the car parallel to the ground. Let C be the point where the tire touches the ground. A line on the ground parallel to the panel passing through C is tangent to the wheel. Create a line segment perpendicular to the line at C passing through the center of the wheel and intersecting the opposite side of the wheel, creating a diameter. The diameter is perpendicular to the bottom panel. Since the bottom of the panel is also a chord of the wheel, Theorem 10.5 can be used to show the diameter bisects the intercepted arc of the chord. Thus C bisects C AB. Answer Transparencies for Checking Homework 305