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1 Answers for Lesson 5-, pp Exercises UW TX; UY VX; YW TV. GJ FK; JL HF; GL HK 3. a. ST PR; SU QR; UT PQ b. 4. FE 5. FG EG 8. AC a. 050 ft b ft. a. 4 ft 9 in. b. Answers may vary. Sample: The highlighted segment is the midsegment of the triangular face of the building a. H(, 0; J(4, b. Slope of HJ ; slope of EF 4 ; therefore HJ. EF c. HJ " "8 "; EF "4 4 "3 4"; therefore HJ EF. m/qpr C x 6; y x 3; DF x 9; EC 6 4 AB CB Geometry Chapter 5 07

2 Answers for Lesson 5-, pp Exercises (cont. 37. Answers may vary. Sample: Draw CA and extend CA to P so that CA AP. Find B, the midpt. of PD. Then, by the Midsegment Thm., AB CD and AB CD. 38. G(4, 4; H(0, ; J(8, UTS; Proofs may vary. Sample: VS SY, YT TZ, and VU UZ because S, T, and U are midpts. of the respective sides; ST VZ so ST VU UZ; SU YZ so SU YT TZ; and TU VY so TU SY SV; therefore YST TUZ SVU UTS by SSS. Geometry Chapter 5 08

3 Answers for Lesson 5-, pp Exercises. AC is the # bis. of BD The set of points equidistant from H and S is the # bis. of HS. 6. x ; JK 7; JM 7 7. y 3; ST 5; TU ; 7 9. is the bis. of KHF because HL a point on HL is equidistant from HK and HF ; Isosceles; it has sides. 6. equidistant; RT RZ 7. A point is on the # bis. of a segment if and only if it is equidistant from the endpts. of the segment isosceles; CS CT and CT CY by the Bis. Thm. 7. Answers may vary. Sample: The student needs to know that QS bisects PR. 8. No; A is not equidistant from the sides of X. 9. Yes; AX bis. TXR. 30. Yes; A is equidistant from the sides of X. Geometry Chapter 5 09

4 Answers for Lesson 5-, pp Exercises (cont. 3. the pitcher s plate 3. a. C E D b. The bisectors intersect at the same point. c. Check students work. 33. a. b. The # bisectors intersect at the same point. c. Check students work Answers may vary. Samples are given. 34. C(0,, D(, ; AC BC, AD BD "5 35. C(3,, D(3, 0; AC BC 3, AD BD "3 36. C(3, 0, D(0, 0; AC BC 3, AD BD 3" 37. C(0, 0, D(, ; AC BC 3, AD BD "5 38. C(,, D(4, 3; AC BC "5, AD BD "0 5 R "6 39. C,,D(5, 3; AC BC, AD BD "3 Q 5 P R Q Geometry Chapter 5 0

5 Answers for Lesson 5-, pp Exercises (cont. * CD 40. AC BC by definition of bisector. # AB, so DCA and DCB are right '. Therefore, DCA DCB because all rt. ' are. DC DC by the Reflexive Property of Congruence. Therefore, CDA CDB by Side-Angle- Side. DA DB because CPCTC, so DA DB. 4. ABP and ABQ are right triangles with a common leg and congruent hypotenuses. Thus, BAP BAQ by the HL Theorem. PB BQ using CPCTC, so AB bisects PQ by the definition of bisector. Hence, AB is the perpendicular bisector of. 4. a. /:y x ; m:x 0 b. (0, PQ 5 c. CA CB 5 d. C is equidist. from and. 43. # OA OB BP AB and PC # AC, thus ABP and ACP are rt.. Since AP bisects BAC, BAP CAP. AP AP by the Reflexive Prop. of. Thus ABP ACP by AAS and PB PC by CPCTC. Therefore, PB PC SP # QP ; SR # QR. Given. QPS and QRS are rt. '.. Def. of # 3. QPS QRS 3. All rt. ' are. 4. SP SR 4. Given 5. QS QS 5. Refl. Prop. of 6. QPS QRS 6. HL 7. PQS RQS 7. CPCTC 8. QS bisects PQR. 8. Def. of bis Geometry Chapter 5

6 Answers for Lesson 5-, pp Exercises (cont. 45. D 46. y 47. y (x 48. y x Line / through the midpoints of sides of ABC is equidistant from A, B, and C. This is because and 3 4 by ASA. C A 3 4 B 50. a. A point on the # bis. of a segment is equidistant from endpoints of the segment (# Bis. Thm., so MA MB and MB MC. b. MA MB MC by part a. EMA, EMB, and EMC are rt. ' by def. of line # plane (page 49, Exercise 36. MG MG by the Refl. Prop. of, so EAM EBM ECM by SAS. Geometry Chapter 5

7 Answers for Lesson 5-3, pp Exercises. (, 3. (0, Q, R Q, R 5. Q3, R 6. Q3, 4 R 7. Q3, 3R 8. C 9. Z 0. Find the # bisectors of the sides of the formed by the tennis court, the playground, and the volleyball court. That point will be equidistant from the vertices of the.. TY 8; TW 7. ZY 4 ;ZU 3 3. VY 6; YX 3 4. Median; A is a midpt. 5. Neither; it s not a segment drawn from a vertex. 6. Altitude; AB is a segment drawn from a vertex of a perp. to the opp. side E D T F U S Geometry Chapter 5 3

8 Answers for Lesson 5-3, pp Exercises (cont. 9. BE 0. FC. CA. DG 3. : or : 4. Find the circumcenter of the formed by the three pines Check students work. 7. D 8. a. bisector; it bisects an. 9. a. b. None of these; it is a midsegment. c. Altitude; AB is # to a side from a vertex. b. c. XC d. # bis. 30. It is given that X is on line / and line m. By the Bisect. Thm., XD XE and XE XF. By the Trans. Prop. of, XD XE XF. X is on ray n by the Conv. of the Bis. Thm. 3. A right triangle; check students explanations. 3. a. L(, 3; M(5, 3; N(4, 0 b. : y x; : y 3x ; : y x c. AB BC * AM Q 0 3, R 3 5 * BN d. 3 7 Q0 3 R e. AM "34; AP 36 "34; BN "40 "0; Ä 9 3 BP 60 4 "0; CL "58; CP 3 3 3"58 Ä 9 Ä * CL Geometry Chapter 5 4

9 Answers for Lesson 5-3, pp Exercises (cont. 33. I-D; II-B; III-C; IV-A 34. I-A; II-C; III-B; IV-D 35. Answers may vary. Sample: Let ABC be isosc. with base B and C. If AD bisects A, then it is # to BC, and * therefore the altitude from A. So, AD contains the circumcenter, incenter, centroid, and orthocenter. 36. bisectors Geometry Chapter 5 5

10 Answers for Lesson 5-4, pp Exercises. Two angles are not congruent.. You are sixteen years old. 3. The angle is obtuse. 4. The soccer game is not on Friday. 5. The figure is not a triangle. 6. m A a. If you don t eat all of your vegetables, then you won t grow. b. If you won t grow, then you don t eat all of your vegetables. 8. a. If a figure is not a square, then at least one of its angles is not a right angle. b. If at least one of the angles is not a right angle, then the figure is not a square. 9. a. If a figure isn t a rectangle, then it doesn t have four sides. b. If a figure doesn t have four sides, then it isn t a rectangle. 0. Assume that it is not raining outside.. Assume that J is a right angle.. Assume that PEN is not isosceles. 3. Assume that none of the angles is obtuse. 4. Assume that XY AB. 5. Assume that m I and II 7. I and II 8. I and III 9. II and III Geometry Chapter 5 6

11 Answers for Lesson 5-4, pp Exercises (cont. 0. a. 0 or more b. the Debate Club and the Chess Club have fewer than 0 members c. the Debate Club has fewer than 0 members. a. right angle b. right angles c. 90 d. 80 e. 90 f. 90 g. 0 h. more than one right angle i. at most one right angle. Assume A B. Then BC AC since if the base are, the sides opp. them are. But this contradicts the given BC AC. Thus A B. 3. Assume one base is a right. Then the other base is also a right since the base of an isosceles are congruent. But a can have at most one right. So neither base is a right. 4. a. If you don t live in El Paso, then you don t live in Texas; false b. If you don t live in Texas, then you don t live in El Paso; true Geometry Chapter 5 7

12 Answers for Lesson 5-4, pp Exercises (cont. 5. a. If four points aren t collinear, then they aren t coplanar; false b. If four points aren t coplanar, then they aren t collinear; true 6 9. Answers may vary. Samples are given. 6. If a figure is a square, then it has four right angles. 7. If today is Sunday, then tomorrow is Monday. 8. Not possible; a conditional and its contrapositive have the same truth value. 9. If two sides of a triangle are congruent, then the triangle is isosceles. 30. Assume that the driver did not apply the brakes. Then there would be no skid marks. This contradicts the fact that fresh skid marks appear. Thus the green car applied the brakes is a true statement. 3. Assume that the temperature outside is more than 3 F. Then ice would not be forming on the sidewalk. This contradicts the fact that ice is forming. Thus the statement that the temperature must be 3 F or less is true. 3. Assume that an obtuse triangle can contain a right angle. Then the sum of the measures of the obtuse angle and the right angle is more than 80. This contradicts the fact that the sum of the 3 angles of a triangle is 80. Thus the statement that an obtuse triangle cannot contain a right angle is true. * XY * XZ * AX * * AX AX 33. Assume and are two different lines # to, with Y and Z on the same side of. If B is on opp. pt. A from X, then m AXY + m YXZ + m ZXB 80. But m AXY m ZXB 90, so m YXZ 0. Thus X,Y, and Z are collinear. Geometry Chapter 5 8

13 Answers for Lesson 5-4, pp Exercises (cont. 34. If the animal is a kitten, then it is a cat. If the animal isn t a cat, then it s not a kitten. 35. If the angle measures 0, then it is obtuse. If the angle isn t obtuse, then it doesn t measure If a number is a whole number, then it is an integer. If a number isn t an integer, then it isn t a whole number. 37. Angie assumed that the inverse of the statement was true, but a conditional and its inverse may not have the same truth value. 38. a. Earl proves that it s later than 5:00. b. He starts with the assumption that it is before 5:00. c. It is not noisy. 39. The culprit entered the room through a hole in the roof; the other possibilities were eliminated. 40. Check students work. 4. Assume XB # AC. Then AXB and CXB are right. Since m ABX m CBX 36, then A C because if two of a are, the third are. Then AB BC since sides opp. are and ABC is an isosceles. But this contradicts the given statement that ABC is scalene. Thus, XB is not # to AC. Geometry Chapter 5 9

14 Answers for Lesson 5-5, pp Exercises. 3 because they are vertical and m m 3 by Corollary to the Ext. Thm. So, m m by subst.. An ext. of a is larger than either remote int.. 3. m m 4 by Corollary to the Ext. Thm. and 4 because if lines, then alt. int. are. 4. M, L, K 5. D, C, E 6. G, H, I 7. A, B, C 8. E, F, D 9. Z, X, Y 0. MN, ON, MO. FH, GF, GH. TU, UV, TV 3. AC, AB, CB 4. EF, DE, DF 5. ZY, XZ, XY 6. No; 3 w Yes; 5; 5 ; No; 8 0 w Yes; 5 5; Yes; 9 0; 9 0 ; No; 4 5 w s 0 3. s 4. 0 s 5. 5 s s 7. 5 s Answers may vary. Sample: If Y is the distance between Wichita and Topeka, then 0 Y Let the distance between the peaks be d and the distances from the hiker to each of the peaks be a and b. Then d a b and d b a. Thus, d b a and d a b. Geometry Chapter 5 0

15 Answers for Lesson 5-5, pp Exercises (cont. 30. a. A D B C E F b. The third side of the st is longer than the third side of the nd. c. See diagram in part (a. d. The included of the first is greater than the included of the second. 3. Answers may vary. Sample: The shortcut across the grass is shorter than the sum of the two paths. 3. AB 33. a. m OTY b. m 3 c. Base of an isosc. are. d. Add. Post. e. Comparison Prop. of Ineq. f. Subst. (step g. An ext. of a is greater than either remote int.. h. Trans. Prop. of Ineq. 34. T is the largest in PTA. Thus PA PT because the longest side of a is opp. the largest. 35. RS 36. CD 37. XY (, 4, (, 5, (, 6, (3, 3, (3, 4, (3, 6, (3, 7, (4, 3, (4, 4, (4, 5, (4, 6, (4, 7, (4, Geometry Chapter 5

16 Answers for Lesson 5-5, pp Exercises (cont. 4. D C A B CD AC was given so ACD is isos. by def. of isos.. This means m D m CAD. Then m DAB m CAD by the Comparison Prop. of Ineq. So by subst., m DAB m D and by Thm. 5- DB AB. Since DC CB DB,by subst. DC CB AB. Using subst. again, AC CB AB. Geometry Chapter 5

Cumulative Test. 101 Holt Geometry. Name Date Class

Cumulative Test. 101 Holt Geometry. Name Date Class Choose the best answer. 1. Which of PQ and QR contains P? A PQ only B QR only C Both D Neither. K is between J and L. JK 3x, and KL x 1. If JL 16, what is JK? F 7 H 9 G 8 J 13 3. SU bisects RST. If mrst