126 Holt McDougal Geometry

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1 test prep 51. m Q = m S 3x + 5 = 5x = x x = J P = = = (5 + 8.) = 6.4 challenge and extend 54. Let given pts. be (0, 5), (4, 0), (8, 5), and possible 4th pts. be X, Y, Z. is horiz., and = 8-0 = 8. So X is horiz., and X = (4-8, 0) = (-4, 0). Similarly, Y is horiz., and Y = (4 + 8, 0) = (1, 0). rom to is rise of 5 and run of 4; rise of 5 from is = 10, run of 4 from is = 4; so Z = (4, 10). 55. Let given pts. be (-, 1), (3, -1), (-1, -4), and possible 4th pts. be X, Y, Z. rom to is rise of 3 and run of 4; rise of 3 from is = 4, run of 4 from is =, so X = (, 4). rom to is rise of -3 and run of -4; rise of -3 from is 1-3 = -, run of -4 from is = -6, so Y = (-6, -). rom to is rise of - and run of 5; rise of - from is -4 - = -6, run of 5 from is = 4; so Z = (4, -6) Let 1 = x and = y. raw È. and are, so ǁ and ǁ by def. So 1 and 3 4 by the lt. Int. Thm. Thus m 1 = m and m 3 = m 4. Then m 1 + m 3 = m + m 4 by the dd. Prop. of =. y the dd. Post., m + m 4 = m. So m 1 + m 3 =. Since and are, with 1 corr. to 3, m 1 = m 3. So m 1 + m 1 = m by subst. So m 1 = m, or y = x Given: is a. ÈÈ bisects. ÈÈ bisects. Prove: ÈÈ ÈÈ Statements Reasons 1. is a. 1. Given ÈÈ bisects. ÈÈ bisects.. 1, 3 4. ef. of bisector 3. ǁ 3. ef. of lt. Int. Thm Trans. Prop. of Reflex. Prop. of S PT 9. 5 and 6 are supp. 9. Lin. Pair Thm and 6 are rt supp. rt. 11. ÈÈ ÈÈ 11. ef. of 6-3 conditions for parallelograms check it out! 1. Think: Show that PQ and RS. are ǁ and. a =.4 PQ = 7(.4) = 16.8, RS = (.4) + 1 = 16.8, so PQ RS b = 9 m Q = 10(9) - 16 = 74, m R = 9(9) + 5 = 106 m Q + m R = 180, so Q and R are supp. y onv. of lt. Int. Thm., PQ ǁ RS. So PQRS is a by Thm since 1 pair of opposite sides is ǁ and. a. Yes; possible answer: the diag. of the quad. forms with pairs of. y 3rd Thm., 3rd pair of in are. So both pairs of opp. of the quad. are, therefore b. No; pairs of cons. sides are, but none of the sets of conditions for a are met. 3. Possible answers: ind slopes of both pairs of opp. sides. slope of KL = 7-0 = slope of MN = = slope of LM = 5-7 = slope of KN = = Since both pairs of opp. sides are ǁ, KLMN is a by definition. 4. Possible answer: Since both pairs of opposite sides are congruent, RS is a. Since is vert. and RS ǁ, RS is vert., so of binoculars stays the same. think and discuss 1. Possible answer: onclusion of each thm. is The quad. is a.. Possible answer: quad. is a, is the hypothesis of each thm., rather than the conclusion. 16 Holt Mcougal Geometry

2 3. Quad. with both pairs of opp. sides exercises Quad. with 1 pair of opp. sides and Quad. with both pairs of opp. sides onditions for Parallelograms Quad. with 1 supp. to both of its consec. (180 - x) x x (180 - x) Quad. with both pairs of opp. Quad. with diags. bisecting each other guided practice 1. Step 1 ind J and JG. J = t + 1 JG = 3t J = = 18 JG = 3(6) = 18 Step ind J and JH. J = s JG = s + 5 J = (5) = 10 JG = = 10 Since J = JG and J = JH, GH is a since its diagonals bisect each other.. L = 5m + 36 L = 5(14) + 36 = 106 P = 6n - 1 P = 6(1.5) - 1 = 74 Q = 4m + 50 Q = 4(14) + 50 = 106 Since = 180, P is supp. to both L and Q. KLPQ is a since an angle is supp. to both its cons. 3. Yes; both pairs of opp. of the quad. are, so 4. No; 1 pair of opp. sides of quad. are. 1 diag. is bisected by other diag. None of the conditions for a are met. 5. Yes; possible answer: a pair of alt. int. are, so 1 pair of opp. sides are ǁ. The same pair of opp. sides are, so quad. is a. 6. Possible answer: ind slopes of both pairs of opp. sides. slope of WX = = 5 slope of YZ = = 5 slope of XY = = 1 3 slope WZ = = 1 3 Since both pairs of opp. sides are ǁ, WXYZ is a by definition. 7. Possible answer: ind slopes of both pairs of opp. sides. slope of RS = = -4 slope of TU = = -4 slope of ST = = 0 slope of RU = = 0 Since both pairs of opp. sides are ǁ, RSTU is a by def. 8. Since ǁ and, is a by Thm , so ǁ by def. of. practice and problem solving 9. = 3(3.) + 7 = 16.6, GH = 8(3.) - 9 = 16.6 H = 3(7) + 7 = 8, G = 6(7) - 14 = 8 GH is a since both pairs of opp. sides are. 10. UV = 10(19.5) - 6 = 189, TW = 8(19.5) + 33 = 189 m V = () + 41 = 85, m W = 7() - 59 = 95 UV TW ; V and W are supp., so by onv. of Same-Side Int. Thm., UV ǁ TW. TUVW is a since one pair of opp. sides are and. 11. Yes; both pairs of opp. sides are, since all sides are, so quad is a. 1. Yes; by dd. Post., 1 pair of opp. are, and by 3rd Thm., nd pair of opp. are. So quad is a since both pairs of opp. angles are. 13. No; by looking at the angles, we can be sure that one pair of sides are ǁ. This is not enough. 14. slope of JK = 7 = - 7 ; slope of LM = -7 = slope of KL = -1 = - 1 ; slope of JM = -1 = both pairs of opp. sides have the same slope, so JK ǁ LM and KL ǁ MJ ; JKLM is a by definition. 15. slope of PQ = 5 ; slope of RS = = 5 3 slope of QR = -6 = -1; slope of PS = = -1. So, PQ ǁ RS and PS ǁ QR. PQRS is a by def. 16. Possible answer: The brackets are always the same length, so it is always true that =. The bolts are always the same dist. apart, so it is always true that =. Since both pairs of opp. sides are, is always a. The side stays horiz. no matter how you move the tray. Since ǁ, stays horiz. Since holds the tray in position, the tray will stay horiz. no matter how it is moved. 17. No; the given measures only indicate that 1 of the quad. is supp. to 1 of its cons.. The other angles can have any measures with sum No; you are only given the measures of the 4 formed by the diags. None of the sets of conditions for a are met. 19. Yes; diags. of the quad. bisect each other. Therefore 17 Holt Mcougal Geometry

3 0. Think: Opp. sides must be. a + 6 = 3a b - 3 = 5(16) = a 6b = 84 b = Think: Middle must be supp. to cons.. 4a a - 10 = 180 1a = 198 a = (16.5) b + 6 = 180 5b = 116 b = 3.. Think: iags. must bisect each other. 5b - 7 = 3b + 6 a = 3(6.5) - 5 b = 13 a = 14.5 b = 6.5 a = Think: 1 pair of opp. sides must be and ǁ. or conditions of onv. of lt. Int. Thm., given must be. 3a = 4a = a 4. Possible answer: If the diags. of a quad. are, you cannot necessarily conclude that 1.4b = b b = 8 b = 0 5. Possible answer: The red and green are isosc. rt., so the measure of each acute of the is 45. ach of the smaller of the yellow stripe is comp. to 1 of the acute of the rt., so the measure of each of the smaller of the yellow stripe is = 45. ach of the larger of the yellow stripe is supp. to 1 of the acute of the rt., so the measure of each of the larger of the yellow stripe is = 135. So the yellow stripe is quad. in which both pairs of opp. are. Therefore the shape of the yellow stripe is a. 6a. Reflex. Prop. of b. c. SSS d. 3 e. f. onv. of lt. Int. Thm. g. def. of 7a. Q b. S c. SP d. RS e. 8. Given: is a, is the mdpt. of, and is the mdpt. of. Prove: and are. Proof: Since is a, ǁ, so ǁ and ǁ. Since opp. sides of a are,. It is given that is the mdpt. of, and is the mdpt. of. ecause these two segs. are, it follows that. Since ǁ and, is a. Similarly, is a. 9. Statements Reasons 1. G, H 1. Given. m = m G, m = m H. ef. of 3. m + m + m G 3. Polygon Sum + m H = 360 Thm. 4. m + m + m 4. Subst. + m = 360, m + m H + m + m H = m + m = 360, 5. istrib. Prop. m + m H = 360 of = 6. m + m = 180, 6. iv. Prop. of = m + m H = is supp. to and H. 7. ef. of supp. 8. ǁ GH, G ǁ H 8. onv. of Same- Side Int. Thm. 9. GH is a. 9. ef. of 30. Statements Reasons 1. JL and KM bisect each 1. Given other.. JN LN, KN MN. ef. of bisect 3. JNK LNM, 3. Vert. Thm. KNL MNJ 4. JNK LNM, 4. SS KNL MNJ 5. JKN LMN, 5. PT KLN MJN 6. JK ǁ LM, KL ǁ MJ 6. onv. of lt Int. Thm. 7. JKLM is a. 7. ef. of 31. Possible answer: Given: and are midsegments of. Prove: is a. Statements Reasons 1. and are 1. Given midsegs. of.. ǁ, ǁ. Midseg. Thm. 3. is a. 3. ef. of 3. Possible answer: quad. is a if and only if both pairs of opp. sides are. quad. is a if and only if both pairs of opp. are. quad. is a if and only if its diags. bisect each other. 18 Holt Mcougal Geometry

4 33. Possible answer: Q P R raw line l. raw P, not on l. raw a line through P that intersects l at Q. onstruct m ǁ to l through P. Place the compass point at Q and mark off a seg. on l. Label the second endpoint of this seg as R. Using the same compass setting, place the compass point at P and mark off a seg. on m. Label the second endpoint of this seg. S. raw RS. Since PS ǁ QR and PS QR, PSRQ is a since one pair of opp. sides are and. 34a. No; none of sets of conditions for a are met. b. Yes; since S and R are supp., PS ǁ QR. Thus PQRS is a since one pair of opp. sides are and. c. Yes; draw PR. QPR SRP (lt. Int. Thm.) and PR PR (Reflex Prop. of ). So QPR SRP (S), and PQ SR (PT). Since PQ ǁ SR and PQ SR, PQRS is a since one pair of opp. sides are and. test prep 35. y onv. of lt. Int. Thm., WX ǁ YZ ; need WX YZ so that one pair of opp. sides are and. 36. G Slope of : rise of 4 and run of, or rise of -4 and run of -; rise of ±4 from is 1 ± 4 = 5 or -3, run of ± from is 6 ± = 8 or 4. So could be at (8, 5) or (4, -3). 37. No; possible answer: slope of RS = 3, slope of 4 TV = 1; RS and TV do not have same slope, so RS TV; RS and TV are opp. sides of RSTV; by def., both pairs of opp. sides of a are ǁ, so RSTV is not a. challenge and extend 38. The top and bottom of each step form a small with the back of the stairs and the base of the railing. The vertices of each have joints that allow the pieces to move. ut the lengths of the sides of stay the same. Since they start out as with opp. sides that are, and the lengths do not change, they remain. Therefore the top and bottom of each step, and thus also the upper platform, remain ǁ to ground regardless of the position of the staircase. S l m 40. Possible answer: raw collinear with and such that. Since is the mdpt. of,. y the Vert. Thm.,. Thus by SS. y PT,. Since is the mdpt. of,. So by the Trans. Prop. of,. lso by PT,. y onv. of lt. Int. Thm., ǁ. Thus is a since 1 pair of opp. sides are ǁ and. Since is a, ǁ by definition. Since opp. sides of a are, and = by the def. of segs. Since, is the mdpt. of, and = 1. y subst., = 1. Ready to go on? Section a Quiz 1. polygon; octagon. not a polygon 3. not a polygon 4. polygon; pentagon 5. (n - )180 (16 - ) (n)m = (n - )180 6m = 70 m = z + 8z + 7z + 11z = z = 360 z = 9 xt. measures are 14(9) = 16, 8(9) = 7, 7(9) = 63, and 11(9) = m(ext. ) = N is mdpt. of KM. m(ext. ) = 36 KM = KN = (13.5) = 7 cm 10. KJ LM 11. MN = KN = 13.5 cm KJ = LM = 17 cm 1. JKL and KJM are supp. m JKL + m KJM = 180 m JKL + 10 = 180 m JKL = JML JKL m JML = m JKL = KLM KJM m KLM = m KJM = slope from to : rise of -5 and run of 1 rise of -5 from is 1-5 = -4; run of 1 from is = -; = (-, -4) 16. WX = YZ 17. YZ = WX = 11 6b - 7 = 10b = 4b b = Let intersection and vertices be P(-, 1.5), (-7, ), (, 6.5), (x, y), and (u, v). P is mdpt. of WX = 6(3) - 7 = 11 and. (-, 1.5) = ( X and W are supp. + x, + y ) m X + m Y = 180 5a a + 7 = 180 x = 3, y = 1; = (3, 1) 8a = 19 (-, 1.5) = ( + u + v, 6.5 ) a = 4 m X = 5(4) - 39 = 81 u = -6, v = -3.5; = (-6, -3.5) 19 Holt Mcougal Geometry

5 15. Think: HG is comp. to GJ and to HG. m HG + m GJ = 90 m HG + m GJ = m GJ = 90 m GJ = 16. m HG = m HJ + m HG = = Think: Use Same-Side Int. Thm., isosc. trap. base. m U + m T = 180 m U + 77 = 180 m U = 103 R U m R = m U = Think: Isosc. trap diags. WY VX WZ + YZ = VX WZ = 53.4 WZ = length of midseg. = 1 (43 + 3) = 33 in. study guide: review 1. vertex of a polygon. convex 3. rhombus 4. base of a trapezoid 6-1 Properties and attributes of polygons 5. not a polygon 6. polygon; triangle 7. polygon; dodecagon 8. irregular; concave 9. irregular; convex 10. regular; convex 11. (n - )180 (1 - ) (n)m(ext. ) = 360 4m(ext. ) = 360 m(ext. ) = (n)m = (n - )180 0m = (18)180 0m = 340 m = m + + m = (6 - )180 8s + 7s + 5s + 8s + 7s + 5s = 70 40s = 70 s = 18 m = m = 8(18) = 144 ; m = m = 7(18) = 16 ; m = m = 5(18) = properties of parallelograms 15. Think: iags. bisect 16. each other. = 1 = = 6.4 = 1 (75) = = = m = m = Think: ons. supp. m + m = m = 180 m = m = m = WX = YZ b + 6 = 5b = 4b 3.5 = b WX = = m W + m X = 180 6a + 14a = 180 0a = 180 a = 9 m W = 6(9) = Y W m Y = m W = 54. YZ = 5(3.5) - 8 = m X = 14(9) = Z X m Z = m X = Slope from R to S is rise of and run of 10; rise of from V to T is -7 + = -5; run of 10 from V to T is = 6; T = (6, -5) 8. Statements Reasons GHLM is a. 1. Given L JMG. G L. opp. 3. G JMG 3. Trans. Prop. of 4. GJ MJ 4. onv. Isosc. Thm. 5. GJM is isosc. 5. ef. of isosc. conditions for parallelograms 9. m = 13 m G = 9(13) = 117 ; n = 7 m = (7) + 9 = 63, m = 3(7) - 18 = 63 Since = 180 G is supp. to and, so one of G is supp. to both of its cons.. G is a by supp. to cons x = 5 m Q = 4(5) + 4 = 104, m R = 3(5) + 1 = 76 ; so Q and R are supp. y = 7 QT = (7) + 11 = 5, RS = 5(7) - 10 = 5 y onv. of Same-Side Int. Thm., QT ǁ RS ; since QT RS, QRST is a by Thm Yes; The diags. bisect each other. y Thm No; y onv. of lt. Int. Thm., one pair of opp. sides is ǁ, but other pair is. None of conditions for a are met. 33. slope of = 10 = 1 ; slope of H = = 1 5 slope of H = -6 = -6; slope of = = -6 oth pairs of opp. sides have the same slope, so ǁ H and H ǁ ; by def., H is a. 6-4 Properties of special parallelograms = = = 18 = (19.8) = Holt Mcougal Geometry

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