mdpt. of TW = ( _ 0 + 1, _ 2 think and discuss Rects.: quads. with 4 rt. exercises guided practice bisect each other TQ = 1_ 2 QS = 1_ (380) = 190 ft

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1 19. m W = 3(4) + 7 = x = 6 RS = 7(6) + 6 = 48, TV = 9(6) - 6 = 48 y = 4.5 RV = 8(4.5) - 8 = 8, ST = 6(4.5) + 1 = 8 RS TV, ST RV RSTV is a (Thm. 6-3-) 1. m = 1 m G = (1) + 31 = 55, m J = 7(1) - 9 = 55 n = 9.5 m K = 1(9.5) + 11 = 15 K supp. to G, and J GHJK is a (Thm ).. Yes; both pairs of opp. sides are ǁ, so quad. is a by definition. 3. No; one pair of opposites of the quad. are. None of the sets of conditions for a are met. 4. No; the diagonals are divided into two segments at their point of intersection, and each segment of one diagonal is to a segment of the other diagonal. None of the sets of conditions for a are met. 5. Slope of CD = 4_ ; slope of EF = _ = 4_ 5 slope of DE = _ - 6 = - 1_ ; slope of FC = _ = - 1_ 3 Both pairs of opp. sides are ǁ, so quad. is a by definition. 6-4 PRoperties of special check it out! 1a. Think: rect. opp. sides HJ GK HJ = GK = 48 in. a. CG = GF 5a = 3a + 17 a = 17 a = 8.5 CD = CG = 5(8.5) = 4.5 b. Think: rect. diags. HK GJ HK = GJ = JL = (30.8) = 61.6 in. b. Think: rhombus cons. supp. m GCD + m CDF = 180 b b - 40 = 180 7b = 17 b = 31 Think: diagonals bisect angles. m GCH = 1_ m GCD = 1_ (31 + 3) = Step 1 Show that SV and TW are. SV = ÇÇÇÇ = ÇÇ 1 TW = ÇÇÇÇ = ÇÇ 1 Since SV = TW, SV TW. Step Show that SV and TW are. slope of SV = 1_ ; slope of TW = _ = -11 Since ( 1_ 11 ) ( -11) = -1, SV and TW are. Step 3 Show that SV and TW bisect mdpt. of SV = ( _-5 + 6, _ -4-3 ) = ( 1_ 7_ ) mdpt. of TW = ( _ 0 + 1, _ - 9 ) = ( 1_ 7_ ) Since SV and TW have same mdpt., they bisect Diags. are bisectors of 4. Possible answer: 1. PQTS is a rhombus. 1. Given. PT bisects QPS.. Thm QPR SPR 3. Def. of bisector 4. PQ PS 4. Def. of rhombus 5. PR PR 5. Reflex. Prop. of 6. QPR SPR 6. SAS 7. RQ RS 7. CPCTC think and discuss 1. If a parallelogram is a rectangle, then its diagonals are. possible answer: when the thm. is written as a conditional statement, it is easier to identify the hypothesis and the conclusion.. Same properties: pairs of ǁ sides, opp. sides, opp., cons. supp., diags. bisect each other; Special properties: 4 sides, diags., each diag. bisects a pair of opp. 3. Quads.: Polygons with 4 Sides : quads. with pairs of sides Rects.: quads. with 4 rt. exercises guided practice Squares: quads. with 4 rt. and 4 sides Rhombuses: quads. with 4 sides 1. rhombus; rectangle; square. rect. diags. 3. PQ RS bisect each other TQ = 1_ PQ = RS = 160 ft QS = 1_ (380) = 190 ft 4. T is mdpt. of QS. 5. rect. diags. are ST = TQ = 190 ft PR QS PR = QS = 380 ft 6. BC = CD 4x + 15 = 7x + 13 = 3x x = 4 1_ 3 AB = BC = 4 ( 4 1_ 3 ) + 15 = 3 1_ Holt McDougal Geometry

2 7. AC BD m AFB = 90 1y = 90 y = 7.5 m ABC + m BCD = 180 m ABC + m FCD = 180 m ABC + (4(7.5) - 1) = 180 m ABC + 58 = 180 m ABC = 1 8. Step 1 Show that JL and KM are. JL = ÇÇÇÇ = ÇÇ 74 KM = ÇÇÇÇ = 74 ÇÇ Since JL = KM, JL KM. Step Show that JL and KM are. slope of JL = 7_ ; slope of KM = _ -5 5_ = Since ( 7_ 5 ) ( 5_ - 7 ) = -1, JL and KM are. Step 3 Show that JL and KM bisect mdpt. of JL = ( _-3 +, _ -5 + ) = ( - 1_ 3_ ) mdpt. of KM = ( _ , _ 1-4 ) = ( - 1_ 3_ ) Since JL and KM have same mdpt., they bisect Diags. are bisectors of 9. Possible answer: 1. RECT is a rect.; RX 1. Given TY. XY XY. Reflex. Prop. of 3. RX = TY, XY = XY 3. Def. of segs. 4. RX + XY = TY + XY 4. Add. Prop. of = 5. RX + XY = RY, 5. Seg. Add. Post. TY + XY = TX 6. RY = TX 6. Subst. 7. RY TX 7. Def. of segs. 8. R and T are rt.. 8. Def. of rect. 9. R T 9. Rt. Thm. 10. RECT is a. 10. Rect. 11. RE CT 11. opp. sides 1. REY TCX 1. SAS practice and problem solving 10. JL = JP 11. KL = JM = 5 in. = (14.5) = 9 in. 1. KM = JL = 9 in. 13. MP = 1_ KM 14. WX = XY 9a - 18 = 3a a = 33 a = 5.5 VW = WX = 9(5.5) - 18 = 31.5 = 1_ (9) = 14 1_ in. 15. m XZW = 90 10b - 5 = 90 10b = 95 b = 9.5 m VWX + m WVY = 180 m VWX + 4(9.5) + 10 = 180 m VWX = 13 m WYX = 1_ m VYX = 1_ m VWX = 1_ (13) = Step 1 Show that PR and QS are. PR = ÇÇÇÇ = ÇÇ 146 QS = ÇÇÇÇ = ÇÇ 146 Since PR = QS, PR QS. Step Show that PR and QS are. slope of PR = _ -5 5_ = - ; slope of QS = _ = 11_ 5 Since ( 5_ - 11 ) ( 11_ 5 ) = -1, PR and QS are. Step 3 Show that PR and QS bisect mdpt. of PR = ( _-4 + 7, _ 0-5 ) = ( 3_ 5_ ) mdpt. of QS = ( _ 4-1, _ 3-8 ) = ( 3_ 5_ ) Since PR and QS have same mdpt., they bisect Diags. are bisectors of 17. Possible answer: 1. RHMB is a rhombus. 1. Given HB is a diag. of RHMB.. MH RH. Def. of rhombus 3. HB bisects RHM. 3. Rhombus each diag. bisects opp. 4. MHX RHX 4. Def. of bisector 5. HX HX 5. Reflex. Prop. of 6. MHX RHX 6. SAS 7. HMX HRX 7. CPCTC 18. m 1 = = 9 (comp. ) m = 61 (Alt. Int. Thm.) m 3 = 90 (def. of rect.) m 4 = m 1 = 9 (Alt. Int. Thm.) m 5 = 90 (def. of rect.) 19. m 1 = = 54 (comp. ) m = 36 (diags. by SSS, by CPCTC) m 3 = 90 - m = 54 (comp. ) m 4 = (m + 36) = 108 ( Sum Thm., Alt. Int. Thm.) m 5 = m 4 = 7 (supp. ) 0. m 1 = 90 (rhombus diag. ) m = m 3, m + m 3 = 90 m = m 3 = 45 (rhombus diag. bisect opp ) m 4 = 45 (same reasoning as, 3) m 5 = m 3 = 45 (rect., Alt. Int. Thm.) 131 Holt McDougal Geometry

3 1. m = 7 (Isosc. Thm.) m 1 = (7 + 7) = 16 ( Sum Thm.) m 3 = m = 7 (Thm ) m 4 = m 1 = 16 (rhombus opp. ) m 5 = 7 (rhombus diag. bisect opp ). m 1 = m, m 1 + m + 70 = 180 m 1 = m = 55 (Isosc. Thm.) m 3 = m = 55 (rhombus diag. bisect opp. ) m 4 = 70 (rhombus opp. ) m 5 = m 1 = 55 (rhombus diag. bisect opp. ) 3. m 1 = 90-6 = 64 (rhombus diag., comp. ) m = m 1 = 64 (rhombus diag. bisect opp. ) m 3 = 6 (rhombus, Alt. Int. Thm.) m 4 = 90 (rhombus diag. ) m 5 = m = 64 (rhombus, Alt. Int. Thm.) 4. always (Thm ) 5. sometimes 6. sometimes 7. sometimes 8. always (all 4 sides ) 9. always (has 4 sides) 30. always (4 rt. ) 31. sometimes 3. No; possible answer: a rhombus with int that measure 70, 110, 70, and 110 is equliateral, but it is not equiangular. A rect. with side lengths 5, 7, 5, and 7 is equiangular, but it is not equilateral. 33a. 1. polygon. polygon 3. polygon 4. polygon 5. Not a polygon b. 1. triangle; reg.. quad.; reg. 3. hexagon; reg. 4. quad.; irreg. c. Shape appears to be a square. Shape 4 appears to be a rhombus. d. Assume polygon is reg. 6m = (6 - )180 = 70 m = You cannot use the final statement because you do not know that JKLM is a. That is what is being proven. However, if both pairs of opp. sides of a quad. are, then the quad. is a. So JKLM is a. 35a. rect. b. HG c. reflex. Prop. of e. GHE d. def. of rect. f. SAS g. CPCTC 36a. slope of AB = _ - = -1; slope of CD = _ - = -1 slope of BC = _ -5 = 1; slope of AD = _ = 1 b. Rect.; adj. sides are. c. The diags. of a rect. are. 37. Possible answer: 1. VWXY is a rhombus. 1. Given. WX YX. Def. of rhombus 3. VWXY is a. 3. Rhombus 4. WZ YZ 4. diags. bisect each other 5. XZ XZ 5. Reflex. Prop. of 6. WZX YZX 6. SSS 7. WZX YZX 7. CPCTC 8. WZX and YZX are 8. Lin. Pair Thm. supp. 9. WZX and YZX are 9. supp. rt. rt m WZX = m YZX 10. Def. of rt. = VX WY 11. Def. of 38. Possible answer: It is given that ABCD is a rect. By def. of a rect., A, B, C, and D are rt.. So A C and B D because all rt. are. Since opp., ABCD is a. 39. Possible answer: 1. ABCD is a rhombus. 1. Given. ABCD is a.. Rhombus 3. B D, A C 3. opp. 4. AB BC CD DA 4. Def. of rhombus 5. E, F, G and H are the 5. Given mdpts. of sides. 6. EB BF HD DG, EA AH FC CG 6. Def. of mdpt. 7. BEF DGH, 7. SAS AEH CGF 8. EF GH, EH GF 8. CPCTC 9. EFGH is a. 9. Quad. with opp. sides = w 41. s = 7 Ç in. w =.5 cm P = 4s l = w Ç 3 =.5 3 Ç cm = 8 Ç in. P = l + w in. = (.5) + (.5 Ç 3 ) A = s = Ç 3 cm = (7 Ç ) = 98 in cm A = lw = (.5 Ç 3 ) (.5) = 6.5 Ç 3 cm cm 4. s = ÇÇÇÇ = 5 cm P = 4s = 0 cm A = 4 ( 1_ (3)(4) ) = 4 cm 43a. By def., a square is a quad. with 4 sides. So it is true that both pairs of opp. sides are. Therefore, a square is a. b. By def., a square is a quad. with 4 rt. and 4 sides. So a square is a rect., because by def., a rect. is a quad. with 4 rt.. 13 Holt McDougal Geometry

4 c. By def., a square is a quad. with 4 rt. and 4 sides. So a square is a rhombus, because by def., a rhombus is a quad. with 4 sides. 44. (1) Both pairs of opp. sides are ǁ. Both pairs of opp. sides are. Both pairs of opp. are. All pairs of cons. are supp. Its diags. bisect () Its diags. are. (3) Its diags. are. Each diag. bisects a pair of opp.. test prep 45. D Since rhombus diag. bisect opp., LKM JKM. JK JM, so JKM is isosc.; by Isosc. Thm., JMK JKM. So m J + m JMK + m JKM =180 m J + x + x = 180 m J = (180 - x) 46. The perimeter of RST is 7. cm. Possible answer: Opp. sides of a rect. are, so RS = QT =.4 and ST = QR = 1.8. Diags. of a rect. bisect each other, so QS = QP = (1.5) = 3. The diags. of a rect. are, so TR = QS = 3. Therefore the perimeter of RST is = H Cons. sides need not be. challenge and extend 48. Think: By Alt. Int. Thm. and Thm , given. 3 x x + x = 90 4 x + x = 0 (4x + 1)(x - 5) = 0 x = 5 or Possible answer: B X A Z Y D Given: ABCD is a rhombus. X is mdpt. of AB. Y is mdpt. of AD. Prove: XY ǁ BD ; XY AC Proof: Since X is the mdpt. of AB and Y is the mdpt. of AD, XY is a midseg. of ABD by def. By the Midsegment Thm., XY ǁ BD. Since ABCD is a rhombus then its diags. are. So AC BD. Since also BD ǁ XY, it follows by the Transv. Thm. that AC XY. C 50. Possible answer: The midseg. of a rect. is a seg. whose endpoints are mdpts. of opp. sides of the rect. B X A D Given: ABCD is a rect. X is mdpt. of AB. Y is mdpt. of CD. Prove: AXYD BXYC Proof: A rect. is a, so ABCD is a. Since opp. sides of a are, AB CD and AD BC. Since X is the mdpt. of AB, AX XB. Since Y is the mdpt. of CD, DY YC. But because AB CD, you can conclude that AX XB DY YC. Opp. sides of a are ǁ by def., so AX ǁ DY. Since also AX DY, AXYD is a. But since ABCD is a rect. A is a rt.. So AXYD contains a rt. and is therefore a rect. By similar reasoning, you can conclude that BXYC is a rect. Since XY XY by the Reflex. Prop. of, all corr. sides are. Also, all rt. are, so all corr. are. Therefore AXYD BXYC by def. of by-1s, 8 1-by-s, 5 1-by-3s, 1-by-4s, 1 1-by-5, 6 -by-1s, 4 -by-s, -by-3s, 3 3-by-1s, 3-by-s, 1 3-by-3 45 rects. construction Check students constructions. 6-5 conditions for special check it out! 1. Both pairs of opp. sides of WXYZ are, so WXYZ is a. The contractor can use the carpenter s square to see if one of WXYZ is a rt.. If so, the frame is a rect. since with one rt. rect.. Not valid; if one of a is a rt., then the is a rect. To apply this thm., you need to know that ABCD is a. 3a. Step 1 Graph KLMN. L(-, 4) y 4 M(3, 1) x K(-5, -1) C Y N(0, 4) Step Determine if KLMN is a rect. KM = ÇÇÇÇ 8 + = ÇÇ 68 = ÇÇ 17 LN = ÇÇÇÇ + 8 = ÇÇ 68 = ÇÇ 17 Since KM = LN, diags. are. KLMN is a rect. 133 Holt McDougal Geometry

5 15. Think: HFG is comp. to FGJ and to FHG. m HFG + m FGJ = 90 m FHG + m FGJ = m FGJ = 90 m FGJ = 16. m EHG = m EHJ + m FHG = = 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 A + + m F = (6 - )180 8s + 7s + 5s + 8s + 7s + 5s = 70 40s = 70 s = 18 m A = m D = 8(18) = 144 ; m B = m E = 7(18) = 16 ; m C = m F = 5(18) = properties of 15. Think: Diags. bisect 16. AD BC BE = 1_ AD = BC = 6.4 BD = 1_ (75) = ED = BE = CDA ABC m CDA = m ABC = Think: Cons. supp. m ABC + m BCD = m BCD = 180 m BCD = BCD DAB m BCD = m DAB = 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) GHLM is a. 1. Given L JMG. G L. opp. 3. G JMG 3. Trans. Prop. of 4. GJ MJ 4. Conv. Isosc. Thm. 5. GJM is isosc. 5. Def. of isosc. conditions for 9. m = 13 m G = 9(13) = 117 ; n = 7 m A = (7) + 9 = 63, m E = 3(7) - 18 = 63 Since = 180 G is supp. to A and E, so one of ACEG is supp. to both of its cons.. ACEG 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 By Conv. of Same-Side Int. Thm., QT ǁ RS ; since QT RS, QRST is a by Thm Yes; The diags. bisect By Thm the quad. is a. 3. No; By Conv. of Alt. Int. Thm., one pair of opp. sides is ǁ, but other pair is. None of conditions for a are met. 33. slope of BD = _ 10 = 1_ ; slope of FH = _ = 1_ 5 slope of BH = _ -6 = -6; slope of DF = _ = -6 Both pairs of opp. sides have the same slope, so BD ǁ FH and BH ǁ DF ; by def., BDFH is a. 6-4 Properties of special 34. AB CD 35. AC = CE AB = CD = 18 = (19.8) = Holt McDougal Geometry

6 36. BD AC BD = AC = WX = WZ 7a + 1 = 9a = a 3.5 = a WX = 7(3.5) + 1 = XV VZ XV = VZ = 3(3.5) = XY WX XY = WX = m TZV = 90 8 n + 18 = 90 8n = 7 n = 9 Think: RT ÈÈ bisects SRV. m TRS = 1_ m SRV = 1_ (9(9) + 1) = m RSV + m TRS = 90 m RSV + 41 = 90 m RSV = STV SRV m STV = m SRV = 9(9) + 1 = m TVR + m STV = 180 m TVR + 8 = 180 m TVR = Think: All 4 are isosc.. m 1 + m = 180 m 3 + m 4 = 180 By Lin. Pair Thm., m + m 4 = 180 By Alt. Int Thm., m 3 = 33 (33) + m 4 = 180 m 4 = 114 m = 180 m = 66 m = 180 m 1 = 114 m 1= 57 By Alt. Int Thm., 1 5 m 5 = m 1 = BE CE BE = CE = XZ = XV = (10.5) = Think: All 4 are rt.. m = m 5 = 53 m 3 = 90 m 4 + m 5 = 90 m = 90 m 4 = m 1 = m 4 = Step 1 Show that RT and SU are congruent. RT = ÇÇÇÇÇÇÇÇÇÇÇ ((-3) - (-5)) + (-6-0) = ÇÇ 10 SU = ÇÇÇÇÇÇÇÇÇÇÇÇÇ ((-7) - (-1)) + ((-4) - (-)) = ÇÇ 10 Since RT = SU, RT SU Step Show that RT and SU are perpendicular. slope of RT: _ (-5) = -3 slope of SU: _-4 - (-) -7 - (-1) = 1_ 3 since -3 ( 1_ 3 ) = -1, RT SU Step 3 Show that RT and SU bisect mdpt. of RT: ( _ + (-3) -5, _ 0 + (-6) ) = (-4, -3) + (-7) mdpt. of SU: ( -1 _, _- + (-4) ) = (-4, -3) Since RT and SU have the same midpoint, they bisect The diagonals are congruent perpendicular biectors of 49. Step 1 Show that EG and FH are congruent. 6-5 EG = ÇÇÇÇÇÇÇÇÇ (5 - ) + (- - 1) = 3 Ç FH = ÇÇÇÇÇÇÇÇÇ ( - 5) + (- - 1) = 3 Ç Since EG = FH, EG FH Step Show that EG and FH are perpendicular. slope of EG: _ = -1 _ slope of FH: = 1 since -1(1) = -1, EG FH Step 3 Show that EG and FH bisect mdpt. of RT: ( _ + 5, _ + (-) 1 ) = ( 7_ 1_ ) mdpt. of SU: ( 5 _ +, _ + (-) 1 ) = ( 7_ 1_ ) Since EG and FH have the same midpoint, they bisect The diagonals are congruent perpendicular biectors of conditions for special 50. Not valid; if the diags. of a are, then the is a rect. If the diags. of a are, then the is a rhombus. If a quad. is a rect. and a rhombus, then it is a square. But to apply this line of reasoning, you must first know that EFRS is a. 51. valid (diags. bisect each other : with diags. rect.) 5. valid (EFRS is a by def.; with 1 pair cons. sides rhombus) 53. BJ = ÇÇÇÇ = 8 Ç ; FN = ÇÇÇÇ = 6 Ç Diags. are, so is not a rect. Therefore is not a square. slope of BJ = 8_ = 1; slope of FN = _ = -1 Diags. are, so is a rhombus. 54. DL = ÇÇÇÇ = 6 Ç 5 ; HP = ÇÇÇÇ = 6 Ç 5 Diags. are, so is a rect. slope of DL = 6_ 1 = 1_ ; slope of HP = _ -1-6 = Diags. are not, so is not a rhombus. Therefore is not a square. 55. QW = 1 ÇÇÇÇ + 8 = 4 13 ÇÇ ; TZ = 8 ÇÇÇÇ + 1 = 4 13 ÇÇ Diags. are, so is a rect. slope of QW = 8_ 1 = _ ; slope of TZ = _ -1 3_ = Diags. are, so is a rhombus. Rect., rhombus is a square. 143 Holt McDougal Geometry