123 Holt McDougal Geometry

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1 y x heck students estimates; possible answer: pentagon is not equiangular; m = 100 ; m = 113 ; m = 113 ; m = 101 ; m = 113 ; yes, pentagon is not equiangular. 45a. heptagon b. (7 - )180 = 900 c. m + m + m + m + m + m + m G = m m + m = 900 3m = 900 3m = 40 m = Let n be number of sides and s (= 7.5) be side length. P = ns 45 = n(7.5) n = Polygon is a (regular) hexagon Possible answer: 49. Possible answer: 50. Possible answer: 51. The figure has sides, so it is a hexagon. The sides are, so the hexagon is equilateral. The are, so the hexagon is equiangular. Since the hexagon is equilateral and equiangular, it is regular. No diagonal contains pts. in the interior, so it is convex. 5. s number of sides increases, isosc. formed by each side become thinner, and dists. from any pt. on base of each triangle to its apex approach same value. or a circle, each pt. is the same dist. from center. So polygon begins to resemble a circle. test prep H (1 -)180 = m + m = (4 - )180 3m = 04 m = 8 m = (8) = 13 challenge and extend 5. measures are a, a + 4,, a + 1, where a is a multiple of 4. a + a a + 1 = (5 - )180 5a + 40 = 540 5a = 500 a = 100 measures are 100, 104, 108, 11, and PQ ST, QR RS, and Q S. So by SS, PRQ SRT. y PT, PR RT, so PRT is isosc. y Isosc. Thm., RTP RPT, so m RTP = m RPT = z. y Sum Thm., z + y = 180 (1) y PT and Isosc. Thm., PRQ SRT QPR RTS m PRQ = m SRT = m QPR = m RTS = x Since PQRST is reg., 5m QRS = (5 - )180 5(x + y) = 540 x + y = 108 () 5m PTS = (5 - )180 5(y + z) = 540 y + z = 108 (3) Subtr. (3) from (1): z = = 7 Subst: in (3): y + 7 = 108 y = 3 Subst. in (): x + 3 = 108 x = 7 x = K ǁ ǁ L. y lt. Int. Thm., L ext. and L ext. m L = m L = _ = 3 m L = m L + m L = Yes, if you allow for measures greater than 180. m + m + m + m + m + m = 70 - PRoperties of parallelograms check it out! 1a. KN LM KN = LM = 8 in. c. O is mdpt. of LN LO = 1_ LN = 1_ () = 13 in. a. J JG J = JG 3w = w + 8 w = 8 w = 4 JG = (4) + 8 = 1 b. NML LKN m NML = m LKN = 74 b. J JH J = JH 4z - 9 = z z = 9 H = JH = (z) = (9) = Holt Mcougal Geometry

2 3. Step 1 Graph given pts. y Step ind slope of PQ by R Q counting units from P to Q. Rise from - to 4 is. x Run from -3 to -1 is. 0 4 S Step 3 Start at S and - count same # of pts. P Rise of from 0 is. Run of from 5 is 7. Step 4 Use slope formula to verify that QR ǁ PS. slope of QR = _ = 1_ 4 slope of PS = _ = 1_ 4 oords. of vertex R are (7, ). 4. Statements Reasons 1. GHJN and JKLM are. 1. Given. N and HJN are supp.; K and MJK are supp.. cons. are supp. 3. HJN MJK 3. Vert. Thm. 4. N K 4. Supps. Thm. think and discuss 1. Measure of opp. is 71. Measure of each cons. is = XY = 1, WZ = 18, and YZ = 18; possible answer:since VWXY is a, opp. sides are, so XY = VW = 1. lso the diagonals of a parallelogram bisect each other, so WZ = YZ = 3 = Properties of Parallelograms Opp. sides : Opp. sides : Opp. : ons. supp.: iags. bisect each other: (180 - x) exercises guided practice x x (180 - x) 1. Only 1 pair of sides is ǁ. y def., a has pairs of ǁ sides.. Possible answer: Q R opp. sides: PQ and RS, QR and SP ; opp. : P and R, Q and S P S 3. is mdpt. of = = (18) = 3 5. is mdpt. of = = 18. and are supp. m = m = = = = m = m = JK LM JK = LM 7x = 3x x = 14 x = 3.5 JK = 7(3.5) = L and M are supp. m L + m M = 180 z z - = 180 7z = 189 z = 7 m L = (7) - 3 = m = m = LM = 3(3.5) + 14 = m M = 5(7) - = Step 1 Graph given pts. y 8 Step ind slope of G by counting units -5 from to G x Rise from 5 to 0 is H 0 G 8 Run from -1 to is Step 3 Start at and -8 count same # of pts. Rise of -5 from 4 is -1. Run of 3 from -9 is -. Step 4 Use slope formula to verify that ǁ GH. slope of = _ = 1_ 8 slope of GH = _ = 1_ 8 oords. of vertex H are (-, -1). 14. Statements Reasons 1. PSTV is a ; PQ RQ 1. Given. STV P. opp. 3. P R 3. Isoc. Thm. 4. STV R 4. Trans. Prop. of practice and problem solving 15. JN = 1_ JL 1. LM = JK = 110 = 1_ (15.8) = LN = JN = m JKL = m JML = m KLM = m JML = = m MJK = m KLM = 130. YW = WV = (10) = 0 4. ZV = (7) = WV = VY b + 8 = 5b 8 = 4b b = WV = () + 8 = XV = ZV 3a - 7 = a a = 7 XZ = ZV = ((7)) = 8 14 Holt Mcougal Geometry

3 5. slope from P to R: rise = 4, run = - slope from V to T: rise of 4 from -1 is 3, run of - from 5 is -1 oords. of T are (-1, 3).. Statements Reasons 1. and GH are. 1. Given., G. opp. 3. G 3. Trans. Prop. of 7. PQ = RS and SP = QR; given PQ = QR, all 4 side lengths are =. P = PQ + QR + RS + SP 84 = 4PQ PQ = QR = RS = SP = 1 8. PQ = RS, SP = QR = 3RS 84 = RS + 3RS + RS + 3RS 84 = 8RS PQ = RS = 10.5 SP = QR = 3(10.5) = PQ = RS = SP - 7, QR = SP 84 = SP SP + SP SP 84 = 4SP = 4SP QR = SP = 4.5, PQ = RS = = PQ = RS, QR = SP = RS 84 = RS + RS + RS + RS 84 = RS + RS 0 = RS + RS = (RS + 7)(RS - ) Since RS > 0, PQ = RS =, and QR = SP = = 3. 31a. 3 1 (orr. Post.) 1 ( opp. ) 8 1 ( opp. ) b. is supp. to 1 ( cons. supp.), 4 is supp. to 1 ( cons. supp.), 5 is supp. to 1 ( cons. supp.), and 7 is supp. to 1 (Subst.). 3. MPR RKM ( opp. ) 33. PRK KMP ( opp. ) 34. MT RT ( diags. bisect each other) 35. PR KM ( opp. sides ) 3. MP ǁ RK (ef. of ) 37. MK ǁ RP (ef. of ) 38. MPK RKP (lt. Int. Thm.) 39. MTK RTP (Vert. Thm.) 40. m MKR + m PRK = 180 ( cons supp.) 41. y props. of, y = 1 x + 1 = 18 x = 119 z = x = y lt. Int. Thm., x = 90 y props. of, z = 53 y def. of comp., y = = y Vert. Thm. and Sum Thm., x = 180 x = 4 y + ( ) + 4 = 180 y = 50 y lt. Int. Thm., z = y = 50 44a. b. c. 4 d. opp. sides of a are e. S f. PT g. bisect 45. Given: is a, Prove: and are supp. and are supp. and are supp. and are supp. Statements Reasons 1. is a. 1. Given. ǁ, ǁ. ef. of. 3. and are supp., 3. Same-side Int. and are supp., Thm. and are supp., and are supp. 4. x = y 4 x = y x = y - 9 3x = 9 x = 3 y = (3) = 48a. m = m x + 1 = 9x = 3x x = 15 m = (15) + 1 = 10 b. m = m = m = = 78 ( cons. supp.) 49. Possible answer: 47. y = x + 3 y + 7 = 3x 7 = x = x x = 5 y = = 8 a. No; possible answer: rawings show a counterexample, since all side pairs are but are. b. No; possible answer: or any given set of side lengths, a could have many different shapes. 50. Possible answer: a quad. is a 4-sided polygon. Since every is a polygon with 4 sides, every is a quad. has pairs of ǁ sides. Since sides of a quad. are not necessarily ǁ, a quad. is not necessarily a. 15 Holt Mcougal Geometry

4 test prep 51. m Q = m S 3x + 5 = 5x = x x = J P = = = (5 + 8.) =.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 = -, so Y = (-, -). rom to is rise of - and run of 5; rise of - from is -4 - = -, run of 5 from is = 4; so Z = (4, -) 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 are supp. 9. Lin. Pair Thm and are rt supp. rt. 11. ÈÈ ÈÈ 11. ef. of -3 conditions for parallelograms check it out! 1. Think: Show that PQ and RS. are ǁ and. a =.4 PQ = 7(.4) = 1.8, RS = (.4) + 1 = 1.8, so PQ RS b = 9 m Q = 10(9) - 1 = 74, m R = 9(9) + 5 = 10 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 the quad. is a. 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 = - 7_ slope of MN = _ = - 7_ 5-3 slope of LM = _ 5-7 = - 1_ slope of KN = _ = - 1_ 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. 1 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 = 1. 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 -1 Properties and attributes of polygons 5. not a polygon. polygon; triangle 7. polygon; dodecagon 8. irregular; concave 9. irregular; convex 10. regular; convex 11. (n - )180 (1 - ) (n)m(ext. ) = 30 4m(ext. ) = 30 m(ext. ) = (n)m = (n - )180 0m = (18)180 0m = 340 m = m + + m = ( - )180 8s + 7s + 5s + 8s + 7s + 5s = 70 40s = 70 s = 18 m = m = 8(18) = 144 ; m = m = 7(18) = 1 ; m = m = 5(18) = 90 - properties of parallelograms 15. Think: iags. bisect 1. each other. = 1_ = =.4 = 1_ (75) = = = m = m = Think: ons. supp. m + m = m = 180 m = m = m = WX = YZ b + = 5b = 4b 3.5 = b WX = = m W + m X = 180 a + 14a = 180 0a = 180 a = 9 m W = (9) = Y W m Y = m W = 54. YZ = 5(3.5) - 8 = m X = 14(9) = 1. Z X m Z = m X = 1 7. 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 = ; T = (, -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 = 3, m = 3(7) - 18 = 3 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 = 7 ; 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 the quad. is a. 3. 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 = _ - = -; slope of = _ = - oth pairs of opp. sides have the same slope, so ǁ H and H ǁ ; by def., H is a. -4 Properties of special parallelograms = = = 18 = (19.8) = Holt Mcougal Geometry

126 Holt McDougal Geometry

126 Holt McDougal Geometry test prep 51. m Q = m S 3x + 5 = 5x - 5 30 = x x = 15 5. J 53. 6.4 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.

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