Pre-Calculus Homework Solutions 183

Pre-Calculus Homework Solutions 183 - 21. PT SR, so P R and...

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Unformatted text preview: 21. PT SR, so P R and T S because they are alternate interior angles. PQT SQR because they are vertical angles. Thus, PQT RQS. PQ RQ 6x 6x IH that HG x) x2 x2 x2 x2 x (6 18 18 18 0 0 x)(3) 3x 3x IH HG IH HG 6 6 QS 3 3 The scale 23. 24. 25. x 0 or 6 x 0 or 3 3 3 factor is TQ 3x 3 0 or 1. SQ Triangles ABC and DFE are isosceles triangles. A D, and BA CA and FD ED so BA CA DFE by SAS FD ED . Thus, ABC Similarity. HI JK, so GHI GJK and GIH GKJ because they are corresponding angles. Thus, GHI GJK by AA Similarity. m L m Q m LMQ ¬180 35 85 m LMQ ¬180 m LMQ ¬60 LMQ NMP, so m NMP 60. m N m P m NMP 180 m N 40 60 180 m N 80 LMQ is not similar to PMN because the angles of the triangles are not congruent. Since AB DE, B E and A D because they are alternate interior angles. By AA Similarity, ABC DEC. Using the definition AC of similar polygons, BC DC . EC x3 11x 2 6(x 6x IK x 4 5 x x or 2x. IH 2x x or 2, and HG 30. In order to IH that HG HG 3x IH HG IK 22 11 or 2. KL IH HG . Thus, show that GL HK, we must show IK KL . Let HI x. Then IG 3x, so x or 2x. IK x 1 2x or 2 , and KL 18 6 or 3. So the sides are not proportional and GL HK. 31. From the Triangle Proportionality Theorem, BC ED AB AE . Substitute the known measures. 4 6 ED 9 4(9) 36 6 32. From BC AB 6(ED) 6(ED) ED the Triangle Proportionality Theorem, ED AE . Substitute the known measures. 16 12 5 AE 5 AE 12 4 12 4(AE) 12(5) 4(AE) 60 AE 15 33. Since BE CD, ABE ACD and AEB ADC by the corresponding angles postulate. Then ABE ACD by the AA Similarity. Using AD the definition of similar polygons, CD BE AE . 2 Substitute the known measures. CD 6 CD 6 4 4) (x 2)(x 5) 16 x2 5x 2x 10 16 x2 7x 10 16 x2 x 10 0 x2 x 6 0 (x 3)(x 2) x 3 0 or x 2 0 x3 x 2 Reject x 2 because otherwise UT 2 2 or 0. So x 3. 28 7 8 or 2 . GL HK. 1 6 3 3x IK KL 1(11x 2) 11x 2 5x 2 5x x 26. V UST and T T, so by AA Similarity, RVT UST. Using the definition of similar RT VT polygons, UT ST . 4 2 4 2 IK 35 7 10 or 2 , and KL IK KL , so GL HK. 29. In order to show that GL HK, we must show IH IK that HG KL . Let KL x. Then IL 3x and 3) 18 18 20 4 2x x 2x x 15 5 9 or 3 . Because the side lengths are not proportional, GL HK. 28. IL IK KL 36 28 KL 8 KL In order to show that GL HK, we must show IH IK that HG KL . PQ 22. IK KL . IK 21 3 14 or 2 , and KL IH HG TQ SQ 3 3x (6 x)(3 6x 3x 18 3x 18 18 27. In order to show that GL HK, we must show 4(2x 8x 8x 8(CD) 8(CD) CD x 8 4 8 12 8 6(12) 72 9 2 185 Chapter 6 ...
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This note was uploaded on 12/19/2011 for the course MAC 1140 taught by Professor Dr.zhan during the Fall '10 term at UNF.

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