{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Pre-Calculus Homework Solutions 183

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online