Pre-Calculus Homework Solutions 175

Pre-Calculus Homework Solutions 175 - 26. TA and WB are...

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Unformatted text preview: 26. TA and WB are medians, so RA AS and UB BV. Then RS 2(RA) and UV 2(UB). TA WB 8 3x 6 8 3x 6 31. Given: RS UV 2(3) 2(x 2) 6 2x 4 AD Prove: RU 32 18x 36 32 2x 36 68 2x 34 x UB x 2 34 2 or 36 27. BF bisects ABC, so by the Angle Bisector C BA BC . Let x represent CF. Then x ¬ 76 .5 67.5 7.5x ¬6x 67.5 ¬13.5x 5 ¬x Thus, CF 5. AC ED, so BED BFC. FBC EBD, so by AA Similarity, EBD FBC. Let y represent BD. 9 BD BC y 7.5 x 30. Given: ABC BD Prove: QS B S Reasons ABC RST AD is a median of ABC. RU is a median of RST. 1. Given 2. Def. of median DB; TU CB TS CD TU DB; US 4. Segment Addition Postulate CD TU DB US 5. Substitution 6. AB RS DB US DB US DB 2(DB) US or 2(US) 6. Substitution 7. AB RS 8. B 9. US 3. Def. of polygons 7. Substitution S ABD AD 10. RU distance from camera ¬ length of camera 12 ¬ 7 15 x z . The cross products yield xy U 5. AB RS 8. Def. of polygons 9. SAS Similarity RSU AB RS 32. Given: CD bisects AE CD. AD AC Prove: DB BC 15x ¬1008 x ¬67.2 The person is 67.2 inches or about 5 feet 7 inches tall. 29. xy z2; ACD CBD by AA Similarity. Thus, AD z CD or y T 3. AB RS 4. CB TS ED ¬ FC ¬9 5 height of person 28. height of image x 12 B D 2. CD 5y ¬67.5 y ¬13.5 Thus, BD 13.5 CD BD R Proof: Statements 1. ABC. AB RS A 16x AF Theorem, CF AF ¬9 x. ABC RST, AD is a median of RU is a median of RST. 10. Def. of polygons ACB. By construction E C 312 z2. AD B Proof: Statements PQR BA QP Reasons 1. CD bisects ACB. By construction, AE CD. 1 2 3. Definition of Angle Bisector 3 1 4. Alternate Interior Angle Theorem 5. 2 E 5. Corresponding Angle Postulate 6. 3 E 6. Transitive Prop. 7. EC AC 8. EC AC 7. Isosceles Th. 8. Def. of congruent segments AD 9. DB 177 2. Triangle Proportionality Theorem 4. CP R D S Proof: Since ABC PQR, A P. BDA QSP because they are both right angles created by the altitude drawn to the opposite side and all right angles are congruent. Thus ABD PQS BA by AA Similarity and BD by the definition QP QS of similar polygons. EC BC 3. A 1. Given AD 2. DB Q AC BC 9. Substitution Chapter 6 ...
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