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Pre-Calculus Homework Solutions 112

# Pre-Calculus Homework Solutions 112 - 28 Given ABC S is the...

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28. Given: ABC S is the midpoint of AC . T is the midpoint of BC . Prove: ST 1 2 AB Midpoint S is b 2 0 , c 2 0 or b 2 , 2 c . Midpoint T is a 2 b , 0 2 c or a 2 b , 2 c . Proof: ST ¬ a 2 b b 2 2 2 c 2 c 2 ¬ a 2 2 0 2 ¬ a 2 2 or a 2 AB ¬ ( a 0) 2 (0 0) 2 ¬ a 2 0 2 or a ST ¬ 1 2 AB 29. Given: ABD , FBD AF 6, BD 3 Prove: ABD FBD Proof: BD ¬ BD by the Reflexive Property. AD ¬ (3 0) 2 (1 1) 2 9 0 or 3 DF ¬ (6 3) 2 (1 1) 2 9 0 or 3 Since AD DF , AD DF . AB (3 0) 2 (4 1) 2 9 9 or 3 2 BF (6 3) 2 (1 4) 2 9 9 or 3 2 Since AB BF , AB BF . ABD FBD by SSS. 30. Given: BPR PR 800, BR 800 Prove: BPR is an isosceles right triangle. Proof: Since PR and BR have the same measure, PR BR . The slope of PR 8 0 00 0 0 or 0. The slope of BR 8 8 0 0 0 0 8 0 00 , which is undefined. PR BR , so PRB is a right angle. BPR is an isosceles right triangle. 31. Given: BPR , BAR PR 800, BR 800, RA 800 Prove: PB BA Proof: PB (800 0) 2 (800 0) 2 or 1,280,000 BA (800 1600) 2 (800 0) 2 or 1,280,000 PB BA , so PB BA . 32. Given: JCT Prove: JCT is a right triangle. Proof: The slope of JC 3 5 0 0 0 0 0 0 or 3 5 . The slope of TC 5 3 0 0 0 0 0 0 or 5 3 . The slope of TC is the opposite reciprocal of the slope of JC. JC TC , so TCJ is a right angle. JCT is a right triangle. 33. Use the Distance Formula to find the distance between J ( 500, 300) and T (300, 500). JT ¬ [300 ( 500)] 2 (500 300) 2 ¬ 680,000 ¬ 824.6 The distance between Tami and Juan is 680,000 or approximately 824.6 feet.
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