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\( ABC, AP = \frac{PQ}{2} = BQ$ $\frac{CR}{RA} = \frac{3}{2}. \frac{[BQC]}{[CRQ]}.$\) in triangle ABC, AP = PQ/2 = BQ and CR/RA=3/2. Find the area of BQC/(the area of)CRQ.

 

 Jun 2, 2020
 #1
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I'm getting [QBC]/[RQC] = 3/5.

 Jun 3, 2020
 #2
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In triangle ABC, AP = PQ/2 = BQ and CR/RA=3/2.
Find the area of BQC/(the area of)CRQ.

In triangle

\(\text{In triangle } ABC,\ AP = \frac{PQ}{2} = BQ,\quad\dfrac{CR}{RA} = \dfrac{3}{2}. \dfrac{[BQC]}{[CRQ]}. \)

 

My answer see here: https://web2.0calc.com/questions/geometry-problem-im-stuck-on#r3

 

laugh

 Jun 3, 2020
edited by heureka  Jun 3, 2020
edited by heureka  Jun 3, 2020
 #3
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Thank you!

 Jun 3, 2020

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