In triangle \(ABC\), \(AP = BQ = \frac{PQ}{2}\) and \(CR/RA = 3/2 \). Find the ratio of the area of triangle \(BQC\) to the area of triangle \(CRQ\).
Since AP = BQ = PQ/2. We can conclude that QB is 1/4 of AB.
Let the area of ABC = x
So BQC = \(\frac{x}{4}\)
Area of triangle CAQ = \(x-\frac{x}{4}=\frac{3x}{4}\)
Since CR/RA = 3/2, that means that CRQ = \(\frac{3x}{4}*\frac{3}{5}=\frac{9x}{20}\)
SO
Area of BQC is \(\frac{x}{4}\)
Area of CRQ is \(\frac{9x}{20}\)
Can you find the ratio now? (Hint: It's simple algebra!)