+14 In triangle ABC, points P and Q are on side \(\overline{AB},\) and point R is on side \(\overline{AC}.\) If \(AP = \frac{PQ}{3} = \frac{QB}{2}\) and \(\frac{AR}{2} = \frac{RC}{3},\) then find \(\frac{[QBC]}{[CRQ]}.\)
I'm getting \(\frac{6}{5}\) but apparently it's incorrect. Please, anyone, I really need help. Thank you!
+1765 Analyzing the Problem
We are given a triangle ABC with points P, Q, and R on sides AB, AB, and AC respectively. Specific ratios are provided for segment lengths related to these points. We need to find the ratio of the areas of triangles QBC and CRQ.
Solution Strategy
Relate Segment Lengths: Use the given ratios to express the lengths of AP, PQ, QB, AR, and RC in terms of a single variable.
Express Triangle Areas: Formulate the areas of triangles QBC and CRQ using the segment lengths obtained in step 1.
Find the Ratio of Areas: Divide the area of triangle QBC by the area of triangle CRQ and simplify the expression.
Steps to Solve
Relating Segment Lengths:
We are given: AP = PQ / 3 = QB / 2 and AR / 2 = RC / 3
Let x be the length of AP (or PQ / 3 or QB / 2).
Then, PQ = 3x and QB = 2x.
Similarly, let y be the length of AR (or 2 * RC / 3).
Then, RC = 3y / 2.
Expressing Triangle Areas:
Area of triangle QBC = (1/2) * Base * Height = (1/2) * QB * RC = (1/2) * 2x * (3y / 2) = 3xy
Area of triangle CRQ = (1/2) * Base * Height = (1/2) * CR * PQ = (1/2) * (3y / 2) * 3x = 4.5xy
Ratio of Areas:
[QBC] / [CRQ] = (3xy) / (4.5xy) = 2 / 3
Answer
Therefore, the ratio of the areas of triangles QBC and CRQ is 2:3.
