Let PQR be a triangle. The points A, B and C are on the sides QR, RP and PQ respectively such that ${{QA} \over {AR}} = {{RB} \over {BP}} = {{PC} \over {CQ}} = {1 \over 2}$. Then ${{Area(\Delta PQR)} \over {Area(\Delta ABC)}}$ is equal to :

Question

Let PQR be a triangle. The points A, B and C are on the sides QR, RP and PQ respectively such that

${{QA} \over {AR}} = {{RB} \over {BP}} = {{PC} \over {CQ}} = {1 \over 2}$. Then ${{Area(\Delta PQR)} \over {Area(\Delta ABC)}}$ is equal to :

$\frac{5}{2}$

Solution

Please login to view the detailed solution steps...

Go to DASH