How to solve geometric question like this? Rectangle ABCD has length 10 breadth 20. P and Q trisect AB. CD is bisected at R. The diagonal AC intersects PR and QR at E and F. Find the area of the quadrilateral PEFQ. This was asked in this year's XAT exam. !enter image description here I changed the question a bit. They just gave the area of the rectangle and asked to select the correct range out of given 5 options for the quadrilateral's area. Since there were a lot of questions in the exam I would prefer if someone showed a shortcut way of cracking this problem. I am hoping to gain some insight.

Based on your graph, I think you mean quadrilateral PEFQ.

The way to hack this problem is that you shall find two pair of "similar" triangulars

triangle AFQ is same shape as FRC by a ratio of 4/3 : 1

triangle AEP is same shape as ERC by a ratio of 1/3 : 1/2

By knowing this, you can calculate the height of AFQ and AEP, and therefore, calculate the area of those two triangular.

height of AFQ = (4/3)/(4/3 + 1)* length of CB

height of AEP = (1/3)/(1/3+1/2)* length of CB

area of quadrilateral PEFQ = area of AFQ - area of AEP