In Fig. 4.236,
and AP : PB = 1 : 2. Find
[CBSE 2008]

Given in the given figure PQ || BC and AP: PB = 1: 2
We know that basic proportionality theorem states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
Since Δ APQ and ΔABC are similar, ![]()
Given ![]()
⇒ PB = 2AP
So, ![]()
we know that the ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

![]()
∴ Area (ΔAPB): Area (ΔABC) = 1: 9