Points P and Q lie on side AB and AC of triangle ABC respectively such that segment PQ is parallel to side BC. If the ratio of AP:PB is 2:3, and area of △APQ is 8sq cm, what is the area of trapezium PQCB?

Given that PQ∥BC, △APQ is similar to △ABC (△APQ∼△ABC).
The ratio of the sides is:

PB
AP
​
=
3
2
​

Therefore, the total length of side AB is:

AB=AP+PB=2+3=5 parts
The ratio of the corresponding sides of △APQ and △ABC is:

AB
AP
​
=
5
2
​

We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides:

Area(△ABC)
Area(△APQ)
​
=(
AB
AP
​
)
2

Area(△ABC)
8
​
=(
5
2
​
)
2
=
25
4
​

Area(△ABC)=8×
4
25
​
=2×25=50sq cm
The area of the trapezium PQCB is the difference between the area of △ABC and △APQ:

Area(PQCB)=Area(△ABC)−Area(△APQ)
Area(PQCB)=50−8=42sq cm