Points P and Q lie on sides AB and AC of triangle ABC respectively such that segment PQ is parallel to side BC. If the ratio of AP:PB is 2:5, and the area of △APQ is 4sq cm, what is the area of trapezium PQCB?

In △ABC, since PQ∥BC, by Basic Proportionality Theorem/AA similarity:
△APQ∼△ABC
The ratio of corresponding sides of these triangles is:

AB
AP
​
=
AP+PB
AP
​

Given
PB
AP
​
=
5
2
​
, we can take AP=2k and PB=5k.

AB=2k+5k=7k
AB
AP
​
=
7k
2k
​
=
7
2
​

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)
4
​
=(
7
2
​
)
2
=
49
4
​

Area(△ABC)=49sq cm
The area of the trapezium PQCB is the difference between the area of △ABC and △APQ:

Area of trapezium PQCB=Area(△ABC)−Area(△APQ)
Area of trapezium PQCB=49−4=45sq cm