Question

P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that area of ΔAPB = area of ΔBQC. [2 MARK]

Answer 1

Statment of the theorem : 1/2 mark
Application : 1/2 mark
Calculation : 1 mark

If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram.

Area of triangle ABP = $\frac{1}{2}$ area of parallelogram ABCD (As they are on the same base and between the same parallels)

Area of triangle BQC = $\frac{1}{2}$ area of parallelogram ABCD (As they are on the same base and between the same parallels)

Hence, Area of ΔABP = Area of ΔBQC