Question

In the given figure, ABCD and ABPQ are two parallelograms and M is a point on AQ and BMP is a triangle.
Then, ar(∆BMP) = $\frac{1}{2}$ ar(||gm ABCD) is
(a) true
(b) false

Answer 1

We know parallelogram on the same base and between the same parallels are equal in area.
Here, AB is the common base and AB || PD
Hence, ar(ABCD) = ar(ABPQ) .....(1)
Also, when a triangle and a parallelogram are on the same base and between the same parallels then the
area of triangle is half the area of the parallelogram.
Here, for the ∆BMP and parallelogrm ABPQ, BP is the common base and they are between the common parallels BP and AQ
So, ar(∆BMP) = $\frac{1}{2}$ ar(||gm ABPQ) .....(2)
From (1) and (2) we have
ar(∆BMP) = $\frac{1}{2}$ ar(||gm ABCD)
Thus, the given statement is true.
Hence, the correct answer is option (a).