Question

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

Answer 1

(a) True

ABCD and ABPQ are on thesame base and between the same parallel lines.
So, ar(parallelogram ABCD) = ​ar(parallelogram ABPQ) ...(i)
Now, parallelogram ABPQ and ∆BMP lie on the same base (BP) and between the same parallel lines.
So, ar(​∆BMP) = $\frac{1}{2}$ar(|| ABPQ)
From equation(i), we have:
ar(​∆BMP) = $\frac{1}{2}$ar(|​|ABCD)

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