Question

Question 5 (ii)
In the given figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that:
$ar\left(\Delta AXS\right)=\frac{1}{2}ar\left(PQRS\right)$

Answer 1

Consider parallelogram PQRS and parallelogram ABRS.
As these lie on the same base SR and are between the same parallel lines PB and SR,
ar(PQRS) = ar(ABRS) .....(1)

Consider $\Delta AXS$ and parallelogram ABRS.
As these lie on the same base AS and are between the same parallel lines AS and BR,
$\therefore ar\left(\Delta AXS\right)=\frac{1}{2}ar\left(ABRS\right)...\left(2\right)$

From equations(1) and (2), we obtain,
$ar\left(\Delta AXS\right)=\frac{1}{2}ar\left(PQRS\right)$