Question

In fig, $PQRS$ and $ABRS$ are parallelogram and $X$ is any point on side $BR$. Show that

(i) $area\left(PQRS\right)=area\left(ABRS\right)$

(ii) $area\left(AXS\right)=\frac{1}{2}area\left(PQRS\right)$

Answer 1

(i) For parallelogram $PQRS$ and $ABRS,$ we notice that they lie on the same base $RS$ and between the same parallel lines $RS$ and $PB$

So, $area\left(PQRS\right)=area\left(ABRS\right)..\left(1\right)$ (since parallelograms lying on the same base $RS$ and between the same parallel lines have equal area)

(ii) For $△AXS$ and parallelogram $ABRS,$ we notice that they lie on the same base $SA$ and between the same parallel lines $SA$ and $BR$.

So, $area\left(△AXS\right)=\frac{1}{2}area\left(ABRS\right)..\left(2\right)$ (since parallelogram and triangle lieing on the same base and between the same parallel lines have equal area)

Subsituting equation $\left(1\right)$ in equation $\left(2\right),$ we get

$area\left(△AXS\right)=\frac{1}{2}area\left(PQRS\right)$