Question

In figure, $PQRS$ and $ABRS$ are parallelogram and $X$ is any point on side $BR$. Show that
(i) $ar\left(PQRS\right)=ar\left(ABRS\right)$
(ii) $ar\left(AXZ\right)=\frac{1}{2}ar\left(PQRS\right)$

Answer 1

$\left(i\right)$ Since PQRS is a parallelogram

$PQ||RX$ [Opposite sides of parallelogram are parallel]

Since ABRS is a parallelogram

$AB||RS$ [Opposite sides of parallelogram are parallel]

Since $PQ||RS\&AB||RS$

We can say that $PB||RS$

Now,

$PQRS\&ABRS$ are two parallelograms with the same base RS and between the same parallels PB and RS

$\therefore ar\left(PQRS\right)=ar\left(ABRS\right)$ [Parallelogram with same base and between the same parallels are equal in area]

$\left(ii\right)$ Since ABRS is a paralleogram

$AS||BR$ [Opposite sides of parallelogram are parallel]

$△AXS$ and parallelogram $ABRS$ lie on the same base AS and are between the same parallel lines AS and BR

$\therefore Area\left(△AXS\right)=\frac{1}{2}Area\left(ABRS\right)$ [Area of triangle is half of parallelogram if they have the same base and parallels]

We proved in part (i)

$ar\left(PQRS\right)=ar\left(ABRS\right)$

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

Hence proved.