Question

In Fig. $ PQRS$and $ ABRS$are parallelograms and $ X$ is any point on side $ BR.$ Show that

$ i) ar(PQRS)=ar(ABRS\left)\phantom{\rule{0ex}{0ex}}ii\right) ar\left(AXS\right)=\frac{1}{2}ar\left(PQRS\right)$

Answer 1

Step $1:$ Show that $ar\left(PQRS\right)=ar\left(ABRS\right).$

From the diagram we notice that parallelogram $PQRS$and $ABRS$lie on the same base $SR$and also, lie in between the same parallel lines $SR$and$PB.$

$\therefore Area\left(PQRS\right)=Area\left(ABRS\right)...\left(1\right)$

Step $2:$Show that $ar\left(AXS\right)=\frac{1}{2}ar\left(PQRS\right).$

When a triangle and a parallelogram lies on the same base and between the same parallels then, the area of a triangle is equal to half the area of a parallelogram.

Consider $△AXS$and parallelogram $ABRS.$

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

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

From equations $\left(1\right)$and $\left(2\right),$we obtain

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

Hence proved.