# In figure below, $PQRS$ and $ABRS$ are parallelograms and $\mathrm{X}$ is any point on side $\mathrm{BR}$. Show that(i) $\operatorname{ar}(\mathrm{PQRS})=\operatorname{ar}(\mathrm{ABRS})$(ii) $\operatorname{ar}(\mathrm{AXS})=\frac{1}{2} \mathrm{ar}(\mathrm{PQRS})$"

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Given:

$PQRS$ and $ABRS$ are parallelograms and $\mathrm{X}$ is any point on side $\mathrm{BR}$.

To do:

We have to show that

(i) $\operatorname{ar}(\mathrm{PQRS})=\operatorname{ar}(\mathrm{ABRS})$
(ii) $\operatorname{ar}(\mathrm{AXS})=\frac{1}{2} \mathrm{ar}(\mathrm{PQRS})$

Solution:

(i) Parallelograms $PQRS$ and $ABRS$ lie on the same base $SR$ and between the same parallels $SR$ and $PB$.

This implies,

$ar(PQRS) = ar (ABRS)$....…(i)

(ii) In parallelogram $ABRS$,

$\triangle AXS$ and parallelogram $ABRS$ lie on the same base $AS$ and between the same parallels $AS$ and $BR$.

This implies,

$ar (\triangle AXS) = \frac{1}{2}$ ar(parallelogram $ABRS$.......…(ii)

From (i) and (ii), we get,

$ar (\triangle AXS) = \frac{1}{2}$ ar (parallelogram $PQRS$)

Updated on 10-Oct-2022 13:41:39