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