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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AcademicMathematicsNCERTClass 9

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 $PQRS$)

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

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