In figure below, $ \mathrm{P} $ is a point in the interior of a parallelogram $ \mathrm{ABCD} $. Show that
(i) $ \operatorname{ar}(\mathrm{APB})+\operatorname{ar}(\mathrm{PCD})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD}) $
(ii) $ \operatorname{ar}(\mathrm{APD})+\operatorname{ar}(\mathrm{PBC})=\operatorname{ar}(\mathrm{APB})+\operatorname{ar}(\mathrm{PCD}) $
[Hint: Through $ \mathrm{P} $, draw a line parallel to $ \mathrm{AB} $.]
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AcademicMathematicsNCERTClass 9

Given:

\( \mathrm{P} \) is a point in the interior of a parallelogram \( \mathrm{ABCD} \). 

To do:

We have to show that
(i) \( \operatorname{ar}(\mathrm{APB})+\operatorname{ar}(\mathrm{PCD})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD}) \)
(ii) \( \operatorname{ar}(\mathrm{APD})+\operatorname{ar}(\mathrm{PBC})=\operatorname{ar}(\mathrm{APB})+\operatorname{ar}(\mathrm{PCD}) \)
Solution:

Draw two lines $EF$ and $GH$ parallel to $AB$ and $BC$ respectively.


(i) $\triangle APB$ and parallelogram $AEFB$ lie on the same base $AB$ and between the same parallels $AB$ and $EF$.

This implies,

$ar (\triangle APB) = \frac{1}{2} ar (AEFB)$...….(i)

$\triangle DPC$ and parallelogram $EFCD$ lie on the same base $DC$ and between the same parallels $DC$ and $EF$.

This implies,

$ar (\triangle DPC) = \frac{1}{2} ar (EFCD)$...….(ii)

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

$ar (\triangle APB)+ ar (\triangle DPC) = \frac{1}{2} ar (AEFB)+\frac{1}{2} ar (EFCD)$

$ar (\triangle APB)+ ar (\triangle DPC) = \frac{1}{2}( ar (AEFB)+ar (EFCD)$

$ar (\triangle APB)+ ar (\triangle DPC) = \frac{1}{2} ar (ABCD)$..........(iii)

(ii) $\triangle APD$ and parallelogram $AGHD$ lie on the same base $AD$ and between the same parallels $AD$ and $GH$.

This implies,

$ar (\triangle APD) = \frac{1}{2} ar (AGHD)$...….(iv)

$\triangle PBC$ and parallelogram $GBCH$ lie on the same base $BC$ and between the same parallels $BC$ and $GH$.

This implies,

$ar (\triangle PBC) = \frac{1}{2} ar (GBCH)$...….(v)

Adding (iv) and (v), we get,

$ar (\triangle APD)+ ar (\triangle PBC) = \frac{1}{2} ar (AGHD)+\frac{1}{2} ar (GBCH)$

$ar (\triangle APD)+ ar (\triangle PBC) = \frac{1}{2}[ ar (AGHD)+ar (GBCH)]$

$ar (\triangle APD)+ ar (\triangle PBC) = \frac{1}{2} ar (ABCD)$............(vi)

(iii) From (iii) and (vi), we get,

$ar (\triangle APB)+ ar (\triangle DPC) =ar (\triangle APD)+ ar (\triangle PBC)$

raja
Updated on 10-Oct-2022 13:41:37

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