The side $ \mathrm{AB} $ of a parallelogram $ \mathrm{ABCD} $ is produced to any point $ \mathrm{P} $. A line through $ \mathrm{A} $ and parallel to $ \mathrm{CP} $ meets $ C B $ produced at $ Q $ and then parallelogram PBQR is completed (see figure below). Show that $ \operatorname{ar}(\mathrm{ABCD})=\operatorname{ar}(\mathrm{PBQR}) $.
[Hint : Join $ \mathrm{AC} $ and $ \mathrm{PQ} $. Now compare ar (ACQ)
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Given:

The side \( \mathrm{AB} \) of a parallelogram \( \mathrm{ABCD} \) is produced to any point \( \mathrm{P} \). A line through \( \mathrm{A} \) and parallel to \( \mathrm{CP} \) meets \( C B \) produced at \( Q \) and then parallelogram PBQR is completed.

To do:

We have to show that \( \operatorname{ar}(\mathrm{ABCD})=\operatorname{ar}(\mathrm{PBQR}) \).

Solution:

Join $AC$ and $PQ$.

$PQ$ and $AC$ are diagonals of parallelograms $PBQR$ and $ABCD$ respectively.

Therefore,

We know that,

Diagonal of a parallelogram divides it into two triangles of equal area.

This implies,

$\operatorname{ar}(\triangle \mathrm{ABC})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD})$......(i)

$\operatorname{ar}(\mathrm{\triangle PBQ})=\frac{1}{2} \operatorname{ar}(\mathrm{PBQR})$........(ii)

$\triangle \mathrm{ACQ}$ and $\triangle \mathrm{AQP}$ are on the same base $AQ$ and between the parallels $\mathrm{AQ}$ and $\mathrm{CP}$.

Therefore,

$ar(\triangle ACQ)=ar(\triangle AQP)$

Subtracting $ar(\triangle ABQ)$ on both sides, we get,

$ar(\triangle ACQ)-ar(\triangle ABQ)=ar(\triangle AQP)-ar(\triangle ABQ)$

$ar(\triangle \mathrm{ABC})=ar(\triangle \mathrm{BPQ})$

$\frac{1}{2} \operatorname{ar}(\mathrm{ABCD})=\frac{1}{2} \operatorname{ar}(\mathrm{PBQR})$

From (i) and (ii),

$\operatorname{ar}(\mathrm{ABCD})=\operatorname{ar}(\mathrm{PBQR})$

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Updated on: 10-Oct-2022

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