# 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),

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

Updated on 10-Oct-2022 13:42:00
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