Question

The side AB of a parallelogram ABCD is produced to any point P.
A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see the following figure).
Show that ar (ABCD)=ar(PBQR). [4 MARKS]

Answer 1

Construction : 1 Mark
Concept : 1 Mark
Proof : 2 Marks


Let us join AC and PQ.
$\Delta ACQ$ and $\Delta AQP$ are on the same base AQ and between the same parallels AQ and CP.
$\therefore ar\left(\Delta ACQ\right)=ar\left(\Delta APQ\right)$

$ar\left(\Delta ACQ\right)-ar\left(\Delta ABQ\right)=ar\left(\Delta APQ\right)-ar\left(\Delta ABQ\right)$
[Subtracting ar $\left(\Delta ABQ\right)$ from both sides]
$\Rightarrow ar\left(\Delta ABC\right)=ar\left(\Delta QBP\right).......\left(1\right)$

Since AC and PQ are diagonals of parallelograms ABCD and PBQR respectively,
$ar\left(\Delta ABC\right)=\frac{1}{2}ar\left(ABCD\right)...\left(2\right)$

$ar\left(\Delta QBP\right)=\frac{1}{2}ar\left(PBQR\right)...\left(3\right)$

From equations (1), (2), and (3), we obtain
$\frac{1}{2}ar\left(ABCD\right)=\frac{1}{2}ar\left(PBQR\right)$

$\Rightarrow ar\left(ABCD\right)=ar\left(PBQR\right)$