Question

Question 4 (i)
In the given figure, P is a point in the interior of a parallelogram ABCD. Show that
$\left(i\right)ar\left(APB\right)+ar\left(PCD\right)=\frac{1}{2}ar\left(ABCD\right)$

Answer 1

(i) Let us draw a line segment EF, passing through point P and parallel to line segment AB.

In parallelogram ABCD,
AB || EF (By construction) ... (1)
ABCD is a parallelogram.
$\therefore$ AD || BC (Opposite sides of a parallelogram)
$\Rightarrow$ AE || BF ... (2)
From equations (1) and (2), we obtain;
AB || EF and AE|| BF
Therefore, quadrilateral ABFE is a parallelogram.
It can be observed that $\Delta APB$ and parallelogram ABFE are lying on the same base AB
and between the same parallel lines AB and EF.
$\therefore Area\left(\Delta APB\right)=\frac{1}{2}Area\left(ABFE\right)...\left(3\right)$
Similarly, for ΔPCD and parallelogram EFCD,
$Area\left(\Delta PCD\right)=\frac{1}{2}Area\left(EFCD\right)...\left(4\right)$
Adding equations (3) and (4), we obtain,
$Area\left(\Delta APB\right)+Area\left(\Delta PCD\right)-\frac{1}{2}\left[Area\right(ABFE)+Area(EFCD\left)\right]$
$Area\left(\Delta APB\right)+Area\left(\Delta PCD\right)=\frac{1}{2}Area\left(ABCD\right)....\left(5\right)$