Question

ABCD is a parallelogram and EFCD is a rectangle AL $\perp$ DC prove that ar(ABCD) = $DC$ x $AL$

Answer 1

Given that:- $ABCD$ is a parallelogram and $EFCD$ is a rectangle.

To prove:- $ar\left(ABCD\right)=DC\times AL$

Proof:-

$\therefore AB\parallel CD.....\left(1\right)[\because \text{Opposite sides of parallelogram are equal}]$

As given that $EFCD$ is a rectangle and we know that rectangle is also a parallelogram.

$\therefore EF\parallel CD.....\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we hahve

$AB\parallel EF$

Now, $ABCD$ and $EFCD$ are two parallelograms with same base $CD$ and between tha same parallels $EB$ and $CD$.

As we know that parallelogram with same base and in between same parallels are equal in area.

$\therefore ar\left(ABCD\right)=ar\left(EFCD\right)$

As from the fig.

$ar\left(EFCD\right)=DC\times FC$

$\Rightarrow ar\left(ABCD\right)=DC\times FC.....\left(3\right)$

$AL\perp CD\left[\text{Given}\right]$

$\therefore AFCL$ is a also a rectangle.

$\therefore AL=FC.....\left(4\right)$

From ${eq}^n\left(3\right)\&\left(4\right)$, we get

$ar\left(ABCD\right)=DC\times AL$

Hence proved that $ar\left(ABCD\right)=DC\times AL$.