Question

In the following figure, $ABCD$ is a parallelogram and $EFCD$ is a rectangle. Also, $AL\perp DC$. Prove that
(i) $ar\left(ABCD\right)=ar\left(EFCD\right)$
(ii) $ar\left(ABCD\right)=DC\times AL$.

Answer 1

$ABCD$ is a parallelogram.

$\therefore$ $AB\parallel CD$ [ Opposite sides of parallelogram are parallel ]

We know that a rectangle is also a parallelogram,

so $EFCD$ is also a parallelogram.

$\therefore$ $EF\parallel CD$ [ Opposite sides of parallelogram are parallel ]

Since, $AB\parallel CD$ and $EF\parallel CD$

$\Rightarrow$$EB\parallel CD$

Now, $ABCD$ and $EFDC$ are two parallelograms with same base $CD$ and between the same parallels $EB$ and $CD$

We know that parallelograms with same base and between the same parallels are equal in area.

$\therefore$ $ar\left(ABCD\right)=ar\left(EFCD\right)$ [ Hence proved ]

$ABCD$ is a parallelogram with base $DC$ and altitude $AL$

Now,

Area of a parallelogram $=$ Base $\times$ Corresponding altitude.

$\therefore$ $ar\left(ABCD\right)=DC\times AL$ [ Hence proved ]