Question

$ABCD$ is a parallelogram and $EFCD$ is a rectangle $AL\bot DC$. Prove that area of$ABCD=DC\times AL$.

Answer 1

Hint:Here, first we have to draw the figure with the given data. We know that a rectangle is also a parallelogram. Therefore, $ABCD$ and $EFCD$ are two parallelograms on the same base and between the same parallels. Now apply the theorem that two parallelograms with the same base and between the same parallels are equal in area.

Complete step-by-step answer:
First, we have to draw the figure with the given data.

According to the question we have, $ABCD$ is a parallelogram and $EFCD$ is a rectangle where $AL\bot DC$.
Now, we have to prove that:
$Area\text{ }of\text{ }ABCD=DC\times AL$
Here, $ABCD$ is a parallelogram, a parallelogram is a quadrilateral whose opposite sides are parallel.
Therefore, we can say that:
$\begin{align}
  & AB\parallel DC\text{ }........\text{ (1)} \\
 & AD\parallel BC \\
\end{align}$
Given that $EFCD$ is a rectangle, but we know that a rectangle is also a parallelogram. Therefore the opposite sides are parallel to each other. i.e. we get
$\begin{align}
  & EF\parallel DC\text{ }.......\text{ (2)} \\
 & ED\parallel FC \\
\end{align}$
From equation (1) and equation (2) we get:
$AB\parallel EF$
Therefore, we can say that $ABCD$ and $EFCD$ are two parallelograms with the same base $CD$ and between the same parallels $EB$ and $DC$.
Hence, by a theorem we can say that the parallelograms with the same base and between the same parallels are equal in area.
So, we will get:
$Area\text{ }of\text{ }ABCD=Area\text{ }of\text{ }EFCD$
We know that $EFCD$ is a rectangle. Therefore, we can write:
$\begin{align}
  & Area\text{ }of\text{ }EFCD=Length\times Breadth \\
 & Area\text{ }of\text{ }EFCD=DC\times FC\text{ }.......\text{ (3)} \\
\end{align}$
We are given that $AL\bot DC$. Therefore, we can say that $AFCL$ is a rectangle.
We know that in a rectangle opposite sides are parallel and equal.
Hence, from the figure we can say that $FC=AL$.
 By substituting this value in equation (3) we get:
$Area\text{ }of\text{ }EFCD=DC\times AL$
Since, $Area\text{ }of\text{ }ABCD=Area\text{ }of\text{ }EFCD$, we can say that:
$Area\text{ }of\text{ }ABCD=DC\times AL$
Hence the proof.

Note: In other way, we can say that area of the parallelogram = Base $\times $ Height. Here the base is $CD$ and we are given that $AL\bot DC$, i.e. $AL$ is the height of the parallelogram. Now, we can write:
$Area\text{ }of\text{ }ABCD=DC\times AL$