Question

\[ABCD\] is a parallelogram and \[EFCD\] is a rectangle. Also, \[AL \bot CD\]. Prove that,
A. \[ar(ABCD) = ar(EFCD)\]
B. \[ar(ABCD) = DC \times AL\]
     

Answer 1

Hint: We know that parallelograms with the same base and between the same parallels are equal in area.
The area of a parallelogram is equal to the product of the base and the altitude. By using the above two hints we solve the question to find the solution.

Complete step-by-step answer:
It is given that, \[ABCD\] is a parallelogram and \[EFCD\] is a rectangle and \[AL \bot CD\]. It means \[AL\] is perpendicular to \[CD\].
Since, \[ABCD\] is a parallelogram, opposite sides are parallel and equal.
Hence \[AB\parallel CD\].
Again, \[EFCD\] is a rectangle, so opposite sides are parallel and equal.
Hence\[EF\parallel CD\].
Since, \[AB\] is a parallel to \[CD\] and \[EF\] is a parallel to \[CD\].
\[AB\parallel EF\]
We can say that \[EB\] is a parallel to \[CD\].
We know that the theorem of equal area states that “parallelograms with the same base and between the same parallels are equal in area”.
Here, \[ABCD\] and \[EFCD\] are two parallelograms with the same parallels \[EB\] and \[CD\].
So, by the theorem of equal area we get,
\[ar(ABCD) = ar(EFCD)\]
Again, from given \[ABCD\] is a parallelogram with base \[CD\] and altitude \[AL.\]
We know that the area of a parallelogram is equal to the product of the base and the altitude.
Hence, by applying the above statement we have,
\[ar(ABCD) = DC \times AL\]
Hence, we have shown that,
\[ar(ABCD) = ar(EFCD)\]
\[ar(ABCD) = DC \times AL\]

Note: Since, a rectangle is also a parallelogram so, we can apply all the properties of a parallelogram on the rectangles. So, the equal area theorem of parallelogram holds for the rectangles also.