Question

Question 7
ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F.
If ar (DFB) = 3 $c{m}^2$, find the area of the parallelogram ABCD.

Answer 1

Given,
ABCD is a parallelogram and CE = BC i.e., C is the midpoint of BE.
Also $ar\left(\Delta DFB\right)=3c{m}^2$

Now, $\Delta ADF$ and $\Delta DFB$ are on the same base DF and between parallels CD and AB.
Then,
$ar\left(\Delta ADF\right)=ar\left(\Delta DFB\right)=3c{m}^2$ ….(i)

In $\Delta ABE$, by the converse of mid-point theorem,
EF = AF [since, C is mid-point of BE] …(ii)

In $\Delta ADF$ and $\Delta ECF$,
$\angle AFD=\angle CFE$ [vertically opposite angles]
AF = EF [from Eq. (ii)]

$\angle DAF=\angle CEF$
[BE || AD and AE is transversal, then alternate interior angles are equal]

$\Delta ADF\cong \Delta ECF$ [by ASA congruence rule]
Then,
$ar\left(\Delta ADF\right)=ar\left(\Delta CFE\right)$ ...(iii)
[since, congruent figures have equal area]

$ar\left(\Delta CFE\right)=ar\left(\Delta ADF\right)=3c{m}^2$ ...(iv)

Now, $ar\left(\Delta BDC\right)=ar\left(\Delta DFB\right)+ar\left(\Delta BFC\right)$
$=3+3=6c{m}^2$ [from Eqs. (i) and (iv)]

We know that, diagonal of a parallelogram divides it into two congruent triangles of equal areas.

$\therefore$ Area of parallelogram ABCD
$=2\times Areaof\Delta BDC$
$=2\times 6=12c{m}^2$

Hence, the area of parallelogram ABCD is $12c{m}^2$.