Question

(i) ABCD is a parallelogram, E and F are the midpoints of AB and CD respectively then what will be the ratio of areas of ($\square$ AEFD) and ($\square$EBCF).

(ii) If A($\square$ABCD) is 72 cm${}^2$, then what will be the area of $\square$EBCF ?

Answer 1

(i) In parallelogram ABCD

AE || DF,

$\Rightarrow AB=DC$ (opposite sides of a parallelogram )

$\Rightarrow \frac{1}{2}\times AB=\frac{1}{2}\times DC$

$\Rightarrow AE=DF$

Therefore, AEFD is a parallelogram.

Similarly we can prove that EBFC is a parallelogram.

We know that parallelograms lying on same/equal bases and between same parallels are equal in area.


AE = EB (Since E is the midpoint of AB)

Parallelograms AEFD and EBCF lie on equal bases AE and EB and same parallels AB and DC, therefore area of AEFD and EBCF are equal.

Ratio of areas of AEFD and EBCF is 1:1.

(ii) It has been proved that A$\left(\square AEFD\right):A\left(\square EBCF\right)=1:1$

$\therefore A\left(\square EBCF\right)=\frac{1}{2}\times A\left(\square ABCD\right)$

$A\left(\square EBCF\right)=\frac{1}{2}\times 72$

$A\left(\square EBCF\right)=36{\text{cm}}^2$