ABCD is a rectangle and E and F are the mid-points of AD and DC respectively. Then the ratio of the areas of EDF and AEFC would be :

Solution (By Examveda Team)

Let AD = x and DC = y

Then,
$$AE = ED = \frac{x}{2}{\text{ and}}$$     $$DE = FC = \frac{y}{2}$$
$${\text{Area }}\left( {\vartriangle EDF} \right) = \left( {\frac{1}{2} \times \frac{x}{2} \times \frac{x}{2}} \right)$$     $$ = \frac{{xy}}{8}$$
$${\text{Area }}\left( {{\text{trap}}{\text{. }}AEFC} \right) = $$     $${\text{Area }}\left( {\vartriangle ADC} \right) - $$     $${\text{Area}}\left( {\vartriangle EDF} \right)$$
$$\eqalign{ & {\text{Area }}\left( {{\text{trap}}{\text{. }}AEFC} \right) = \frac{{xy}}{2} - \frac{{xy}}{8} \cr & {\text{Area }}\left( {{\text{trap}}{\text{. }}AEFC} \right) = \frac{{3xy}}{8} \cr} $$
∴ Required ratio :
$$\eqalign{ & = \frac{{xy}}{8}:\frac{{3xy}}{8} \cr & = 1:3 \cr} $$