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 :
A. 1 : 2
B. 1 : 3
C. 1 : 4
D. 2 : 3
Answer: Option B
Solution(By Examveda Team)
Let AD = x and DC = yThen,
$$AE = ED = \frac{x}{2}{\text{ and}}$$ $$DE = FC = \frac{y}{2}$$
$${\text{Area }}\left( {\vartriangle EDF} \right) = \left( {\frac{1}{2} \times \frac{x}{2} \times \frac{x}{2}} \right)$$ $$ = \frac{{xy}}{8}$$
$${\text{Area }}\left( {{\text{trap}}{\text{. }}AEFC} \right) = $$ $${\text{Area }}\left( {\vartriangle ADC} \right) - $$ $${\text{Area}}\left( {\vartriangle EDF} \right)$$
$$\eqalign{ & {\text{Area }}\left( {{\text{trap}}{\text{. }}AEFC} \right) = \frac{{xy}}{2} - \frac{{xy}}{8} \cr & {\text{Area }}\left( {{\text{trap}}{\text{. }}AEFC} \right) = \frac{{3xy}}{8} \cr} $$
∴ Required ratio :
$$\eqalign{ & = \frac{{xy}}{8}:\frac{{3xy}}{8} \cr & = 1:3 \cr} $$
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