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 :

Answer: Option B

Let AD = x and DC = y

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