ABCD is a square. E is the mid-point of BC and F is the mid-point of CD. The ratio of the area of triangle AEF to the area of the square ABCD is :

Answer: Option D

Let the length of side of the square be a units
Then,
$$BE=EC=DF=FC=\frac{a}{2}$$
$$\begin{array}{rl}& AE=\sqrt{{\left(AB\right)}^2+{\left(BE\right)}^2}\\ & =\sqrt{a^2+{\left(\frac{a}{2}\right)}^2}\\ & =\sqrt{a^2+\frac{a^2}{4}}\\ & =\sqrt{\frac{5a^2}{4}}\\ & =\frac{\sqrt{5}a}{2}\end{array}$$
Similarly, $$AF=\frac{\sqrt{5}a}{2}$$
$$\begin{array}{rl}& EF=\sqrt{{\left(CE\right)}^2+{\left(CF\right)}^2}\\ & =\sqrt{{\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{2}\right)}^2}\\ & =\sqrt{\frac{2a^2}{4}}\\ & =\frac{a}{\sqrt{2}}\\ & EX=\frac{1}{2}EF=\frac{a}{2\sqrt{2}}\\ & AX=\sqrt{{\left(AE\right)}^2-{\left(EX\right)}^2}\\ & =\sqrt{{\left(\frac{\sqrt{5}a}{2}\right)}^2-{\left(\frac{a}{2\sqrt{2}}\right)}^2}\\ & =\sqrt{\frac{5a^2}{4}-\frac{a^2}{8}}\\ & =\sqrt{\frac{9a^2}{8}}\\ & =\frac{3a}{2\sqrt{2}}\end{array}$$
$$\begin{array}{rl}& \therefore \text{Area of }\left(△AEF\right):\\ & =\frac{1}{2}\times EF\times AX\\ & =\frac{1}{2}\times \frac{a}{\sqrt{2}}\times \frac{3a}{2\sqrt{2}}\\ & =\frac{3a^2}{8}\end{array}$$
$$\begin{array}{rl}& \text{Required ratio :}\\ & =\frac{3a^2}{8}:a^2\\ & =3:8\end{array}$$