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 :

Solution(By Examveda Team)

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