In square ABCD above, if DE = EB and DF = FC, then the area of the sha

Let the side of square be \(= 2\). Therefore, \(AB = BC = CD = AD = 2\)

Area of square \(ABCD = 2^2 = 4\)

Given; \(DF = FC\). Therefore; \(DF = FC = 1\)

Diagonal of square \(= (side) \sqrt{2}\). Therefore, \(BD = 2\sqrt{2}\)

\(DE = EB = \frac{(2\sqrt{2})}{ 2}\) \(=> DE = EB = \sqrt{2}\)

Joining \(EC\) to \(A\), we get \(AC\) as another diagonal. Therefore, \(EC = DE = EB\). Hence \(EC = \sqrt{2}\)

Using Pythagorean theorem in \(\triangle EFC\), we get;

\(EF^2 + FC^2 = EC^2\)

\(EF^2 = EC^2 - FC^2\)

\(EF^2 =\)\(\sqrt{2^2}\) \(- 1^2\)

\(EF^2 = 2 - 1 = 1\)

\(EF = \sqrt{1} = 1\)

Shaded region is \(\triangle EFC\), Which is right angled triangle.

Area of shaded region \(= \frac{1}{2} EF * FC = \frac{1}{2} * 1*1 = \frac{1}{2}\)

Required fraction = Area of the shaded region / Area of square

Required fraction \(= \frac{(1/2)}{4} = \frac{1}{8}\)

Answer (B)...