In the square ABCD above, each side is 10 , so the area of half of the square i.e. area of triangle $\(A B C=\frac{1}{2}(\text { Side })^2=\frac{1}{2}(10)^2=50\)$

As $\(\mathrm{E} \& \mathrm{~F}\)$ are the midpoints of $\(\mathrm{AC} \& \mathrm{BC}\)$ respectively, so we get $\(\mathrm{EF}=\frac{1}{2}(\mathrm{AB})=\frac{1}{2}(10)=5\)$ (The line joining the mid-points of the two

sides of a triangle is parallel to the third side \& is half of it) Now, the area of triangle $\(\mathrm{ECF}=\frac{1}{2} \times\)$ Base $\(\times\)$ Height $\(=\frac{1}{2} \times \mathrm{EC} \times \mathrm{EF}=\frac{1}{2} \times 5 \times 5=12.5\)$

(As C is a mid - point of AC, EC must be half of AC)

Finally the area of the shaded portion $\(=\)$ area triangle $\(\mathrm{ABC}-\)$ area of triangle $\(\mathrm{ECF}=50-12.5=\)$ 37.5

Hence the answer is (B).