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).