As the area of the shaded portion is $\(75 \%\)$ of the area of the square, the area of the remaining square i.e. the un-shaded portion must be $\(100 \%-75 \%=25 \%\)$ of the area of the square i.e. it should be $\(25 \%$ of $(15)^2=\frac{1}{4} \times 225=\frac{225}{4}\)$..

Now, the un-shaded portion is nothing but two identical right triangles having perpendicular sides $\((15-\mathrm{x})\)$ each, so the area of the un-shaded portion is $\(2\left(\frac{1}{2} \times(15-\mathrm{x}) \times(15-\mathrm{x})\right)=(15-\mathrm{x})^2\)$


Using (1) \& (2), we get $\((15-x)^2=\frac{225}{4}\)$ i.e. $\((15-x)=\frac{15}{2} \Rightarrow x=\frac{15}{2}=7.5\)$

Hence the answer is (C).