squre

Let the side length of the bigger square be $1$ . Then the diagonal is $\sqrt{2}$ . Due to symmetry, the length of the smaller square is $\dfrac{1}{4}$ of the diagonal. So the length is $\dfrac{\sqrt{2}}{4}$ . Hence, the required percentage is $\dfrac{\left(\frac{\sqrt{2}}{4}\right)^2}{1^2} \times 100\% = \boxed{12.5\%}$

number of shaded triangle is 2

number of unshaded triangle is 16

so the ratio is $\frac{2}{16}$ = 12.5%

• Assuming the side length of external square is $2a$ , the area of external square is $4a^2$ .

• Then the diagonal length of little square is $a$ , with the area of the little square is $\frac{a^2}{2}$ .

• Now, $\frac{\frac{a^2}{2}}{4a^2} = \frac{1}{8} = 0.125 = \boxed{12.5}$ %.