In the figure,a circle is inscribed in a square of side $5 \, cm$ and another circle is circumscribing the square. Is it true to say that the area of the outer circle is two times the area of the inner circle? Give reasons for your answer.

(A) Yes,it is true.
Let the side of the square be $a = 5 \, cm$.
The diameter of the inner circle is equal to the side of the square,so $d_{inner} = a = 5 \, cm$. The radius is $r = \frac{5}{2} \, cm$.
The area of the inner circle is $A_{inner} = \pi r^2 = \pi \left(\frac{5}{2}\right)^2 = \frac{25\pi}{4} \, cm^2$.
The diameter of the outer circle is equal to the diagonal of the square,so $d_{outer} = a\sqrt{2} = 5\sqrt{2} \, cm$. The radius is $R = \frac{5\sqrt{2}}{2} \, cm$.
The area of the outer circle is $A_{outer} = \pi R^2 = \pi \left(\frac{5\sqrt{2}}{2}\right)^2 = \pi \left(\frac{25 \times 2}{4}\right) = \frac{50\pi}{4} = \frac{25\pi}{2} \, cm^2$.
Comparing the two areas: $\frac{A_{outer}}{A_{inner}} = \frac{25\pi / 2}{25\pi / 4} = \frac{4}{2} = 2$.
Thus,the area of the outer circle is indeed two times the area of the inner circle.