In the figure,a square is inscribed in a circle of diameter $d$ and another square is circumscribing the circle. Is the area of the outer square four times the area of the inner square? Give reasons for your answer.

(N/A) The statement is False.
Given the diameter of the circle is $d$.
$1$. For the inner square inscribed in the circle,the diagonal of the square is equal to the diameter of the circle.
Diagonal $= d$.
Let the side of the inner square be $x$.
Using the property of a square,the diagonal is $\sqrt{2} \times \text{side}$.
So,$\sqrt{2}x = d \Rightarrow x = \frac{d}{\sqrt{2}}$.
Area of the inner square $= x^2 = (\frac{d}{\sqrt{2}})^2 = \frac{d^2}{2}$.
$2$. For the outer square circumscribing the circle,the side of the square is equal to the diameter of the circle.
Side $= d$.
Area of the outer square $= d^2$.
$3$. Comparing the areas:
Ratio $= \frac{\text{Area of outer square}}{\text{Area of inner square}} = \frac{d^2}{d^2/2} = 2$.
Thus,the area of the outer square is twice the area of the inner square,not four times.