Solution:
The given statement is False
Explanation:
From the fig, Let the Diameter of the circle = d
Therefore,
Diagonal of inner square (EFGH) = Side of the outer square (ABCD) = Diameter of circle = d
Let us take side of inner square EFGH be a
Now assume EFG is a right angled triangle,
Then, \[{{(EG)}^{2}}={{(EF)}^{2}}+{{(FG)}^{2}}\]
By Pythagoras theorem)
i.e., \[{{d}^{2}}={{a}^{2}}+{{a}^{2}}\]
⇒ \[{{d}^{2}}=2{{a}^{2}}\]
⇒ \[{{a}^{2}}={{d}^{2}}/2\]
Therefore, Area of inner circle = \[{{a}^{2}}={{d}^{2}}/2\]
Also, Area of outer square (ABCD) = \[{{d}^{2}}\]
Therefore, the area of the outer circle(ABCD) is only two times the area of the inner circle(EFGH).
Thus, the given statement is false.