In $Fig.$, 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.
In $Fig.$, 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.
False
Given diameter of circle is $d.$
$\therefore$ Diagonal of inner square $=$ Diameter of circle $= d$
Let side of inner square $EFGH$ be $x.$
$\therefore$ In right angled $\triangle EFG$
$E G^{2}=E F^{2}+F G^{2} \quad$ [by Pythagoras theorem]
$\Rightarrow d^{2}=x^{2}+x^{2}$
$\Rightarrow d^{2}=2 x^{2} \Rightarrow x^{2}=\frac{d^{2}}{2}$
But side of the outer square $ABCS$ $=$ Diameter of circle $= d$
$\therefore \quad$ Area of outer square $= d ^{2}$
Hence, area of outer square is not equal to four times the area of the inner square.
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