Question

Question 3
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.

Answer 1

Given diameter of circle is d

$\therefore$ Diagonal of inner square = Diameter of circle = d

Let side of inner square EFGH be x

$E{G}^2=E{F}^2+F{G}^2$ [ by Pythagoras theorem]

$\Rightarrow d^2=x^2+x^2$

$\Rightarrow d^2=2x^2\Rightarrow x^2=\frac{d^2}{2}$

$\therefore$ Area of inner square EFGH $=(side{)}^2=x^2=\frac{d^2}{2}$

But side of the outer square ABCD = Diameter of circle = d

$\therefore$ Area of outer square $=d^2$

Hence, area of outer squares is not equal to four times the area of the inner square.