Question

Find the area of a circle which encloses a square of side 10 cm.

Answer 1

Let ABCD be the square of side 10 cm.

Now in right-angled triangle ABC,
Using Pythagoras Theorem,
we get,

$A{B}^2+B{C}^2=A{C}^2\text{, AC being the diagonal.}$

$A{C}^2={10}^2+{10}^2$
$\Rightarrow A{C}^2=100\Rightarrow AC=10\sqrt{2}cm$

Now, AC is also the diameter of the circle

So $D=10\sqrt{2}cm$

$2r=10\sqrt{2}\Rightarrow r=5\sqrt{2}cm$

So, the radius of circle is $5\sqrt{2}cm$

Hence, the area of circle = $\pi r^2=\pi (5\sqrt{2}{)}^2$ = $50\pi c{m}^2$