In the adjoining figure, ABCD is a square. Find the ratio between

(i) the circumferences

(ii) the areas of the incircle and the circumcircle of the square.

(i) Let side of the square be 2a units.

From figure,

AD = Diameter of incircle.

Radius of incircle (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{2a}{2}$ = a units.

In right angle triangle ABC,

Using pythagoras theorem,

AC² = AB² + BC²

AC² = (2a)² + (2a)²

AC² = 4a² + 4a²

AC² = 8a²

AC = $\sqrt{8a^2} = 2\sqrt{2}a$ units.

From figure,

AC is the diameter of circumcircle and AO is radius.

AO (R) = $\dfrac{\text{Diameter}}{2} = \dfrac{2\sqrt{2}a}{2} = \sqrt{2}$a units.

$\text{Ratio between circumference } = \dfrac{\text{Circumference of incircle}}{\text{Circumference of circumcircle}} \\[1em] = \dfrac{πr}{πR} \\[1em] = \dfrac{r}{R} \\[1em] = \dfrac{a}{\sqrt{2}a} \\[1em] = \dfrac{1}{\sqrt{2}} = 1 : \sqrt{2}.$

Hence, ratio between circumferences = $1 : \sqrt{2}$.

(ii)

$\text{Ratio between areas } = \dfrac{\text{Area of incircle}}{\text{Area of circumcircle}} \\[1em] = \dfrac{πr^2}{πR^2} \\[1em] = \dfrac{r^2}{R^2} \\[1em] = \dfrac{a^2}{(\sqrt{2}a)^2} \\[1em] = \dfrac{a^2}{2a^2} = 1 : 2.$

Hence, ratio between areas = 1 : 2.