Mensuration Practice Test 7

R → circumradius of the circle
r → inradius of the circle

Let us suppose that the side of a square = 4 cm.
In ΔACD by Pythagoras theorem,
R = $\frac{1}{2}\sqrt{2{\mathit{x}}^2}$ = $\frac{1}{2}\mathit{S}\sqrt{2}$
R = $\frac{4}{\sqrt{2}}$
r = $\sqrt{{\left(\frac{\mathit{S}}{\sqrt{2}}\right)}^2-{\left(\frac{\mathit{S}}{2}\right)}^2}$ = $\sqrt{\frac{16}{2}-\frac{16}{4}}$
= $\sqrt{8-4}$ = $\sqrt{4}$ = 2
Ratio of inradius and circumradius
= $2:\frac{4}{\sqrt{2}}$
2 : $2\sqrt{2}$
1 : $\sqrt{2}$