Question 107

The ratio of inradius and circumradius of a square is :

Solution

Let the side of square be $$a$$

=> Inradius(OA) = $$\frac{a}{2}$$ = AB

In $$\triangle$$OAB

=> OB = $$\sqrt{(OA)^2 + (AB)^2}$$

=> OB = $$\sqrt{(\frac{a}{2})^2 + (\frac{a}{2})^2}$$

=> OB = $$\sqrt{\frac{a^2}{4} + \frac{a^2}{4}} = \sqrt{\frac{a^2}{2}}$$

=> OB = $$\frac{a}{\sqrt{2}}$$

To find : $$\frac{OA}{OB}$$

= $$\frac{\frac{a}{2}}{\frac{a}{\sqrt{2}}}$$

= $$\frac{1}{\sqrt{2}}$$

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