Question

The inradius of a regular tetrahedron of side length $a$ is

• A. $\sqrt{\frac{2}{3}}a$
• B. $\sqrt{\frac{3}{2}}a$
• C. $\frac{a}{\sqrt{6}}$
• D. $\frac{a}{2\sqrt{6}}$

Answer 1

The correct option is D $\frac{a}{2\sqrt{6}}$

Area of one face $=\frac{\sqrt{3}}{4}{a}^2$ ....... ($a$ is each side of tetrahedron)

$R=\frac{\sqrt{6}}{4}a$

$r=\frac{R}{3}=\frac{a}{\sqrt{24}}$

Hence, $\frac{a}{2\sqrt{6}}$ is the correct answer.