Question

If the distance between incenter and one of the excenter of an equilateral triangle is $4$ units, then find the inradius of the triangle.

• A. $1$ units
• B. $2$ units
• C. $3$ units
• D. $4$ units

Answer 1

Answer: (A)

$1$ units

Let $I$ be the incenter of the equilateral triangle and $I_1$ be the excenter of the equilateral triangle.

Given that the distance between incenter and one of the excenter of a equilateral triangle is $D=4$ units.

We know that the distance between between incenter and one of the excenter is given by $D=4Rsin\left(\frac{A}{2}\right)$ where $R$ is the circumradius.

Since we are given that the triangle is equilateral, we get $A={60}^{\circ }$.

$D=4Rsin\left(\frac{{60}^{\circ }}{2}\right)$

$\Rightarrow 4=4Rsin\left({30}^{\circ }\right)$

$\Rightarrow 1=R\times \frac{1}{2}$

$\Rightarrow R=2$ units

We know that $R=2r$ where $R$ is circumradius and $r$ is inradius.

$\Rightarrow 2=2r$

$\Rightarrow r=1$ units.

Hence the inradius of the triangle is $1units.$