Chapter 9.2, Problem 23WE

To determine

The radius of the circle inscribed in the circle

Answer to Problem 23WE

The radius of the circle is $2\sqrt{2}$

Explanation of Solution

Given information:

The given figure is,

  

In any triangle

  $s=\frac{a+b+c}{2}$ where a, b, and c are the sides of the triangle

  $\begin{array}{c}s=\frac{12+12+8}{2}\\ s=\frac{32}{2}\\ s=16\end{array}$

Let the radius of circle be r.

The radius of the circle inscribed in the triangle is given by the formula

  $\begin{array}{c}{r}^2=\frac{\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s}\\ r^2=\frac{\left(16-12\right)\left(16-12\right)\left(16-8\right)}{16}\\ r^2=\frac{4\times 4\times 8}{16}\\ r=\sqrt{8}\\ r=2\sqrt{2}\end{array}$

Since length is positive so neglecting the negative value.

Therefore the radius of the circle is $2\sqrt{2}$