Chapter 7.7, Problem 32E

To determine

To find: the radius of the inscribed circle for the triangle with vertices.

Answer to Problem 32E

  $\frac{60}{2\sqrt{10}+3\sqrt{5}+5}$

Explanation of Solution

Given information:

Let the vertices be $A,B\text{ and }C$ .

  $A\left(0,0\right),B\left(10,0\right)\text{ and }C\left(4,12\right)$

The intersection of angle bisectors is the center of the inscribed circle of the triangle.

The inscribed circle is tangent to each side of triangle.

Calculation:

Evaluate the equation for line $AB$ as shown below.

  $\begin{array}{c}y-y_B=\frac{y_B-y_A}{x_B-x_A}\left(x-x_B\right)\\ y-0=\frac{0-0}{10-0}\left(x-10\right)\left[\text{Put the point coordinates}\right]\\ y=0\end{array}$

So, the value of $A,B\text{ and }C$ is $0,0\text{ and 0}$ .

Evaluate the equation for line $AC$ as shown below.

  $\begin{array}{c}y-y_C=\frac{y_C-y_A}{x_C-x_A}\left(x-x_C\right)\\ y-\left(12\right)=\frac{12-0}{4-0}\left(x-\left(4\right)\right)\left[\text{Put the point coordinates}\right]\\ y-12=3x-12\\ 3x-y=0\end{array}$

So, the value of $A,B\text{ and }C$ is $\text{3,}-1\text{ and 0}$ .

Evaluate the equation for line $BC$ as shown below.

  $\begin{array}{c}y-y_B=\frac{y_C-y_B}{x_C-x_B}\left(x-x_B\right)\\ y-0=\frac{12-0}{4-10}\left(x-10\right)\left[\text{Put the point coordinates}\right]\\ y=-2\left(x-10\right)\\ y=-2x+20\\ -2x-y+20=0\end{array}$

So, the value of $A,B\text{ and }C$ is $-2\text{,}-1\text{ and 20}$ .

Use the distance formula to evaluate bisector of $\angle A$ as follows:

  $\begin{array}{c}\frac{y}{-\sqrt{0^2+1^2}}=\frac{3x-y}{-\sqrt{3^2+1^2}}\\ y=\frac{3x-y}{\sqrt{10}}\\ \sqrt{10}y=3x-y\\ 3x-\left(\sqrt{10}+1\right)y=0\\ or\\ x=\frac{\left(\sqrt{10}+1\right)y}{3}\mathrm{...}\left(1\right)\end{array}$

Use the distance formula to evaluate bisector of $\angle B$ as follows:

  $\begin{array}{c}\frac{y}{-\sqrt{0^2+1^2}}=\frac{-2x-y+20}{-\sqrt{3^2+1^2}}\\ y=\frac{-2x-y+20}{\sqrt{5}}\\ \sqrt{5}y=-2x-y+20\\ 2x+\left(\sqrt{5}+1\right)y-20=0\\ or\\ x=\frac{-\left(\sqrt{5}+1\right)y+20}{2}\mathrm{...}\left(2\right)\end{array}$

Compare equation (1) and (2) and find the point of intersection as follows:

  $\begin{array}{c}\frac{\left(\sqrt{10}+1\right)y}{3}=\frac{-\left(\sqrt{5}+1\right)y+20}{2}\\ 2\left(\sqrt{10}+1\right)y=3\left(-\left(\sqrt{5}+1\right)y+20\right)\\ \left(2\sqrt{10}+2\right)y=60-\left(3\sqrt{5}+3\right)y\\ \left(2\sqrt{10}+2+3\sqrt{5}+3\right)y=60\\ \left(2\sqrt{10}+3\sqrt{5}+5\right)y=60\\ y=\frac{60}{2\sqrt{10}+3\sqrt{5}+5}\end{array}$

The distance between the $y$ -coordinate and $AB$ gives the inradius i.e. the radius of the inscribed circle. Thus, the radius is $\frac{60}{2\sqrt{10}+3\sqrt{5}+5}$ .