Chapter 16.6, Problem 3PS

To determine

To find: the inradius and the circumradius of triangle having three sides.

Answer to Problem 3PS

Inradius= $\frac{A}{s}=\frac{\sqrt{445}}{13}$ and Circumradius $=\frac{abc}{4A}=\frac{6\times 8\times 12}{4\left(\sqrt{455}\right)}=\frac{144\sqrt{455}}{455}.$

Explanation of Solution

Given:

The side of the triangle are:

  $6,8\text{ and 12}$ .

Concept used:

Heron’s formula:

  $A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}.$

Where $s$ is semi perimeter:

  $s=\frac{\text{perimeter}}{2}$ .

Inradius= $\frac{A}{s}$ .

Circumradius= $\frac{abc}{4A}$ .

Calculation:

According to the given side of the triangle are:

  $6,8\text{ and 12}$ .

Semi perimeter can be calculated as:

  $\begin{array}{l}s=\frac{\text{perimeter}}{2}.\\ s=\frac{\text{6+8+12}}{2}=13.\end{array}$

Using heron’s formula area of triangle can be calculated as:

  $A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}.$

Where $a,b\text{ and }c$ are the three side of the triangle.

  $A=\sqrt{13\left(13-6\right)\left(13-8\right)\left(13-12\right)}=\sqrt{455}.$

Inradius= $\frac{A}{s}=\frac{\sqrt{445}}{13}$ .

Circumradius $=\frac{abc}{4A}=\frac{6\times 8\times 12}{4\left(\sqrt{455}\right)}=\frac{144\sqrt{455}}{455}.$

Hence, Inradius= $\frac{A}{s}=\frac{\sqrt{445}}{13}$ and Circumradius $=\frac{abc}{4A}=\frac{6\times 8\times 12}{4\left(\sqrt{455}\right)}=\frac{144\sqrt{455}}{455}.$