Chapter 16.6, Problem 1PS

To determine

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

Answer to Problem 1PS

Inradius= $\frac{A}{s}=\frac{2\sqrt{14}}{7}$ and Circumradius $=\frac{abc}{4A}=\frac{3\times 5\times 6}{4\left(2\sqrt{14}\right)}=\frac{45\sqrt{14}}{56}.$

Explanation of Solution

Given:

The side of the triangle are:

  $3,5\text{ and }6$ .

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:

  $3,5\text{ and }6$ .

Semi perimeter can be calculated as:

  $\begin{array}{l}s=\frac{\text{perimeter}}{2}.\\ s=\frac{\text{3+5+6}}{2}=7.\end{array}$

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.

  $A=\sqrt{7\left(7-3\right)\left(7-5\right)\left(7-6\right)}=2\sqrt{4}.$

Inradius= $\frac{A}{s}=\frac{2\sqrt{14}}{7}$ .

Circumradius $=\frac{abc}{4A}=\frac{3\times 5\times 6}{4\left(2\sqrt{14}\right)}=\frac{45\sqrt{14}}{56}.$

Hence, Inradius= $\frac{A}{s}=\frac{2\sqrt{14}}{7}$ and Circumradius $=\frac{abc}{4A}=\frac{3\times 5\times 6}{4\left(2\sqrt{14}\right)}=\frac{45\sqrt{14}}{56}.$