Chapter 16.6, Problem 2PS

To determine

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

Answer to Problem 2PS

Inradius= $\frac{A}{s}=\frac{180}{45}=4.$ and Circumradius $=\frac{abc}{4A}=\frac{9\times 40\times 41}{4\left(180\right)}=\frac{41}{2}.$

Explanation of Solution

Given:

The side of the triangle are:

  $9,40\text{ and 41}$ .

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:

  $9,40\text{ and 41}$ .

Semi perimeter can be calculated as:

  $\begin{array}{l}s=\frac{\text{perimeter}}{2}.\\ s=\frac{\text{9+40+41}}{2}=45.\end{array}$

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.

  $A=\sqrt{45\left(45-9\right)\left(45-40\right)\left(45-41\right)}=180.$

Inradius= $\frac{A}{s}=\frac{180}{45}=4.$

Circumradius $=\frac{abc}{4A}=\frac{9\times 40\times 41}{4\left(180\right)}=\frac{41}{2}.$

Hence, Inradius= $\frac{A}{s}=\frac{180}{45}=4.$ and Circumradius $=\frac{abc}{4A}=\frac{9\times 40\times 41}{4\left(180\right)}=\frac{41}{2}.$