Chapter 5.6, Problem 32E

Solution

To find:The area of a regular nonagon inscribed in a circle.

The area of a regular nonagon inscribed in a circle is $289.2$ square inch.

Given information:

The radius of the circle is $10\text{inch}$ .

Formula used:

Law of Cosines is

  $c^2=a^2+b^2-2\left(a\right)\left(b\right)\cos C$

Area of triangle using Heron’s formula is

  $\text{Area}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$ where $s$ is the semi-perimeter such that $s=\frac{a+b+c}{2}$ .

Calculation:

An inscribed nonagon is made of $9$ isosceles triangle. The angle of vertex angle of each angle is $\frac{360}{9}=40°$ . And equal sides of the triangle is equal to the radius of the circle. The length of the third side can be determined using the Cosine Law.

Substitute $10$ for $a$ , $10$ for $b$ , and included angle $C$ is $40°$ in Cosine Law

  $\begin{array}{c}{c}^2=a^2+b^2-2\left(a\right)\left(b\right)\cos C\\ c=\sqrt{{10}^2+{10}^2-2\left(10\right)\left(10\right)\cos 40}\\ =6.84\end{array}$

Calculate the semi-perimeter

Substitute $10$ for $a$ , $10$ for $b$ , $6.84$ for $c$ .

  $\begin{array}{c}s=\frac{a+b+c}{2}\\ =\frac{10+10+6.84}{2}\\ =13.42\end{array}$

Calculate the area of triangle using Heron’s formula

Substitute $10$ for $a$ , $10$ for $b$ , $6.84$ for $c$ , and $13.42$ for $s$

  $\begin{array}{c}{\text{Area}}_{\text{triangle}}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\\ =\sqrt{13.42\left(13.42-10\right)\left(13.42-10\right)\left(13.42-6.84\right)}\\ =32.13\end{array}$

Calculate the area of the nonagon

  $\begin{array}{c}{\text{Area}}_{\text{nonagon}}{\text{=9Area}}_{\text{triangle}}\\ \text{=9×32}\text{.13}\\ \text{=289}\text{.2}\end{array}$

Therefore, the area of a regular nonagon inscribed in a circle is $289.2$ square inch.