Chapter 5.6, Problem 33E

Solution

To find: The area of the regular hexagon circumscribed in a circle.

The area of the regular hexagon circumscribed in a circle is $498.8$ square inch.

Given information:

Radius of the circum circle is $12\text{inch}$ .

Formula used:

The equation for the area of a hexagon is:

  $\text{Area}=6\times \frac{1}{2}\times \left(\text{product}\text{of}\text{two}\text{sides}\right)\times \left(\text{sine}\text{of}\text{included}\text{angle}\right)$

Calculation:

An inscribed hexagon is made of $6$ equilateral triangle. The angle of vertex angle of each angle is $\frac{360}{6}=60°$ . Let $x$ be the side of the isosceles triangle where altitude is $12\text{in}$ and angle of vertex is $60°$ .

  

Calculate the length of the side $x$ using cosine function:

  $\cos 30=\frac{\text{adjacent}\text{side}}{\text{hypotenuse}\text{side}}$

Substitute $12$ for adjacent side and $x$ for hypotenuse side

  $\begin{array}{c}\cos 30=\frac{12}{x}\\ x=8\sqrt{3}\end{array}$

Calculate area of the hexagon as follows:

Substitute $8\sqrt{3}$ for the side of the equilateral triangle, and $60°$ for included angle

  $\begin{array}{c}\text{Area}=6\times \frac{1}{2}\times \left(\text{product}\text{of}\text{two}\text{sides}\right)\times \left(\text{sine}\text{of}\text{included}\text{angle}\right)\\ =6\times \frac{1}{2}\times 8\sqrt{3}\times 8\sqrt{3}\times \sin 60°\\ =498.8\end{array}$

Therefore, the area of the regular hexagon circumscribed in a circle is $498.8$ square inch.