Chapter 12.5, Problem 20WE

a.

To determine

To determine the perimeter of a regular pentagon inscribed in a unit circle.

Answer to Problem 20WE

Perimeter is approximately 5.88 units.

Explanation of Solution

Given information :

  

First we shall consider the right angle triangle OFB.

Here hypotenuse is of unit length.

Each angle of any polygon can be calculated by the formula,

  $\frac{\left(n-2\right){180}^{\circ }}{n}$ Where $n$ is the number of sides of polygon.

Here it is given regular pentagon. Hence sides are 5.

Angle is $\frac{\left(5-2\right){180}^{\circ }}{5}={108}^{\circ }$

Now angles are,

  $\begin{array}{l}\angle OBF=\frac{{108}^{\circ }}{2}={54}^{\circ }\\ \angle FOB={90}^{\circ }-{54}^{\circ }={46}^{\circ }\end{array}$

Now we can write,

  $\begin{array}{l}\cos {54}^{\circ }=\frac{BF}{OB}=\frac{BF}{1}\\ BF=0.588\end{array}$

Now perimeter is $5\times \left(2BF\right)=10\times 0.588=5.88$

b.

To determine

To determine the perimeter of a regular pentagon circumscribed about a unit circle.

Answer to Problem 20WE

Perimeter is approximately 13.76 units.

Explanation of Solution

Given information :

  

First we shall consider the right angle triangle OFB.

Here perpendicular length is of unit length.

Each angle of any polygon can be calculated by the formula,

  $\frac{\left(n-2\right){180}^{\circ }}{n}$ Where $n$ is the number of sides of polygon.

Here it is given regular pentagon. Hence sides are 5.

Angle is $\frac{\left(5-2\right){180}^{\circ }}{5}={108}^{\circ }$

Now angles are,

  $\begin{array}{l}\angle OBF=\frac{{108}^{\circ }}{2}={54}^{\circ }\\ \angle FOB={90}^{\circ }-{54}^{\circ }={46}^{\circ }\end{array}$

Now we can write,

  $\begin{array}{l}\tan {54}^{\circ }=\frac{OF}{FB}=\frac{1}{FB}\\ FB=1.376\end{array}$

Now perimeter is $5\times \left(2BF\right)=10\times 1.376=13.76$