Chapter 12.5, Problem 21WE

a.

To determine

To determine the perimeter of a regular Polygon having 10 sides is inscribed in a unit circle.

Answer to Problem 21WE

Perimeter is approximately 6.18 units.

Explanation of Solution

  

Here AB is one of the 10 sides of the regular polygon of 10 sides. We have drawn a fraction of figure to understand.

First we shall consider the right angle triangle OFB.

Here hypotenuse is of unit length.(given, radius of circle 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 $\frac{\left(10-2\right){180}^{\circ }}{10}={144}^{\circ }$

Now angles are,

  $\angle OBF=\frac{{144}^{\circ }}{2}={72}^{\circ }$

Now we can write,

  $\begin{array}{l}\cos {72}^{\circ }=\frac{BF}{OB}=\frac{BF}{1}\\ BF=0.309\end{array}$

Now perimeter is $10\times \left(2BF\right)=20\times 0.309=6.18$

b.

To determine

To determine the perimeter of a regular Polygon having 10 sides is inscribed in a unit circle.

Answer to Problem 21WE

Perimeter is approximately 6.5 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 $\frac{\left(10-2\right){180}^{\circ }}{10}={144}^{\circ }$

Now angles are,

  $\angle OBF=\frac{{144}^{\circ }}{2}={72}^{\circ }$

Now we can write,

  $\begin{array}{l}\tan {72}^{\circ }=\frac{OF}{FB}=\frac{1}{FB}\\ FB=0.325\end{array}$

Now perimeter is $10\times \left(2BF\right)=20\times 0.325=6.5$