Chapter 11.3, Problem 89E

Solution

a.

To prove that the area of an 8-sided regular polygon is $A=4hb$ and the area of an $n$ -sided regular polygon is $A=\frac{1}{2}nhb$ , where $h$ is height and $b$ is the base of each isosceles triangle.

Given information:

  $h$ is height and $b$ is the base of each isosceles triangle.

Proof:

Area of each isosceles triangle is $\frac{1}{2}hb$ . Since there 8 isosceles triangles in an 8-sided regular polygon, the total area will be 8 times the area of each triangle. Therefore, the area of 8-sided regular polygon is $A=8\times \frac{1}{2}hb=4hb$ .

When the polygon is $n$ -sided, the total area of the polygon will be $n$ times the area of each triangle. Therefore, the area of the $n$ -sided regular polygon is $A=n\times \frac{1}{2}hb=\frac{1}{2}nhb$ .

b.

To prove that the base $b$ of an $n$ -sided regular polygon is $b=2h\tan \left(\frac{{180}^{\circ }}{n}\right)$ , where $h$ is height and $b$ is the base of each isosceles triangle.

Given information:

  $h$ is height and $b$ is the base of each isosceles triangle.

Proof:

The angle subtended by each isosceles triangle of an $n$ -sided regular polygon is $\frac{{360}^{\circ }}{n}$ .

The angle bisector of $\frac{{360}^{\circ }}{n}$ forms two right triangles each of height $h$ and base $\frac{b}{2}$ , and $\theta =\frac{{180}^{\circ }}{n}$ .

  

Now, $\tan \theta =\frac{\left(b/2\right)}{h}$ . Then,

  $\begin{array}{c}\tan \left(\frac{{180}^{\circ }}{n}\right)=\frac{\left(b/2\right)}{h}\\ \tan \left(\frac{{180}^{\circ }}{n}\right)=\frac{b}{2h}\\ b=2h\tan \left(\frac{{180}^{\circ }}{n}\right)\end{array}$

Therefore, the base of the isosceles triangle is $2h\tan \left(\frac{{180}^{\circ }}{n}\right)$ .

c.

To prove that the area of an $n$ -sided regular polygon is $A=n{h}^2\tan \left(\frac{{180}^{\circ }}{n}\right)$ , where $h$ is height and $b$ is the base of each isosceles triangle.

Given information:

  $h$ is height and $b$ is the base of each isosceles triangle.

Proof:

From part (a), the area of the n-sided regular polygon is $A=\frac{1}{2}nhb$ and from part (b), the base of each isosceles triangle is $b=2h\tan \left(\frac{{180}^{\circ }}{n}\right)$ .

Substitute $b=2h\tan \left(\frac{{180}^{\circ }}{n}\right)$ in $A=\frac{1}{2}nhb$ and simplify.

  $\begin{array}{c}A=\frac{1}{2}nh\left(2h\tan \left(\frac{{180}^{\circ }}{n}\right)\right)\\ =n{h}^2\tan \left(\frac{{180}^{\circ }}{n}\right)\end{array}$

Therefore, the area of an $n$ -sided regular polygon is $A=n{h}^2\tan \left(\frac{{180}^{\circ }}{n}\right)$ .

d.

To construct a table of values for $h=1$ and $n=4,8,16,100,500,1000,10000,100000$ , and find if $A$ has limit as $n\to \infty$ .

  $\underset{n\to \infty }{\lim }A=\pi$

Given information:

  $h$ is height and $b$ is the base of each isosceles triangle.

Calculation:

Substitute $h=1$ in $A=n{h}^2\tan \left(\frac{{180}^{\circ }}{n}\right)$ . Then, $A=n\tan \left(\frac{{180}^{\circ }}{n}\right)$ .

Now, construct a table of values for $n=4,8,16,100,500,1000,10000,100000$ .

  

Observe that as the values of $n$ increases $A$ approaches the value of $\pi$ .

Therefore, $\underset{n\to \infty }{\lim }A=\pi$ .

e.

To construct a table of values for $h=3$ and $n=4,8,16,100,500,1000,10000,100000$ , and find if $A$ has limit as $n\to \infty$ .

  $\underset{n\to \infty }{\lim }A=9\pi$

Given information:

  $h$ is height and $b$ is the base of each isosceles triangle.

Calculation:

Substitute $h=3$ in $A=n{h}^2\tan \left(\frac{{180}^{\circ }}{n}\right)$ . Then, $A=9n\tan \left(\frac{{180}^{\circ }}{n}\right)$ .

Now, construct a table of values for $n=4,8,16,100,500,1000,10000,100000$ .

  

Observe that as the values of $n$ increases $A$ approaches the value of $9\pi$ .

Therefore, $\underset{n\to \infty }{\lim }A=9\pi$ .

f.

To explain that $\underset{n\to \infty }{\lim }A=\pi h^2$ , the area of a circle with radius $h$ .

Given information:

  $h$ is height and $b$ is the base of each isosceles triangle.

Calculation:

Observe that when $h=1$ , the limit is $\underset{n\to \infty }{\lim }A=\pi$ or $\underset{n\to \infty }{\lim }A=\pi {\left(1\right)}^2$ which is the area of a circle of radius 1.

When $h=3$ , the limit is $\underset{n\to \infty }{\lim }A=9\pi$ or $\underset{n\to \infty }{\lim }A=\pi {\left(3\right)}^2$ which is the area of a circle of radius 3.

This shows that the as the number of sides of the polygon tends to $\infty$ , the base of the isosceles triangle approaches 0 and the regular polygon takes the shape of a circle of radius $h$ .

Thus, $\underset{n\to \infty }{\lim }A=\pi h^2$ , the area of a circle of radius $h$ .