Chapter 11.4, Problem 33PPS

To determine

To find: The area of the equilateral triangle with the perimeter of 3 inches. The area of square, regular pentagon and the regular hexagon. Determine how the perimeter changes as the number of side increases.

Answer to Problem 33PPS

The calculation shows that as the perimeter changes the area of the polygon increases.

Explanation of Solution

Given:

The perimeter of the triangle is 3 inches.

Calculation:

Consider the formula for the area of the regular polygon with the side length of s is,

  $S=\frac{n}{4\tan \left(\frac{180°}{n}\right)}{s}^2$

Consider the isosceles triangle with the vertex at the centre of the pentagon with the segment that bisects the central angle and the sides of the polygon to form the right triangle is shown in Figure 2

  

Figure 2

Find the tangent ratio with the help of the apothem as,

  $\begin{array}{l}\tan \left(\frac{180°}{n}\right)=\frac{\frac{s}{2}}{a}\\ a=\frac{\frac{s}{2}}{\tan \left(\frac{180°}{n}\right)}\end{array}$

The area with the help of side length and the apothem is obtained as,

  $\begin{array}{c}A=\frac{1}{2}aP\\ =\frac{1}{2}\left(\frac{\frac{s}{2}}{\tan \left(\frac{180°}{n}\right)}\right)ns\\ =\frac{n}{4\tan \left(\frac{180°}{n}\right)}{s}^2\end{array}$

Replace $s=\frac{p}{n}$ in the above equation.

  $S=\frac{P^2}{4n\tan \left(\frac{180°}{n}\right)}$

Then, for $P=3$ .

  $\begin{array}{c}S=\frac{{\left(3\right)}^2}{4n\tan \left(\frac{180°}{n}\right)}\\ =\frac{9}{4n\tan \left(\frac{180°}{n}\right)}\end{array}$

Then, the value of $n=3$ is,

  $\begin{array}{c}S=\frac{9}{4\left(3\right)\tan \left(\frac{180°}{3}\right)}\\ \approx 0.43\end{array}$

Then, the value of $n=4$ is,

  $\begin{array}{c}S=\frac{9}{4\left(4\right)\tan \left(\frac{180°}{4}\right)}\\ \approx 0.56\end{array}$

Then, the value of $n=4$ is,

  $\begin{array}{c}S=\frac{9}{4\left(4\right)\tan \left(\frac{180°}{4}\right)}\\ \approx 0.56\end{array}$

Then, the value of $n=5$ is,

  $\begin{array}{c}S=\frac{9}{4\left(5\right)\tan \left(\frac{180°}{5}\right)}\\ \approx 0.62\end{array}$

Then, the value of $n=6$ is,

  $\begin{array}{c}S=\frac{9}{4\left(6\right)\tan \left(\frac{180°}{6}\right)}\\ \approx 0.65\end{array}$

Thus, the calculation shows that as the perimeter changes the area of the polygon increases.