Chapter 8.4, Problem 4GP

To determine

To calculate: To compare the rate of change of perimeter of regular hexagon with the rate of change for a perimeter of square.

Answer to Problem 4GP

The rate of change of perimeter of hexagon $1.5$ times the rate of change of perimeter of square

Explanation of Solution

Given information: The side of a regular hexagon increases by $1$ inch and the side of square increases by $1$ inch

Formula Used:

Perimeter is defined as the sum of all the sides of the figure.

Perimeter of square $=4\times Side$

Perimeter of hexagon $=6\times Side$

Calculation:

Let the length of the side of hexagon be $a$ and the length of the side of square be $b$

Now, perimeter of regular hexagon $=6a$

Perimeter of square $=4b$

Given that the side of a regular hexagon increases by $1$ inch and the side of square increases by $1$ inch

Thus,

New Perimeter of regular hexagon $=6\left(a+1\right)$

New Perimeter of square $=4\left(b+1\right)$

Now,

Rate of change of perimeter of regular hexagon is calculated as

  $\begin{array}{l}{R}_H=\frac{6\left(a+1\right)-6a}{\left(a+1\right)-a}\\ R_H=\frac{6a+6-6a}{a+1-a}\\ R_H=\frac{6}{1}\\ R_H=6\end{array}$

Also, rate of change of perimeter of square is calculated as

  $\begin{array}{l}{R}_S=\frac{4\left(b+1\right)-4b}{\left(b+1\right)-b}\\ R_S=\frac{4b+4-4b}{b+1-b}\\ R_S=\frac{4}{1}\\ R_S=4\end{array}$

On comparing the rate of change of perimeter of hexagon with rate of change of perimeter of square,

  $\begin{array}{l}\frac{R_H}{R_S}=\frac{6}{4}\\ \frac{R_H}{R_S}=\frac{3}{2}\\ R_H=1.5R_S\end{array}$

Conclusion:

Hence, the rate of change of perimeter of hexagon $1.5$ timesthe rate of change of perimeter of square