Chapter 1.5, Problem 41PS

Solution

To write an equation that can be used to find the value of the radius $r$ of the circular slot.

The equation that can be used to find the radius $r$ of the circular plot:

  $\pi r^2=100$

Given:

Given that the square plot has its side length $10\text{ft}$ .

Formula Used:

Area of a square:

If the side length , $s\text{ units}$ , of a square. Then, its area $A_S$ is given by:

  $A_S=s^2{\text{ unit}}^2$ .

Area of a circle:

If the radius, $r\text{units}$ , of a circle. Then, its area $A_C$ is given by:

  $A_C=\pi r^2{\text{ unit}}^2$ .

Calculation:

Since the side length of the square plot is $10\text{ft}$ , its area:

  $A_S={10}^2{\text{ ft}}^2$ .

The square plot is now converted into a circular one without reducing its area.

Thus, the circular plot and the square plot will have the same area as that of the square one.

Suppose that $A_C$ denotes the area of the circular plot.

Then it follows that:

  $\begin{array}{l}{A}_C=A\\ A_C={10}^2{\text{ft}}^2\\ A_C=100{\text{ft}}^2\end{array}$ .

Substitute $A_C=\pi r^2{\text{ft}}^2$ into the equation $A_C=100{\text{ft}}^2$ :

  $\pi r^2{\text{ft}}^2=100{\text{ft}}^2$ .

Cancel the units from the equation $\pi r^2{\text{ft}}^2=100{\text{ft}}^2$ :

  $\pi r^2=100$ .

Thus, the equation that can be used to find the radius of the circular plot:

  $\pi r^2=100$ .

Conclusion:

The equation that can be used to find the radius $r$ of the circular plot:

  $\pi r^2=100$ .

b.

To solve the equation found in part a. to find the radius $r$ of the circular plot.

The radius $r$ of the circular plot is found while solving $\pi r^2=100$ :

  $r=\frac{10}{\sqrt{\pi }}\text{ft}$ .

Given Information:

  $\pi r^2=100$

Calculation:

Solve the equation $\pi r^2=100$ :

  $\begin{array}{c}\pi r^2=100\\ r^2=\frac{100}{\pi }\\ r=\pm \sqrt{\frac{100}{\pi }}\end{array}$ .

Now, the radius of a circle cannot be negative.

Then, it follows that:

  $r=\sqrt{\frac{100}{\pi }}\text{ft}$ .

Conclusion:

The radius $r$ of the circular plot is found while solving $\pi r^2=100$ :

  $r=\sqrt{\frac{100}{\pi }}\text{ft}$ .

c.

To determine a general rule for the radius $r$ of the circular plot if the side length of the square plot is $s\text{ units}$ .

The equation $r=\frac{s}{\sqrt{\pi }}\text{units}$ can be used to find radius of the circular plot.

Given:

Given the side length of the square:

  $s\text{ units}$

Calculation:

Since the side length of the square plot is $s\text{ units}$ , its area is given by:

  $A_S=s^2{\text{ unit}}^2$ .

Now, the square plot is converted into a circular one.

So, both the square plot and the circular plot will have the same are.

Suppose that $A_C$ denotes the area of the circular plot.

Then, it follows that;

  $A_C=s^2{\text{ unit}}^2$ .

Now, Substitute $A_C=\pi r^2{\text{ unit}}^2$ into the equation $A_C=s^2{\text{ unit}}^2$ :

  $\pi r^2{\text{ unit}}^2=s^2{\text{ unit}}^2$ .

Solve the equation $\pi r^2{\text{ unit}}^2=s^2{\text{ unit}}^2$ for $r$ :

  $\begin{array}{c}\pi r^2{\text{ unit}}^2=s^2{\text{ unit}}^2\\ \pi r^2=s^2\\ {\text{r}}^2=\frac{s^2}{\pi }\\ \text{r}=\pm \sqrt{\frac{s^2}{\pi }}\end{array}$ .

Now, the radius of a circle cannot be negative.

Then, it follows that:

  $\text{r=}\sqrt{\frac{s^2}{\pi }}\text{units}$ .

Now, the equation $\text{r=}\frac{s}{\sqrt{\pi }}\text{units}$ can be used to find radius of the circular plot.

Conclusion:

The equation $\text{r=}\sqrt{\frac{s^2}{\pi }}\text{units}$ can be used to find radius of the circular plot.