Chapter 7.2, Problem 60PPS

a.

To determine

To calculate: The ratio of the area of the circle to that of the square for the given figure.

Answer to Problem 60PPS

The ratio of the area of the circle to that of the square $=\frac{\pi }{4}$ .

Explanation of Solution

Given information: Area of the circle is given by $A_1=\pi r^2$ and the area of square is given by $A_2=s^2$ .

The ratio of the area of circle to area of square will be given by

  $\frac{A_1}{A_2}=\frac{\pi r^2}{s^2}$

Also from the figure in the question, it can be observed that $s=2r$ .Therefore the ratio will be

  $\begin{array}{c}\frac{A_1}{A_2}=\frac{\pi r^2}{{\left(2r\right)}^2}\\ =\frac{\pi r^2}{4r^2}\\ =\frac{\pi }{4}\end{array}$

Hence the ratio of the area of the circle to that of the square $=\frac{\pi }{4}$ .

b.

To determine

To calculate: The ratio of the area of the circle to that of the square for the given figure if both the radius of circle and the length of square is doubled.

Answer to Problem 60PPS

The ratio of the area of the circle to that of the square remains same and $=\frac{\pi }{4}$ .

Explanation of Solution

Given information: Area of the circle is given by $A_1=\pi r^2$ and the area of square is given by $A_2=s^2$ .

From the figure given in the question, it can be observed that $s=2r$ .

Hence the ratio of the area of circle to area of square will be given by

  $\begin{array}{c}\frac{A_1}{A_2}=\frac{\pi r^2}{{\left(2r\right)}^2}\\ =\frac{\pi }{4}\end{array}$

Now if new radius and square length are doubled ( $r^{\text{'}}=2r\&s^{\text{'}}=2s$ ).Then the ratio will be

  $\begin{array}{c}\frac{A_1{}^{\text{'}}}{A_2{}^{\text{'}}}=\frac{\pi {\left(2r\right)}^2}{{\left(2s\right)}^2}\\ =\frac{4\pi r^2}{4s^2}=\frac{\pi r^2}{s^2}\\ =\frac{\pi r^2}{{\left(2r\right)}^2}\\ =\frac{\pi }{4}\end{array}$

Hence The ratio of the area of the circle to that of the square remains same and $=\frac{\pi }{4}$ .

c.

To determine

To calculate: The ratio of the area of the circle to that of the square for the given figure for different values of r.

Answer to Problem 60PPS

The ratio of the area of the circle to that of the square remains same and $=\frac{\pi }{4}$ .

Explanation of Solution

Given information: Area of the circle is given by $A_1=\pi r^2$ and the area of square is given by $A_2=s^2$ .

From the figure given in the question, it can be observed that $s=2r$ …. (1)

Hence the ratio of the area of circle to area of square will be given by

  $\begin{array}{c}\frac{A_1}{A_2}=\frac{\pi r^2}{{\left(2r\right)}^2}\\ =\frac{\pi }{4}\end{array}$

Now if new radius is doubled ( $r^{\text{'}}=2r$ ) then square length will also be doubled by using equation (1)

  $\begin{array}{c}{s}^{\text{'}}=2r^{\text{'}}\\ =2\left(2r\right)\\ =2s\end{array}$

And the new ratio will be

  $\begin{array}{c}\frac{A_1{}^{\text{'}}}{A_2{}^{\text{'}}}=\frac{\pi {\left(2r\right)}^2}{{\left(2s\right)}^2}\\ =\frac{4\pi r^2}{4s^2}=\frac{\pi r^2}{s^2}\\ =\frac{\pi r^2}{{\left(2r\right)}^2}\\ =\frac{\pi }{4}\end{array}$

Similarly if $r^{\text{'}}=3r$ then

  $\begin{array}{c}{s}^{\text{'}}=2r^{\text{'}}\\ =2\left(3r\right)\\ =3\left(2r\right)\\ =3s\end{array}$

And the new ratio will be

  $\begin{array}{c}\frac{A_1{}^{\text{'}}}{A_2{}^{\text{'}}}=\frac{\pi {\left(3r\right)}^2}{{\left(3s\right)}^2}\\ =\frac{9\pi r^2}{9s^2}=\frac{\pi r^2}{s^2}\\ =\frac{\pi r^2}{{\left(2r\right)}^2}\\ =\frac{\pi }{4}\end{array}$

Hence it can be observed that

  $\begin{array}{l}\text{if}{r}^{\text{'}}=2r\text{then}{\text{s}}^{\text{'}}=2s,\text{ratio=}\frac{\pi }{4}\\ r^{\text{'}}=3r\text{then}{\text{s}}^{\text{'}}=3s,\text{ratio=}\frac{\pi }{4}\\ r^{\text{'}}=4r\text{then}{\text{s}}^{\text{'}}=4s,\text{ratio=}\frac{\pi }{4}\\ r^{\text{'}}=5r\text{then}{\text{s}}^{\text{'}}=5s,\text{ratio=}\frac{\pi }{4}\\ r^{\text{'}}=6r\text{then}{\text{s}}^{\text{'}}=6s,\text{ratio=}\frac{\pi }{4}\end{array}$

So it can be observed that the ratio will always remain the same for all values of r .

Hence The ratio of the area of the circle to that of the square remains same and $=\frac{\pi }{4}$ .