Chapter 6.8, Problem 2E

To determine

To calculate: The radius of the hole,

A metal washer are made by cutting a hole in a circular piece of metal and a washer is made by removing center piece of metal with a $1.8\text{ inch}$ in diameter. The area of washer is $0.65\pi$.

Answer to Problem 2E

The radius of the hole is: $0.40\text{ inch}$ , option $H$ is right answer.

Explanation of Solution

Given information:

A metal washer are made by cutting a hole in a circular piece of metal and a washer is made by removing center piece of metal with a $1.8\text{ inch}$ in diameter. The area of washer is $0.65\pi$.

Here the options are:

  $\begin{array}{l}\text{F 0}\text{.35 inch}\\ \text{G 0}\text{.38 inch}\\ \text{H 0}\text{.40 inch}\\ \text{I 0}\text{.42 inch}\end{array}$

Formula used:

To find the Radius of circle:

  $\frac{\pi }{4}\left(d_1{}^2-d_2{}^2\right)=\text{area of washer }$ ,

  ${\text{where d}}_1{\text{ is a diameter of wsaher and d}}_2\text{ is a diameter of holl}$ .

Calculation:

Consider the information,

A metal washer are made by cutting a hole in a circular piece of metal and a washer is made by removing center piece of metal with a $1.8\text{ inch}$ in diameter. The area of washer is $0.65\pi$.

Recall To find the Radius of circle:

  $\frac{\pi }{4}\left(d_1{}^2-d_2{}^2\right)=\text{area of washer }$ ,

Here diameter of the washer is: $1.8$.

Area of the washer is: $0.65\pi$

Apply these values in the formula:

  $\begin{array}{c}\frac{\pi }{4}\left(d_1{}^2-d_2{}^2\right)=\text{area of washer }\\ \frac{\pi }{4}\left({1.8}^2-d_2{}^2\right)=0.65\pi \\ \left({1.8}^2-d_2{}^2\right)=0.65\times 4\\ 3.24-d^2=2.60\end{array}$

Simplify further the expression as:

  $\begin{array}{c}3.24-d^2=2.60\\ -d^2=2.60-3.24\\ -d^2=-0.64\\ d^2=0.64\end{array}$

The value of $d$ is:

  $\begin{array}{c}{d}^2=0.64\\ d=\sqrt{0.64}\\ d=0.8\end{array}$

Then the radius of the hole is:

  $\begin{array}{c}r=\frac{d}{2}\\ d=0.8\\ r=\frac{0.8}{2}\\ r=0.4\end{array}$

Then the radius is: $0.40$

Thus,the radius of the hole is: $0.40\text{ inch}$ , option $H$ is right answer.