Chapter 6, Problem 28CT

To determine

The number of square feet of paving blocks will Logan need to build the walk, if she has a garden in the shape of a sector of a circle, and the outer rim of the garden is $25$ feet long and the central angle of the sector is ${50}^0$ . She wants to add a $3$ -foot- wide walk to the outer rim.

Answer to Problem 28CT

Solution:

The area of walk is $78.93$ square feet.

Given information:

Logan has a garden in the shape of a sector of a circle; the outer rim of the garden is $25$ feet long and the central angle of the sector is ${50}^0$ degree. She wants to add a $3$ -foot- wide walk to the outer rim.

Explanation of Solution

Logan has a garden in the shape of a sector of a circle.

Now, Logan is planning to make a path around the outside of this sector of $3$ -foot- wide, the diagrammatic representation is as shown below

  

Notice that there is a sector $OAD$ inside a sector $OBC$ .

To find the area of walk want to find the area between the sector $OBC\text{and OAD}$ .

Let $r$ be the radius of circle then area $A_1$ of sector $OAD$ is

  $A_1=\frac{1}{2}{r}^2\theta$

And the radius of sector $OBC$ is $r+3$ then area $A_2$ of sector $OBC$ is

  $A_2=\frac{1}{2}{(r+3)}^2\theta$

Therefore, the area of walk is

  $A=A_2-A_1=\frac{1}{2}{(r+3)}^2\theta -\frac{1}{2}{r}^2\theta$

  $\Rightarrow A=\frac{1}{2}\left(r^2+6r+9-r^2\right)\theta$

  $\Rightarrow A=\frac{1}{2}\left(6r+9\right)\theta$

The outer rim of the garden is $25$ feet long and the central angle of the sector is ${50}^0$ degree.

Using the formula for length of the arc of the sector with central angle $\theta$ is,

  $S=r\theta$ , where $\theta$ is in radian

The outer rim of the garden is $25$ feet long and the central angle of the sector is ${50}^0$ degree.

That is, the length of the arc of sector is $S=25$ feet and $\theta ={50}^0=\frac{5\pi }{18}$ radian.

  $\Rightarrow 25=r\cdot \frac{5\pi }{18}$

  $\Rightarrow r=\frac{25\cdot 18}{5\pi }$

  $r=\frac{90}{\pi }$

Therefore, the area of walk is

  $A=\frac{1}{2}\left(6\left(\frac{90}{\pi }\right)+9\right)\left(\frac{5\pi }{18}\right)$

  $\Rightarrow A=\frac{5\pi }{36}\left(\frac{540}{\pi }+9\right)$

  $\Rightarrow A=75+\frac{5\pi }{4}\approx 78.93$ square feet

Thus, the area of the walk is $78.93$ square feet