Q. 28 - Precalculus Enhanced with Graphing Utilities

Question

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 $50°$. She wants to add a $3$-foot wide walk to the outer rim; how many square feet of paving blocks will she need to build the walk?

Step-by-Step Solution

Answer

Logan needs to build $78.93$ square feet of walk.

Step 1. Given Information

Length of outer rim of the garden is $25$ feet.

Central angle of the sector is $50°$.

Width of walk to the outer rim is $3$ feet.

We need to find the area of the paving blocks that Logan needs to build. 

Step 2. Finding the radius of the outer rim

Converting the angle of the sector into radian, we get:

$50\times \frac{\mathrm{\pi }}{180}$

$=\frac{5\mathrm{\pi }}{18}$radian

We know, $\theta =\frac{arc length}{radius}$

Here, $\theta =\frac{5\mathrm{\pi }}{18}, s=25$

$$\begin{gathered} \therefore \theta =\frac{s}{r} \\ \Rightarrow r=\frac{s}{\theta } \\ \Rightarrow r=\frac{25}{{\displaystyle \frac{5\mathrm{\pi }}{18}}} \\ \Rightarrow r=\frac{90}{\mathrm{\pi }} \end{gathered}$$

Step 3. Finding the outer radius of the pavement

Outer radius of pavement = radius of outer rim + width of the walk

$$\begin{gathered} \therefore R=r+w \\ \Rightarrow R=\frac{90}{\mathrm{\pi }}+3 \end{gathered}$$

Step 4. Area of outer circle

The given sector is $\frac{{\displaystyle \frac{5\mathrm{\pi }}{18}}}{2\mathrm{\pi }}=\frac{5}{36}$ times the full circle.

Area of outer circle

$$\begin{gathered} A_1=\frac{5}{36}{\mathrm{\pi R}}^2 \\ =\frac{5}{36}\mathrm{\pi }{\left(\frac{90}{\mathrm{\pi }}+3\right)}^2 \\ =437.03 \end{gathered}$$

Step 5. Area of inner circle

Area of inner circle

$$\begin{gathered} A_1=\frac{5}{36}\pi r^2 \\ =\frac{5}{36}\mathrm{\pi }{\left(\frac{90}{\mathrm{\pi }}\right)}^2 \\ =358.1 \end{gathered}$$

Step 6. Area of walk

$$\begin{gathered} A_1-A_2 \\ =437.03-358.1 \end{gathered}$$

$=78.93$square feet