Chapter 9, Problem 19TP

Solution

a.

To write an equation that relates $\theta$ and $x$ for the smaller sector.

  $x\theta =5\pi$

Given:

The arc length of smaller sector is $5\pi {\text{ inch}}^2\text{,}$ radius is $x\text{ inch}$ and the angle of sector is $\theta .$

Formula Used:

The Arc length is given by:

  $S=r\theta$

Calculation:

Writing an equation that relates $\theta$ and $x$ as:

  $\begin{array}{l}\text{Arc length}\left(S\right)=r\theta \\ S=x\theta \\ 5\pi =x\theta \\ x\theta =5\pi \end{array}$

Hence the equation is $x\theta =5\pi .$

b.

To write an equation that relates $\theta$ and $x+10$ for the larger sector.

  $\left(x+10\right)\theta =14\pi$

Given:

The arc length of larger sector is $14\pi {\text{ inch}}^2\text{,}$ radius is $\left(x+10\right)\text{ inch}$ and the angle of sector is $\theta .$

Formula Used:

The arc length is given by:

  $S=r\theta$

Calculation:

Writing an equation that relates $\theta$ and $x+10$ as:

  $\begin{array}{l}\text{Arc length}\left(S\right)=r\theta \\ S=\left(x+10\right)\theta \\ 14\pi =\left(x+10\right)\theta \\ \left(x+10\right)\theta =14\pi \end{array}$

Hence, the equation is $\left(x+10\right)\theta =14\pi .$

c.

To solve the system of equations from part (a) and part (b) to find $x$ and $\theta .$

  $x=5.56\text{ inches and }\theta =161.82°.$

Calculation:

The equations relating $\theta$ and $x$ are:

  $\begin{array}{l}x\theta =5\pi \\ \left(x+10\right)\theta =14\pi \end{array}$

The equation $x\theta =5\pi$ can be rewritten as:

  $\theta =\frac{5\pi }{x}$

Substitute the value of $\theta =\frac{5\pi }{x}$ in the equation $\left(x+10\right)\theta =14\pi$ as:

  $\begin{array}{l}\left(x+10\right)\frac{5\pi }{x}=14\pi \\ 5\pi x+50\pi =14\pi x\\ 14\pi x-5\pi x=50\pi \\ 9\pi x=50\pi \\ 9x=50\\ x=\frac{50}{9}\\ x=5.555\\ x\approx 5.56\text{ inches}\end{array}$

Now, substitute the value of $x=5.56\text{ in}$ the equation $\theta =\frac{5\pi }{x}$ as:

  $\begin{array}{l}\theta =\frac{5\pi }{x}\\ \theta =\frac{5\pi }{5.56}\\ \theta =0.899\times 180\\ \theta =161.82°\end{array}$

Hence, the value of $x=5.56\text{ inches}$ and $\theta =161.82°.$

d.

To determine the amount of fabric using the formula of the area of a sector.

  $17088.19{\text{ in}}^2$

  $$

Formula Used:

  $\text{Area of a sector}\left(A\right)=\frac{1}{2}{r}^2\theta$

where,

  $\begin{array}{l}r=\text{ radius}\\ \theta =\text{ Angle of a sector}\end{array}$

Calculation:

The angle of sector is $\theta =161.82°$ and radius of smaller rector $x=5.56\text{ inches}$ . And the radius of larger sector is:

  $x+10=5.56+10=15.56\text{ inches}\text{.}$

Now, calculating the smaller sector area as:

  $\begin{array}{l}{A}_1=\frac{1}{2}{r}_1^2\theta \\ A_1=\frac{1}{2}{\left(5.56\right)}^2\left(161.82\right)\\ A_1=2501.22{\text{ in}}^2\end{array}$

And, calculating the larger sector area as:

  $\begin{array}{l}{A}_2=\frac{1}{2}{r}_2^2\theta \\ A_2=\frac{1}{2}{\left(15.56\right)}^2\left(161.82\right)\\ A_2=19589.41{\text{ in}}^2\end{array}$

Next, the area of the fabric is:

  $\begin{array}{l}\text{Area of fabric}=\text{Area of larger sector}-\text{Area of smaller sector}\\ \text{Area of fabric}=A_2-A_1\\ \text{Area of fabric}=19589.41-2501.22\\ \text{Area of fabric}=17088.19{\text{ in}}^2\end{array}$

Hence, the area of fabric will be $17088.19{\text{ in}}^2.$