Chapter 11.3, Problem 43PPS

(a)

To determine

The equation for the area A of the circle.

Answer to Problem 43PPS

The equation is $\text{A}\left(\text{segment}\right)=\frac{x\pi r^2}{360}-r^2\left[\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)\right]$

Explanation of Solution

Given information:

The area A of a segment of a circle with a radius r and a central angle of x°. (Hint: Use trigonometry to find the base and height of the triangle.)

  

Formula used:

  $\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}$

Calculation:

Draw a perpendicular from the center to the chord to get two congruent triangles whose hypotenuse is r units long.

Let the height of the triangle be h and the length of the chord, which is a base of the triangle be l. 

  

Use trigonometry to find l and h in terms of r and x.

  $\begin{array}{l}\cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}\\ \cos \frac{x}{2}=\frac{h}{r}\\ r\cos \frac{x}{2}=h\\ \sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}\\ \sin \frac{x}{2}=\frac{\frac{l}{2}}{r}\\ r\sin \frac{x}{2}=\frac{l}{2}\\ 2r\sin \frac{x}{2}=l\end{array}$

The area of the triangle is determined as below,

  $\begin{array}{l}A=\frac{1}{2}lh\\ =\frac{1}{2}\left[2r\sin \left(\frac{x}{2}\right)\right]\left[r\cos \left(\frac{x}{2}\right)\right]\\ A=r^2\left[\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)\right]\end{array}$

Lastly, find the area of the segment.

  $\begin{array}{l}\text{A}\left(\text{segment}\right)=\text{A}\left(\text{sector}\right)-\text{A}\left(\text{triangle}\right)\\ \text{A}\left(\text{segment}\right)=\frac{x\pi r^2}{360}-r^2\left[\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)\right]\end{array}$

Conclusion:

The equation is $\text{A}\left(\text{segment}\right)=\frac{x\pi r^2}{360}-r^2\left[\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)\right]$

(b)

To determine

The table ten values of A for x ranging from 10 to 90

Answer to Problem 43PPS

The value is $\text{A}\left(\text{segment}\right)=\frac{2\pi x}{5}-144\left[\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)\right]$

Explanation of Solution

Given information:

Calculate and record in a table ten values of A for x-values ranging from 10 to 90 if r is 12 inches. Round to the nearest tenth

Formula used:

  $\text{A}\left(\text{segment}\right)=\text{A}\left(\text{sector}\right)\text{-A}\left(\text{traingle}\right)$

Calculation:

If r = 12, then the new formula is:

  $\begin{array}{l}\text{A}\left(\text{segment}\right)=\text{A}\left(\text{sector}\right)\text{-A}\left(\text{traingle}\right)\\ =\frac{x\pi {\left(12\right)}^2}{360}-{12}^2\left[\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)\right]\\ =\frac{144\pi x}{360}\times 144\pi -144\left[\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)\right]\\ \text{A}\left(\text{segment}\right)=\frac{2\pi x}{5}-144\left[\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)\right]\end{array}$

Enter this formula into Y1 of your calculator. The select the table function and set the range for 10 to 90 by 10.

  

Conclusion:

The value is $\text{A}\left(\text{segment}\right)=\frac{2\pi x}{5}-144\left[\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)\right]$

(c)

To determine

To sketch: The graph based on the x-values in the horizontal axis.

Answer to Problem 43PPS

The graph gets plotted against x and y values and hence the value get increases continuously.

Explanation of Solution

Given information:

Graph the data from your table with the x-values on the horizontal axis and the A-values on the vertical axis

Formula used:

The central angle measure is plotted against x axis and area of segment is plotted against y axis.

Calculation:

One other option would be to enter 10 through 90 by 10 in L1 and enter the formula for L2, replacing x with L1. Then, you can select STATPLOT L1, L2.

  

Conclusion:

The graph gets plotted against x and y values and hence the value get increases continuously.

(d)

To determine

The value of A.

Answer to Problem 43PPS

The area of the segment is about 15.0 when x is 63°

Explanation of Solution

Given information:

Use your graph to predict the value of A when x is 63. Then use the formula you generated in part a to calculate the value of A when x is 63. How do the values compare?

Formula used:

  $\text{A}\left(\text{segment}\right)=\text{A}\left(\text{sector}\right)\text{-A}\left(\text{triangle}\right)$

Calculation:

From the graph, it looks like the area would be about 15.5 when x is 63°.

  $\begin{array}{l}\text{A}\left(\text{segment}\right)=\text{A}\left(\text{sector}\right)\text{-A}\left(\text{triangle}\right)\\ =\frac{2\pi x}{5}-144\left[\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)\right]\\ =\frac{2\pi \left(63\right)}{5}-144\left[\sin \left(\frac{63}{2}\right)\cos \left(\frac{63}{2}\right)\right]\\ =25.2\pi -144\left[\sin \left(31.5\right)\cos \left(31.5\right)\right]\\ \text{A}\left(\text{segment}\right)=15.02\end{array}$

Therefore, the area of the segment is about 15.0 when x is 63°

Conclusion:

The area of the segment is about 15.0 when x is 63°