Exercise 55 Page 802

We are given the equations for two circles and two lines. We will start by graphing the given equations. Recall that the standard form for the equation of a circle with center ( h, k) and radius r is (x- h)^2+(y- k)^2= r^2. Let's look at the first equation. (x-3)^2+(y-5)^2=64 ⇕ (x- 3)^2+(y- 5)^2=( sqrt(64))^2 We see above that h= 3, k= 5, and r= sqrt(64)= 8. Note that we only consider the positive square root of 64 because a radius must be positive. The position of this circle is at the center ( 3, 5). The range is 8 units. Let's look at the second equation. (x-3)^2+(y-5)^2=25 ⇕ (x- 3)^2+(y- 5)^2=( sqrt(25))^2

We see above that h= 3, k= 5, and r= sqrt(25)= 5. The position of this circle is at the center ( 3, 5). The radius is 5 units. Looking at the equation of the first line, we see it is in slope-intercept form. y=mx+b ⇕ y=2/3x+3 The slope of this line is m=23. This means for every vertical change of 2 units there is a horizontal change of 3 units. The y-intercept is b=3. This means the line will cross the y-axis at y=3. The final equation y=5 will be a horizontal line that crosses the y-axis at y=5. Let's graph the four equations.

We can see that the areas of sections 1 and 3 are equal. Similarly, the areas of sections 2 and 4 are equal. The area of the four sections collectively makes up the difference between the area of the large circle and the area of the small circle. Therefore, we can write the following equation. A_1+A_2+A_3+A_4= A_(Big)- A_(Small) ⇕ A_1+A_2+A_3+A_4= π R^2 - π r^2 Let's substitute the radius of the large and small circles to solve for the right-hand side of the equation.

A_1+A_2+A_3+A_4=π R^2 - π r^2

SubstituteII

R= 8, r= 5

A_1+A_2+A_3+A_4=π ( 8)^2 - π ( 5)^2

CalcPow

Calculate power

A_1+A_2+A_3+A_4=64π - 25π

FactorOut

Factor out π

A_1+A_2+A_3+A_4=π(64-25)

SubTerm

Subtract term

A_1+A_2+A_3+A_4=39π

To find the area of each section, we will need to find the proportion each section makes up of the whole area. To do this, we will start by determining the angle made by the two lines. This give us the angle measure opposite of sections 1 and 3.

tan θ = 2/3

tan^(-1)(LHS) = tan^(-1)(RHS)

θ = tan^(-1) 2/3

UseCalc

Use a calculator

θ ≈ 33.69 °

Because an entire circle is 360 degrees, we will take the ratio of this angle measure over 360 degrees. Then we will multiply it by the total area of the section to find the area of sections 1 and 3. A_1=A_3= (33.69 °/360 °) * 39 π ≈ 11.5 units^2 The sum of the angle measures opposite of sections 1 and 2 (or sections 3 and 4) makes up a horizontal line, which equals 180 degrees. We found earlier that the measure of the angle opposite of section 1 is 33.69 degrees. Therefore, to find the measure of the angle opposite of section 2 we will subtract 33.69 from 180 degrees. A_2=A_4= (180 ° - 33.69 °/360 °) * 39 π ≈ 49.8 units^2