b Place the arc measures along the x-axis and the areas of the sectors along the y-axis.
C
c Can you connect the points in the graph using a straight line? Does the formula show a linear relationship?
D
d Does the formula used in Part A change? Does it show a linear relationship between x and y?
A
aTable:
x
30^(∘)
60^(∘)
90^(∘)
120^(∘)
150^(∘)
180^(∘)
y (in^2)
2.4
4.7
7.1
9.4
11.8
14.1
B
bGraph:
C
c Yes.
The rate of change is constant.
D
dYes, the areas in the table change.
No, the answer to Part C does not change. Explanation: See solution.
Practice makes perfect
a Let's consider a circle with a radius of 3 inches.
To complete the given table, we will use the formula to find the area of a sector of a circle with arc measure x.
Area of a sector_y = x/360* π r^2
Let's substitute r=3 into the formula above.
Finally, we will substitute the corresponding value of x to find each value of y.
x
30^(∘)
60^(∘)
90^(∘)
120^(∘)
150^(∘)
180^(∘)
y
π/40^(∘)* 30^(∘) = 3π/4
π/40^(∘)* 60^(∘) = 3π/2
π/40^(∘)* 90^(∘) = 9π/4
π/40^(∘)* 120^(∘) = 3π
π/40^(∘)* 150^(∘) = 15π/4
π/40^(∘)* 180^(∘) = 9π/2
Rounded Value (in^2)
2.4
4.7
7.1
9.4
11.8
14.1
b Let's graph the data in the table we completed in the previous part. To do that, we place the arc measures along the x-axis and the areas of the sectors along the y-axis.
c As we can see in the graph we did in Part B, the relationship between x and y is linear.
To find the values of y in Part A, we used the formula below.
y = π/40* x
As we can see, this formula shows a linear relationship between x and y with a rate of change of π40.
d Since we now are considering a circle with a radius of 5 inches, we have that the formula to find the value of y changes as follows.
y = x/360*π(5)^2
⇓
y = 5π/72x
This new formula is different from the one we used in Part A. Therefore, if we repeat the three previous parts with this formula, the areas will change but the answer from Part C will not, because it is still a linear relationship.
