Problem 68 - Algebra and Trigonometry

Question

Problems \(68-77\) are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Find the area of the sector of a circle of radius 4 feet and central angle \(\theta\) if the arc length subtended by \(\theta\) is 5 feet.

Step-by-Step Solution

Answer

The area of the sector is 10 square feet.

Step 1 - Recall the formula for arc length

The arc length, s, of a sector is given by the formula: \[ s = r \theta \] where \( r \) is the radius and \( \theta \) is the central angle in radians.

Step 2 - Solve for \( \theta \)

Given that the arc length \( s = 5 \) feet and the radius \( r = 4 \) feet, we can substitute these values into the arc length formula to solve for \( \theta \): \[ 5 = 4 \theta \] Divide both sides by 4: \[ \theta = \frac{5}{4} \, \text{radians} \]

Step 3 - Recall the formula for the area of a sector

The area, A, of a sector is given by the formula: \[ A = \frac{1}{2} r^2 \theta \]

Step 4 - Substitute the known values into the formula

Substitute \( r = 4 \) feet and \( \theta = \frac{5}{4} \) radians into the area formula: \[ A = \frac{1}{2} (4^2) \left( \frac{5}{4} \right) \] Simplify inside the parentheses first: \[ A = \frac{1}{2} (16) \left( \frac{5}{4} \right) \]

Step 5 - Simplify the expression

Now, simplify the expression further: \[ A = \frac{1}{2} (4) (5) \] \[ A = 10 \, \text{square feet} \]

Key Concepts

Arc Length

The arc length is the distance along the curved line making up the arc. It is crucial in understanding sectors because it directly relates to the central angle and radius of the circle.
The formula to calculate arc length is given by: \( s \ = \ r \theta \), where:

• s is the arc length
• r is the radius
• \( \theta \) is the central angle in radians

In the given problem, if we know the arc length and radius, we can determine the central angle by rearranging the formula. Thus, understanding how to manipulate the arc length formula helps solve multiple problems involving circles.

Radius

The radius of a circle is the distance from its center to any point on the circumference. It plays a fundamental role in various calculations related to circles, including those involving sectors and arcs.
To find the area of a sector, knowing the radius is crucial because it appears in both the arc length and area formulas. For example, if the radius is given as 4 feet, it simplifies solving problems by substituting this value into appropriate formulas.
In the given problem, knowing that the radius is 4 feet allows us to use the given arc length to find the central angle and subsequently the area of the sector.

Central Angle

The central angle \( \theta \) is the angle subtended at the center of the circle by two radii. This angle can be measured in degrees or radians, but using radians simplifies many mathematical formulas.
Central angle in radians is particularly useful for finding arc length and area of the sector. The formula for arc length, \( s \ = \ r \theta \), directly links the radius and central angle to the arc length.
Given the arc length and radius, as in the example problem, you can determine the central angle using: \( \theta = \frac{s}{r} \). So, if \( s \) is 5 feet and \( r \) is 4 feet, then \( \theta = \frac{5}{4} \) radians. This value of \( \theta \) can then be used to find the area.

Radians

Radians provide a natural way to measure angles based on the radius of the circle. One radian is the angle made when the arc length is equal to the radius of the circle.
Using radians simplifies many formulas involving circles, such as those for arc length \( s \ = \ r \theta \) and the area of a sector \( A \ = \ \frac{1}{2} r^2 \theta \).
For instance, if the central angle is \( \frac{5}{4} \) radians with a radius of 4 feet, inserting these into the area formula gives: \( A \ = \ \frac{1}{2} (4^2) \left( \frac{5}{4} \right) = 10 \).
Notice how using radians ensures that the resulting calculations are straightforward and efficient.