Lesson Explainer: Areas of Sectors and Segments Mathematics

In this explainer, we will learn how to find the areas of circular sectors and segments when the angles are given in radians.

Before we discuss the areas of circular sectors and segments, we can start by recalling a very similar result involving the length of the arc of a circle. We recall that if an arc subtends an angle of πœƒ rad in a circle of radius π‘Ÿ, then the length 𝑙 of the arc is given by 𝑙=π‘Ÿπœƒ.

We proved this result by considering the proportion of the entire circumference of the circle as the arc; we can follow the same process by considering the entire area of the circle to find a formula for the area of a portion or a sector of a circle.

Formally, a sector of a circle is a part of a circle enclosed by two radii and an arc between them. There are two sectors of a circle between two radii of a circle, so we refer to the sectors as major, minor, or semicircular based on the arc or the angle subtended by the arc between the radii.

We call the angle at the center of the circle the angle subtended by the arc or sector. The sector that subtends a smaller angle is called the minor sector and the sector that subtends a larger angle is called the major sector. The smaller angle will be less than πœ‹ rad and the larger angle will be greater than πœ‹ rad. If both sectors subtend the same angle, then we call them semicircular arcs and the angles will both be equal to πœ‹ rad.

We can define this formally as follows.

Definition: A Sector of a Circle

A sector of a circle is a part of a circle enclosed by two radii and the arc between them. We name the sectors based on the type of arc between the radii.

If the sector subtends an angle less than πœ‹ rad, we refer to this as a minor sector, if the sector subtends an angle greater than πœ‹ rad, we refer to this as a major sector, and if the sector subtends an angle equal to πœ‹ rad, then the sector is a semicircle.

We want to determine the area of a sector of a circle, and the best way to do this is by considering an example.

Consider a sector of a circle of radius π‘Ÿ that subtends an angle of πœ‹2 rad.

This is a quarter of the circle, so we can calculate the area of this sector by multiplying the area of the whole circle ο€Ήπœ‹π‘Ÿο…οŠ¨ by 14. The reason this works is that the whole circle subtends an angle of 2πœ‹, so we can determine the proportion of the circle that is part of the sector by dividing the angles. We have 2πœ‹=14.οŽ„οŠ¨

More generally, a sector that subtends an angle of πœƒ rad will be a proportion of πœƒ2πœ‹ of the circle. So, the area of this sector is πœƒ2πœ‹ο€Ήπœ‹π‘Ÿο…=12π‘Ÿπœƒ.

This then gives us the following formula for finding the area of circular sectors.

Formula: The Area of a Circular Sector

The area 𝐴 of a sector of a circle of radius π‘Ÿ that subtends an angle of πœƒ rad is given by 𝐴=12π‘Ÿπœƒ.

Let’s see an example of using this formula to determine the area of a sector of a circle from the measure of the arc it subtends and the radius of the circle.

Example 1: Finding the Area of a Sector given Its Radius and Angle

An arc has a measure of πœ‹3 radians and a radius of 5. Give the area of the sector, in terms of πœ‹, in its simplest form.

Answer

We begin by recalling that the area 𝐴 of a sector of a circle of radius π‘Ÿ that subtends an angle of πœƒ rad is given by 𝐴=12π‘Ÿπœƒ.

In this question, we are told that an arc has a measure of πœ‹3 radians and a radius of 5; this is equivalent to saying that the sector subtends an angle of πœ‹3 rad and the radius is 5. Substituting π‘Ÿ=5 and πœƒ=πœ‹3 into the formula yields 𝐴=12(5)ο€»πœ‹3=12(25)ο€»πœ‹3=25πœ‹6.

In our next example, we will use the area of a sector and the central angle to determine the radius of the circle.

Example 2: Finding the Radius of a Sector given Its Area and Angle

The area of a circular sector is 1β€Žβ€‰β€Ž790 cm2 and the central angle is 1.5 rad. Find the radius of the circle giving the answer to the nearest centimetre.

Answer

We begin by recalling that the area 𝐴 of a sector of a circle of radius π‘Ÿ with a central angle of πœƒ rad is given by 𝐴=12π‘Ÿπœƒ.

We are told that the area of the sector is 1β€Žβ€‰β€Ž790 cm2 and that the central angle is 1.5 rad. Substituting 𝐴=1790cm and πœƒ=1.5rad into the equation gives 1790=12π‘Ÿ(1.5).

We note that 12(1.5)=34, so we have 1790=34π‘Ÿ.

Multiplying both sides of the equation by 43 yields π‘Ÿ=1790Γ—43=71603.

We now take the square roots of both sides of the equation, where we note that π‘Ÿ is nonnegative since it is a length: π‘Ÿ=ο„ž71603=48.85….cm

Rounding this value to the nearest centimetre gives us a radius of approximately 49 cm.

Now, we have seen some examples of how we can find the areas of circular sectors; let’s look at how we can find the areas of circular segments that are regions of a circle enclosed by an arc and a chord.

We can use the formula for the area of a sector combined with the formula for the area of a triangle to determine the area of a circular segment. Let’s start by adding the radii of length π‘Ÿ to the diagram and let’s say that the arc 𝐴𝐡 subtends an angle of πœƒ rad.

We can note that the shaded region is sector 𝑂𝐴𝐡 with △𝑂𝐴𝐡 removed. We can determine the areas of these two shapes separately.

First, sector 𝑂𝐴𝐡 has radius π‘Ÿ and it subtends an angle of πœƒ rad, so its area is 12π‘ŸπœƒοŠ¨. Second, △𝑂𝐴𝐡 has area 12π‘ŸπœƒοŠ¨sin. Therefore, if we call the area of the shaded region 𝐴, we have 𝐴=12π‘Ÿπœƒβˆ’12π‘Ÿπœƒ.sin

Taking out the shared factor of 12π‘ŸοŠ¨ gives 𝐴=12π‘Ÿ(πœƒβˆ’πœƒ).sin

We have shown the following result.

Formula: The Area of a Circular Segment

The area 𝐴 of a segment of a circle of radius π‘Ÿ that subtends an angle of πœƒ rad is given by 𝐴=12π‘Ÿ(πœƒβˆ’πœƒ).sin

Let’s now see an example of using this formula to determine the area of a region in a circle.

Example 3: Finding the Area of a Shaded Region in a Circle

Find the area of the shaded region of the circle given in the diagram. Give your answer to one decimal place.

Answer

We begin by recalling that the area 𝐴 of a segment of a circle of radius π‘Ÿ that subtends an angle of πœƒ rad is given by 𝐴=12π‘Ÿ(πœƒβˆ’πœƒ)sin. We can use this to determine the area of the unshaded region in the circle. To find the angle subtended by this sector, we recall that the sum of angles around a point is 2πœ‹. Thus, this sector subtends an angle of 2πœ‹βˆ’13πœ‹8=3πœ‹8.

Substituting π‘Ÿ=8cm and πœƒ=3πœ‹8rad into the formula for the segment area gives us 𝐴=128ο€Ό3πœ‹8βˆ’ο€Ό3πœ‹8=32ο€Ό3πœ‹8βˆ’ο€Ό3πœ‹8.sinsincm

This is the area of the unshaded region. We need to subtract this value from the area of the whole circle to find the area of the shaded region. The area of a circle of radius π‘Ÿ is given by πœ‹π‘ŸοŠ¨, so this circle has an area of πœ‹(8)=64πœ‹οŠ¨οŠ¨cm.

Therefore, the area of the shaded region is given by 64πœ‹βˆ’32ο€Ό3πœ‹8βˆ’ο€Ό3πœ‹8=192.92….sincm

We round this value to one decimal place to get that the area of the shaded region is approximately 192.9 cm2.

In our next example, we will find the area of a circular segment using its radius and chord length.

Example 4: Finding the Area of a Segment given Its Radius and Chord Length

Find the shaded area, giving your answer to three decimal places.

Answer

We are asked to find the area of a circular segment; we can do this by recalling that the area 𝐴 of a segment of a circle of radius π‘Ÿ that subtends an angle of πœƒ rad is given by 𝐴=12π‘Ÿ(πœƒβˆ’πœƒ).sin

We are given that the circle has a radius of 5 cm; however, we do not know βˆ π΄π‘‚π΅. We know all three side lengths of △𝐴𝑂𝐡 since 𝑂𝐴 is a radius and therefore must have a length of 5 cm. We can find βˆ π΄π‘‚π΅ by using the cosine rule: 𝑐=π‘Ž+π‘βˆ’2π‘Žπ‘πΆ.cos

Substituting π‘Ž=5cm, 𝑏=5cm, 𝑐=8cm, and 𝐢=πœƒ into the cosine rule yields 8=5+5βˆ’2(5)(5)πœƒ.cos

We can simplify to get 64=50βˆ’50πœƒ.cos

We can then rearrange as shown: 64βˆ’50=βˆ’50πœƒ14=βˆ’50πœƒβˆ’725=πœƒ.coscoscos

We now take the inverse cosine of both sides of the equation. We note that πœƒ is less than πœ‹ rad, so we only need to consider the solution in the second quadrant: πœƒ=ο€Όβˆ’725=1.85….cosrad

We can now substitute this exact value for πœƒ into the formula for the segment area along with π‘Ÿ=5cm. We obtain 𝐴=12π‘Ÿ(πœƒβˆ’πœƒ)=12(5)(1.85β€¦βˆ’(1.85…))=11.1823….sinsincm

To three decimal places, we have that the area of the shaded region is 11.182 cm2.

In our next example, we will use the area and radius of a sector to determine its perimeter.

Example 5: Solving a Word Problem Involving the Area and Perimeter of a Segment

A circular sector has an area of 108 cm2 and radius 6 cm. Find the perimeter of the sector.

Answer

We are given the area and radius of a sector and are asked to determine its perimeter. We can recall that the area 𝐴 of a sector of a circle of radius π‘Ÿ that subtends an angle of πœƒ rad is given by 𝐴=12π‘ŸπœƒοŠ¨. We are told that 𝐴=108cm and π‘Ÿ=6cm, so we can substitute these values into the formula to get 108=12(6)πœƒ.

We can then solve the equation for πœƒ: 108=18πœƒπœƒ=10818=6.rad

We can use this to sketch the major sector.

We can then recall that the length 𝑙 of an arc of a circle that subtends an angle of πœƒ rad is given by 𝑙=π‘Ÿπœƒ. Therefore, we can substitute π‘Ÿ=6cm and πœƒ=6rad into the formula to find the length of the arc of this sector. We get 𝑙=6Γ—6=36.cm

The perimeter of the sector is equal to the sum of the two radii and the arc length. Hence, we get 𝑙=36+6+6=48.cm

In our final example, we will determine the area of tiling required for a sector-shaped pool.

Example 6: Solving a Real-World Problem Using the Area of a Segment

A pool is in the shape of a circular sector and the tiles surrounding the front of the pool form a triangle as shown.

Calculate the area of the tiles needed to tile the front of the pool to one decimal place.

Answer

We want to determine the area of the unshaded region in the diagram. We can find this area by subtracting the area of the sector from the area of the triangle.

We can recall that the area 𝐴 of a sector of a circle of radius π‘Ÿ that subtends an angle of πœƒ rad is given by 𝐴=12π‘ŸπœƒοŠ¨. We can see that π‘Ÿ=12m; however, we are not given the value of πœƒ. Since we have all of the side lengths of the triangle, we can determine this angle by using the cosine rule: π‘Ž=𝑏+π‘βˆ’2𝑏𝑐𝐴.cos

We substitute π‘Ž=22m, 𝑏=12+5=17m, 𝑐=12+7=19m, and 𝐴=πœƒ into the cosine rule to get 22=17+19βˆ’2(17)(19)πœƒ.cos

We can evaluate this, giving us 484=289+361βˆ’646πœƒ.cos

Rearranging the equation yields βˆ’166=βˆ’646πœƒ83323=πœƒ.coscos

We can then solve for πœƒ by taking the inverse cosine of both sides of the equation and noting that πœƒ is acute: πœƒ==1.31….cosrad

We can now find the area of the triangle and sector separately.

First, the area of a triangle is given by the formula 12π‘Žπ‘πΆsin, where π‘Ž and 𝑏 are side lengths and 𝐢 is the included angle. We have π‘Ž=17m, 𝑏=19m, and 𝐢=1.31…rad. Substituting these values into the formula gives areaoftrianglesinm=12(17)(19)(1.31…)=156.07….

It is important to use the exact values since we do not want to round until the end of the calculation.

The area of a sector of radius π‘Ÿ subtended by an angle of πœƒ rad is given by 12π‘ŸπœƒοŠ¨. We have π‘Ÿ=12m and πœƒ=1.31…rad. Substituting these values into the formula gives areaofsectorm=12(12)(1.31…)=94.38….

We can now find the difference in these areas to calculate the area of the tiles needed: areaoftilesareaoftriangleareaofsectorm=βˆ’=156.07β€¦βˆ’94.38…=61.69…, where we note that it is important to use the exact values in this calculation.

We can now round this to one decimal place to get 61.7 m2.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • The area 𝐴 of a sector of a circle of radius π‘Ÿ that subtends an angle of πœƒ rad is given by 𝐴=12π‘ŸπœƒοŠ¨.
  • The area 𝐴 of a segment of a circle of radius π‘Ÿ that subtends an angle of πœƒ rad is given by 𝐴=12π‘Ÿ(πœƒβˆ’πœƒ)sin.

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