Geometry · G.C.B.5

Sectors, segments and their areas

A slice of pie takes some fraction of the full turn. How much of the pie’s area does it take, and what is left when the slice is cut straight across instead of to the center?

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Arcs and central anglesGeometry Area of a circleMath 7 Area of trianglesMath 6 Chords, secants and tangentsGeometry +1 more

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Radian measurePrecalculus

Where you are 15 in Circles

Geometry11

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▴ 6 above Area of a circle Arc length

Sectors, segments and their areas

The inscribed angle theorem Angles made by chords, secants and tangents ▾ Show all 15 (4 more)

When you reach for it

The area of a whole circle is \pi r^2. Most circular things you meet are not whole circles: a slice of pie, the patch of lawn a sprinkler reaches as it turns through part of a turn, the band a windscreen wiper clears, a shaded piece of a diagram. Each is a piece of a circle, and the question is how much of the circle’s area that piece holds.

Two kinds of piece come up again and again.

A sector is the region bounded by two radii OA and OB and one of the arcs between A and B. It is the shape of a slice of pie. The angle \angle AOB at the center is the sector’s central angle, written \theta. Two radii always cut a circle into two sectors: the minor sector, whose angle is less than 180^\circ, and the major sector, whose angle is the rest of the turn, 360^\circ - \theta. They sit on the minor and major arcs AB.

A segment is the region bounded by a Chords, secants and tangents AB and one of the two arcs between A and B. It is what you get by cutting a pie straight across instead of to the center. Again there are two: the minor segment on the minor arc and the major segment on the major arc.

Figure 1A sector is cut out by two radii; a segment is cut off by a chord. The segment is the sector with the triangle OAB taken away.

Reach for this lesson whenever a region is bounded partly by an arc. Every such region is built from sectors, triangles and whole circles, added or taken away.

The procedure

The area of a sector

A sector takes the same fraction of the circle’s area as its central angle takes of the full turn. With \theta measured in degrees, that fraction is \displaystyle \frac{\theta}{360}, so

A_{\text{sector}} = \frac{\theta}{360}\,\pi r^2

The reason is that equal central angles cut off congruent sectors. Turn one sector about the center O until one of its radii lies along a radius of the other, flipping it over that radius if the two open in opposite directions. A turn or a flip about O keeps every distance from O, so it carries the circle onto itself and radii onto radii; since the two central angles are equal, the second radii land on each other too, and the two sectors match exactly. Congruent shapes have equal areas.

Now cut the circle into 360 sectors of 1^\circ each. They are all congruent, they do not overlap and together they fill the circle, so each one has area \displaystyle \frac{1}{360}\pi r^2. A sector of 80^\circ is made of 80 of them, so its area is \displaystyle \frac{80}{360}\pi r^2, and the same counting works for any whole number of degrees. For an angle such as 22.5^\circ cut half-degree slices instead, of which there are 720; and an angle that no finite slicing fits exactly is still squeezed between two counts of slices as thin as you like, so its sector’s area is pinned to the same fraction.

Figure 2Three of eight equal slices make a 135^\circ sector, holding \displaystyle \frac{135}{360} = \frac{3}{8} of the area.

The picture with eight equal slices shows the idea at a glance. Each slice has a central angle of \displaystyle \frac{360^\circ}{8} = 45^\circ; three of them make a 135^\circ sector holding \displaystyle \frac{3}{8} of the area, and \displaystyle \frac{135}{360} is indeed \displaystyle \frac{3}{8}.

Sector area from the arc length

The arc length of the sector takes the same fraction of the circumference: \displaystyle s = \frac{\theta}{360} \cdot 2\pi r. Put the two formulas side by side:

A = \frac{\theta}{360}\,\pi r^2 = \frac{1}{2}\, r \cdot \left(\frac{\theta}{360} \cdot 2\pi r\right) = \frac{1}{2}\, r s

So the area of a sector is half the radius times the arc length, whatever the angle. This is the area of a triangle with base s and height r, and that is no accident: a thin sector is very nearly a tall thin triangle with its apex at O, which is the same picture the area of a circle uses when it lays wedges into a near-rectangle. Reach for \displaystyle A = \frac{1}{2}rs when the arc length is given and the angle is not.

Sector area with the angle in radians

If \theta is measured in radians, a full turn is 2\pi, not 360, so the fraction of the turn is \displaystyle \frac{\theta}{2\pi} and

A = \frac{\theta}{2\pi}\,\pi r^2 = \frac{1}{2}\,r^2 \theta

This is the same formula in other units, and it is the form used in arc length and sector area in radians. The two must never be mixed: the 360 goes with degrees and the 2\pi goes with radians.

The area of a segment

Draw the chord AB inside the minor sector OAB. It splits the sector into two pieces that do not overlap: the triangle \triangle OAB and the minor segment. Areas of pieces that do not overlap add, so

A_{\text{segment}} = A_{\text{sector}} - A_{\triangle OAB}

The triangle is isosceles, with two sides equal to r, and its area is half its base times its height. For three angles that area comes straight from facts you already have.

\theta = 90^\circ. The angle at O is a right angle, so the two radii are the legs, and the area is \displaystyle \frac{1}{2} r \cdot r = \frac{1}{2}r^2.

\theta = 60^\circ. The two base angles are equal and the three angles add to 180^\circ, so each base angle is 60^\circ too. The triangle is equilateral with side r, and its area is \displaystyle \frac{\sqrt{3}}{4}r^2.

\theta = 120^\circ. Drop the perpendicular OM from the center to the chord. A perpendicular from the center bisects the chord, and in an isosceles triangle it also bisects the angle at the apex, so it cuts \triangle OAB into two 30^\circ-60^\circ-90^\circ triangles with hypotenuse r. The side opposite the 30^\circ angle is half the hypotenuse, so \displaystyle OM = \frac{r}{2}, and \displaystyle AM = \frac{\sqrt{3}}{2}r. The base is \displaystyle AB = \sqrt{3}\,r, and the area is \displaystyle \frac{1}{2}\cdot\sqrt{3}\,r \cdot \frac{r}{2} = \frac{\sqrt{3}}{4}r^2, the same as for 60^\circ.

For other angles the triangle’s area needs trigonometry, where it turns out to be \displaystyle \frac{1}{2}r^2\sin\theta; this lesson stays with the three angles above.

The major segment is everything else in the circle, so

A_{\text{major segment}} = \pi r^2 - A_{\text{minor segment}}

Looked at directly, the major segment contains the center: it is the major sector with the triangle \triangle OAB added on, not taken away. Both routes give the same number, and Example 5 works both. When the chord is a diameter, the triangle flattens to nothing and each segment is a semicircle, \displaystyle \frac{1}{2}\pi r^2.

Worked examples

Example 1

A sector from its angle

A sector of a circle of radius 6 cm has a central angle of 80^\circ. Find its area, exactly and to two decimal places.

Solution

The sector takes \displaystyle \frac{80}{360} = \frac{2}{9} of the full turn, so it takes \displaystyle \frac{2}{9} of the circle’s area. The whole circle has area \pi \cdot 6^2 = 36\pi square centimeters, so

A = \frac{2}{9} \cdot 36\pi = 8\pi \approx 25.13 \text{ cm}^2

Simplifying the fraction first kept the numbers small. The exact answer 8\pi is the one to carry into any later working; the decimal is for reading off a size.

Example 2

The major sector

Points A and B on a circle of radius 9 m cut off a minor arc of 100^\circ. Find the area of the major sector AOB.

Solution

The major sector sits on the major arc, whose angle is the rest of the turn: 360^\circ - 100^\circ = 260^\circ. Its fraction of the circle is \displaystyle \frac{260}{360} = \frac{13}{18}, and the whole circle has area 81\pi, so

A = \frac{13}{18} \cdot 81\pi = \frac{117\pi}{2} \approx 183.78 \text{ m}^2

Check. The minor sector has area \displaystyle \frac{100}{360} \cdot 81\pi = \frac{45\pi}{2}, and the two sectors together fill the circle: \displaystyle \frac{45\pi}{2} + \frac{117\pi}{2} = \frac{162\pi}{2} = 81\pi.

Example 3

Working backwards to the angle or the radius

(a) A sector of a circle of radius 6 has area 15\pi. Find its central angle.

(b) A sector with central angle 40^\circ has area 10\pi. Find the radius.

Solution

The formula ties four quantities together, so any three fix the fourth.

(a) Put the known values into \displaystyle A = \frac{\theta}{360}\pi r^2:

15\pi = \frac{\theta}{360} \cdot 36\pi = \frac{\theta}{10}\,\pi

Dividing both sides by \pi gives \displaystyle \frac{\theta}{10} = 15, so \theta = 150^\circ. As a check, the circle has area 36\pi and \displaystyle \frac{15\pi}{36\pi} = \frac{5}{12}, which is \displaystyle \frac{150}{360}.

(b) Now the radius is the unknown:

\begin{aligned} 10\pi &= \frac{40}{360}\,\pi r^2 = \frac{1}{9}\,\pi r^2 \\ r^2 &= 90 \\ r &= \sqrt{90} = 3\sqrt{10} \approx 9.49 \end{aligned}

The negative square root is discarded because a radius is a length. A 40^\circ sector is \displaystyle \frac{1}{9} of its circle, so the whole circle has area 90\pi, which is another way to see r^2 = 90.

Example 4

Area from the arc length

A sector of a circle of radius 8 cm has an arc of length 6\pi cm. Find its area, its central angle and its perimeter.

Solution

The arc length is given and the angle is not, so use \displaystyle A = \frac{1}{2}rs:

A = \frac{1}{2} \cdot 8 \cdot 6\pi = 24\pi \approx 75.40 \text{ cm}^2

For the angle, the arc is \displaystyle \frac{6\pi}{16\pi} = \frac{3}{8} of the circumference 2\pi \cdot 8, so the angle is \displaystyle \frac{3}{8} of the turn: \displaystyle \theta = \frac{3}{8} \cdot 360^\circ = 135^\circ. The area agrees: \displaystyle \frac{3}{8} \cdot 64\pi = 24\pi.

The perimeter of a sector is the distance all the way round it: two radii and the arc.

P = 8 + 8 + 6\pi = 16 + 6\pi \approx 34.85 \text{ cm}

Example 5

A segment on a quarter circle

A chord AB of a circle of radius 10 subtends a right angle at the center O. Find the areas of the minor and the major segments.

Solution

The minor sector is a quarter of the circle, since \displaystyle \frac{90}{360} = \frac{1}{4}, so its area is \displaystyle \frac{1}{4} \cdot 100\pi = 25\pi. The triangle \triangle OAB has a right angle at O and legs OA = OB = 10, so its area is \displaystyle \frac{1}{2} \cdot 10 \cdot 10 = 50. The chord splits the sector into the triangle and the minor segment, so

A_{\text{minor}} = 25\pi - 50 \approx 28.54

For the major segment, take the minor segment from the whole circle:

A_{\text{major}} = 100\pi - (25\pi - 50) = 75\pi + 50 \approx 285.62

The direct route agrees. The major sector has angle 270^\circ and area \displaystyle \frac{3}{4} \cdot 100\pi = 75\pi, and the major segment is that sector with the triangle added, 75\pi + 50.

Example 6

A segment on a 120^\circ arc

In a circle with center O and radius 6, the chord AB cuts off an arc of 120^\circ. Find the area of the minor segment.

Figure 3In Example 6, \angle AOB = 120^\circ. The perpendicular OM halves it into two 30^\circ-60^\circ-90^\circ triangles.

Solution

The sector is \displaystyle \frac{120}{360} = \frac{1}{3} of the circle, whose area is 36\pi, so the sector has area 12\pi.

For the triangle, draw OM perpendicular to AB. It meets the chord at its midpoint M and halves \angle AOB, so \angle AOM = 60^\circ and \triangle AOM is a 30^\circ-60^\circ-90^\circ triangle with hypotenuse OA = 6. The side OM is opposite the 30^\circ angle at A, so it is half the hypotenuse, OM = 3, and then \displaystyle AM = \sqrt{6^2 - 3^2} = \sqrt{27} = 3\sqrt{3} by the Pythagorean theorem. So the chord is \displaystyle AB = 6\sqrt{3} and the height is OM = 3:

A_{\triangle OAB} = \frac{1}{2} \cdot 6\sqrt{3} \cdot 3 = 9\sqrt{3}

Taking the triangle from the sector,

A_{\text{segment}} = 12\pi - 9\sqrt{3} \approx 22.11

The answer is exact as \displaystyle 12\pi - 9\sqrt{3}; a \pi and a square root do not combine, so this is as simple as it gets.

Example 7

The band a wiper clears

A windscreen wiper’s blade runs from 15 cm to 60 cm out from its pivot, and the wiper swings through 120^\circ. What area of glass does it clear?

Figure 4The wiper in Example 7 clears the band between the two arcs, not the small sector near the pivot.

Solution

The whole arm, pivot to tip, sweeps a sector of radius 60 cm. The part of the arm with no blade on it sweeps a smaller sector of radius 15 cm with the same center and the same angle, and that glass is not cleared. The cleared band is the large sector with the small one taken away. Both take \displaystyle \frac{120}{360} = \frac{1}{3} of their circles, so the fraction can be pulled out:

\begin{aligned} A &= \frac{1}{3}\pi \cdot 60^2 - \frac{1}{3}\pi \cdot 15^2 \\ &= \frac{1}{3}\pi\,(3600 - 225) \\ &= 1125\pi \approx 3534.29 \text{ cm}^2 \end{aligned}

That is roughly a third of a square meter. Note that 60^2 - 15^2 is not (60 - 15)^2: the band is not a sector of radius 45, because its inner edge is 15 cm from the pivot, not at it.

Boundaries

The angle must be the central angle. The fraction \displaystyle \frac{\theta}{360} uses the angle with its vertex at the center. An angle with its vertex on the circle that opens onto the same arc is half as large, by the inscribed angle theorem, so it has to be doubled before it goes into the formula.

Degrees with 360, radians with 2\pi. \displaystyle \frac{\theta}{360}\pi r^2 needs \theta in degrees and \displaystyle \frac{1}{2}r^2\theta needs it in radians. Putting 60 into the radian form gives an area more than nine times the whole circle.

\theta runs from 0^\circ to 360^\circ. At 360^\circ the sector is the whole circle. Past that there is no new region: the sweep has gone round onto itself.

r is the radius. A diameter given in the problem must be halved first.

Sector minus triangle gives the minor segment only. When the arc is more than 180^\circ, the center lies inside the segment and the triangle is added to the major sector, not taken away. At exactly 180^\circ the chord is a diameter, the triangle has no area, and each segment is a semicircle.

The triangle’s area needs its own tools. This lesson finds it for 90^\circ, 60^\circ and 120^\circ. Any other central angle needs trigonometry, \displaystyle \frac{1}{2}r^2\sin\theta.

A band between two arcs is a difference of sectors only when they share a center and an angle. That is what made the wiper’s band \displaystyle \frac{\theta}{360}\pi(R^2 - r^2).

Where this goes next

The area formula and the arc length formula carry the same fraction of a turn, and that shared fraction is what makes radians natural: measure the angle by arc length over radius, as radian measure does, and the sector’s area becomes \displaystyle \frac{1}{2}r^2\theta with no 360 in sight. That form is the working one in arc length and sector area in radians, and in calculus the area swept out by a curve in polar coordinates is added up from thin sectors of area \displaystyle \frac{1}{2}r^2\,\Delta\theta, the start of area in polar coordinates.

A circle graph works because of this lesson: a category that is 30\% of the total gets 30\% of the turn, and therefore 30\% of the area, so the eye reads the share correctly. Shaded regions built from arcs, squares and triangles are the subject of area of composite figures, and the curved surface of a cone unrolls flat into a sector, whose area \displaystyle \frac{1}{2}rs gives the cone’s curved surface area.

Exercises

Area of a sector

Give each area exactly and to two decimal places.

A sector of a circle of radius 10 with central angle 72^\circ.
A sector of a circle of radius 4 with central angle 225^\circ.
A sector of a circle of diameter 14 with central angle 45^\circ.
Points A and B cut off a minor arc of 40^\circ on a circle of radius 9. Find the area of the major sector.
A sector of a circle of radius 4 with central angle \displaystyle \frac{3\pi}{4} radians.
On a circle of radius 6 with center O, an inscribed angle of 35^\circ opens onto the arc AB. Find the area of the sector AOB on that arc.

Working backwards

A sector of a circle of radius 6 has area 6\pi. Find its central angle.
A sector with central angle 80^\circ has area 32\pi. Find the radius.
A sector of a circle of radius 5 has an arc of length 4\pi. Find its area and its central angle.
A sector has area 24\pi and arc length 4\pi. Find the radius and the central angle.

Segments

Give each area exactly and to two decimal places.

A chord of a circle of radius 8 subtends a right angle at the center. Find the area of the minor segment.
A chord of a circle of radius 6 subtends 60^\circ at the center. Find the area of the minor segment.
For the circle and chord of exercise 12, find the area of the major segment.
A chord of a circle of radius 10 cuts off an arc of 120^\circ. Find the area of the minor segment.
A chord of a circle of radius 5 passes through the center. Find the area of each segment.

Regions and problems

A square has side 10. Two quarter circles of radius 10 are drawn inside it, centered at two opposite corners, as in the figure. Find the area of the leaf between the two arcs.

Figure 5Exercise 16. Each arc is a quarter of a circle of radius 10 centered at a corner of the square.

A fan-shaped flower bed lies between two arcs of radius 4 m and 10 m with the same center, and between two radii that meet at 150^\circ. Find its area.
A pizza 12 inches across is cut into 8 equal slices, and a pizza 10 inches across into 6 equal slices. Which slice has more area, and how long is the crust edge of a slice of the first pizza?
A sector of a circle of radius 12 cm has central angle 75^\circ. Find its area and its perimeter.
In a circle graph of radius 5 cm, one category is 30\% of the total. Find the central angle of its sector and the sector’s area.

Reasoning

Show that the area of any sector is \displaystyle \frac{1}{2}rs, where s is its arc length, and explain why a triangle with base s and height r is a fair picture of a sector.
Explain what happens to the area of a sector when its central angle is doubled and the radius kept, and when its radius is doubled and the angle kept.
Explain why the area of a major segment is the area of the major sector plus the area of \triangle OAB, and not minus it.
Show that in a circle of radius r, the triangles \triangle OAB for central angles of 60^\circ and 120^\circ have the same area. Their segments do not; explain why.

Answers

20\pi \approx 62.83. The fraction is \displaystyle \frac{72}{360} = \frac{1}{5} of 100\pi.
10\pi \approx 31.42. The fraction is \displaystyle \frac{225}{360} = \frac{5}{8} of 16\pi.
\displaystyle \frac{49\pi}{8} \approx 19.24. The radius is 7, and \displaystyle \frac{45}{360} = \frac{1}{8} of 49\pi.
72\pi \approx 226.19. The major sector has angle 320^\circ, and \displaystyle \frac{320}{360} = \frac{8}{9} of 81\pi.
6\pi \approx 18.85. In radians, \displaystyle A = \frac{1}{2} \cdot 4^2 \cdot \frac{3\pi}{4} = 6\pi.
7\pi \approx 21.99. The central angle on the same arc is twice the inscribed angle, 70^\circ, and \displaystyle \frac{70}{360} \cdot 36\pi = 7\pi.
60^\circ. \displaystyle \frac{\theta}{360} \cdot 36\pi = 6\pi gives \displaystyle \frac{\theta}{10} = 6.
r = 12. \displaystyle \frac{80}{360}\pi r^2 = \frac{2}{9}\pi r^2 = 32\pi, so r^2 = 144.
Area 10\pi, angle 144^\circ. \displaystyle A = \frac{1}{2} \cdot 5 \cdot 4\pi, and the arc is \displaystyle \frac{4\pi}{10\pi} = \frac{2}{5} of the circumference, so \displaystyle \theta = \frac{2}{5} \cdot 360^\circ.
r = 12, angle 60^\circ. \displaystyle 24\pi = \frac{1}{2} \cdot r \cdot 4\pi = 2\pi r, and then the arc is \displaystyle \frac{4\pi}{24\pi} = \frac{1}{6} of the circumference.
16\pi - 32 \approx 18.27. The quarter sector has area \displaystyle \frac{1}{4} \cdot 64\pi = 16\pi, and the right triangle has area \displaystyle \frac{1}{2} \cdot 8 \cdot 8 = 32.
\displaystyle 6\pi - 9\sqrt{3} \approx 3.26. The sector is \displaystyle \frac{1}{6} \cdot 36\pi = 6\pi; the triangle is equilateral with side 6, area \displaystyle \frac{\sqrt{3}}{4} \cdot 36 = 9\sqrt{3}.
\displaystyle 30\pi + 9\sqrt{3} \approx 109.84. The whole circle less the minor segment: \displaystyle 36\pi - (6\pi - 9\sqrt{3}).
\displaystyle \frac{100\pi}{3} - 25\sqrt{3} \approx 61.42. The sector is \displaystyle \frac{1}{3} \cdot 100\pi; the perpendicular from the center gives OM = 5 and \displaystyle AB = 10\sqrt{3}, so the triangle is \displaystyle \frac{1}{2} \cdot 10\sqrt{3} \cdot 5 = 25\sqrt{3}.
Each is \displaystyle \frac{25\pi}{2} \approx 39.27. The chord is a diameter, so the segments are two semicircles of a circle of area 25\pi.
50\pi - 100 \approx 57.08. The diagonal splits the leaf into two segments, each a quarter sector 25\pi less a right triangle 50.
35\pi \approx 109.96 square meters. \displaystyle \frac{150}{360}\pi(10^2 - 4^2) = \frac{5}{12} \cdot 84\pi.
The slice of the 12-inch pizza: \displaystyle \frac{9\pi}{2} \approx 14.14 square inches against \displaystyle \frac{25\pi}{6} \approx 13.09. Its crust is \displaystyle \frac{3\pi}{2} \approx 4.71 inches long. The slices are \displaystyle \frac{36\pi}{8} and \displaystyle \frac{25\pi}{6}, and the crust is \displaystyle \frac{1}{8} of 12\pi.
Area 30\pi \approx 94.25 cm^2, perimeter 24 + 5\pi \approx 39.71 cm. The fraction is \displaystyle \frac{75}{360} = \frac{5}{24}, so the area is \displaystyle \frac{5}{24} \cdot 144\pi; the arc is \displaystyle \frac{5}{24} \cdot 24\pi = 5\pi, and the perimeter adds two radii.
Angle 108^\circ, area \displaystyle \frac{15\pi}{2} \approx 23.56 cm^2. 30\% of 360^\circ is 108^\circ, and 30\% of 25\pi is 7.5\pi.
Both A and s carry the same fraction \displaystyle \frac{\theta}{360}, so \displaystyle A = \frac{\theta}{360}\pi r^2 = \frac{1}{2}r\left(\frac{\theta}{360} \cdot 2\pi r\right) = \frac{1}{2}rs. Cut the sector into many thin sectors; each is nearly a triangle with height r and a short piece of arc as its base, and their areas add to \displaystyle \frac{1}{2}r times the total arc, s.
Doubling the angle doubles the area, because \displaystyle \frac{\theta}{360} doubles and \pi r^2 does not change. Doubling the radius multiplies the area by 4, because the area depends on r^2 and (2r)^2 = 4r^2; the fraction of the turn is unchanged.
When the arc is more than 180^\circ the center O lies inside the major segment, on the far side of the chord from the minor arc. The triangle \triangle OAB lies between the chord and O, so it is part of the major segment but outside the major sector, whose edges are the two radii. Major sector and triangle do not overlap and together make the segment, so their areas add.
For 60^\circ the triangle is equilateral with side r, area \displaystyle \frac{\sqrt{3}}{4}r^2. For 120^\circ the perpendicular from O gives height \displaystyle \frac{r}{2} and base \displaystyle \sqrt{3}\,r, area \displaystyle \frac{\sqrt{3}}{4}r^2 as well. The sectors differ, \displaystyle \frac{1}{6}\pi r^2 against \displaystyle \frac{1}{3}\pi r^2, so the segments, sector less triangle, differ by \displaystyle \frac{1}{6}\pi r^2.