The Area of a Segment

A sector, cut by its chord

A sector is the slice of a circle between two radii and the arc joining their ends. Call the center O and the ends of the arc A and B, so the two radii are OA and OB. The angle AOB between them, at the center, is the angle of the sector, written $\theta$θ.

Now join the two ends of the arc with a straight line, a chord. The chord cuts the sector into two pieces. On the side of the center is a triangle, made by the two radii and the chord. On the other side is a segment, the region between the chord and the arc. The two pieces together are the whole sector, so the area of the segment is the area of the sector minus the area of the triangle.

A sector with an angle of 100° at the center.

The same 100° sector with its chord drawn. The triangle with its corner at the center and the shaded segment make up the sector.

The area of the sector

A sector with an angle of $\theta$θ is $\frac{\theta }{360}$θ/360 of the whole circle, so its area is $\frac{\theta }{360}$θ/360 of the area of the circle: $\left(\frac{\theta }{360}\right)\times \pi r^2$(θ/360) × πr².

The area of the triangle

The area of any triangle is ½ × base × height. Take one radius, OA, as the base, so the base is r. The height is the distance from B, the end of the other radius, to the line OA, measured along a line that meets OA at a right angle.

When $\theta =90^{\circ }$θ = 90°, the two radii are at right angles, so the height is the other radius itself, and the area is $\frac{1}{2}\times r\times r=\frac{1}{2}{r}^2$½ × r × r = ½r².

For any other angle the height is shorter than the radius, and it depends on $\theta$θ. Take a circle of radius 1 first. The height of its triangle is a number fixed by $\theta$θ alone, and it is called the sine of $\theta$θ, written $\sin \theta$sin θ. A calculator gives it with its sin key, set to degrees.

A triangle with two sides of length r and the angle $\theta$θ between them is the triangle with two sides of 1 enlarged by a scale factor of r: two pairs of sides in the same ratio, with the same angle between them, make similar triangles. Every length is multiplied by r, the height too, so the height is $r\sin \theta$r sin θ. The area of the triangle is $\frac{1}{2}\times r\times r\sin \theta =\frac{1}{2}{r}^2\sin \theta$½ × r × r sin θ = ½r² sin θ.

The triangle made by two radii of length r with the angle $\theta$θ between them. Its height h, from the end of one radius to the other radius, meets that radius at a right angle, and $h=r\sin \theta$h = r sin θ.

Checking the sine at 90° and at 60°

At 90° the height is the whole radius, so for a radius of 1 it is 1: sin 90° = 1. Then $\frac{1}{2}{r}^2\sin 90^{\circ }=\frac{1}{2}{r}^2$½r² sin 90° = ½r², which agrees with the right-angled case.

At 60°, the triangle has two equal sides, the radii, so its other two angles are equal: each is (180° − 60°) ÷ 2 = 60°. All three angles are 60°, so the triangle is equilateral, and every side is 1 when the radius is 1. The height from one corner meets the opposite side at its middle, so it cuts off a right-angled triangle with a hypotenuse of 1 and a base of 0.5. By Pythagoras' theorem the height is $\sqrt{1^2-0.5^2}=\sqrt{0.75}=0.866$√(1² − 0.5²) = √0.75 = 0.866 to three decimal places. That is sin 60°, and the calculator gives the same.

Angles over 90°

When $\theta$θ is more than 90°, the height from B falls outside the triangle, onto the line OA continued past O. The angle between OB and that continued line is $180^{\circ }-\theta$180° − θ, because the two angles make a straight line. So the height is a side of a right-angled triangle with the radius OB as its hypotenuse and an angle of $180^{\circ }-\theta$180° − θ at O, which is exactly the right-angled triangle that gives the height for an angle of $180^{\circ }-\theta$180° − θ. For $\theta =120^{\circ }$θ = 120°, the height is the same as for 180° − 120° = 60°: sin 120° = sin 60° = 0.866. The 60° triangle and the 120° triangle, on the same radius, have the same area.

The formula

Put the two pieces together. The area of the segment is the area of the sector minus the area of the triangle: $\left(\frac{\theta }{360}\right)\times \pi r^2-\frac{1}{2}{r}^2\sin \theta$(θ/360) × πr² − ½r² sin θ. The angle at the center and the radius are all it needs.

A quarter circle of radius 10

Take r = 10 and $\theta =90^{\circ }$θ = 90°. The sector is a quarter of the circle: $\left(\frac{90}{360}\right)\times \pi \times 10^2=25\pi$(90/360) × π × 10² = 25π. The triangle is $\frac{1}{2}\times 10^2\times \sin 90^{\circ }=\frac{1}{2}\times 100\times 1=50$½ × 10² × sin 90° = ½ × 100 × 1 = 50. So the segment is $25\pi -50$25π − 50, exactly.

As a decimal, with the $\pi$π key, $25\pi =78.5398\dots$25π = 78.5398…, and 78.5398… − 50 = 28.54 to two decimal places. With 3.14 for $\pi$π, the sector is 78.5 and the segment is 78.5 − 50 = 28.5. The two answers differ only because 3.14 is a little less than $\pi$π.

A quarter circle of radius 10. The triangle is half of a 10 by 10 square, 50, and the segment is what is left of the quarter circle, $25\pi -50$25π − 50.

A 60° segment and a 120° segment

Take a circle of radius 9 cm and a sector of 60°. The sector is $\frac{60}{360}=\frac{1}{6}$60/360 = 1/6 of the circle: $\left(\frac{1}{6}\right)\times \pi \times 81=13.5\pi$(1/6) × π × 81 = 13.5π cm². The triangle is ½ × 81 × sin 60° = 40.5 × 0.8660… = 35.07 cm². The segment is $13.5\pi -40.5\sin 60^{\circ }=42.41-35.07=7.34$13.5π − 40.5 sin 60° = 42.41 − 35.07 = 7.34 cm² to two decimal places.

Now open the sector to 120° on the same circle. The sector doubles, to $27\pi$27π cm², but the triangle has the same area as before, because sin 120° = sin 60°. The segment is $27\pi -40.5\sin 120^{\circ }=84.82-35.07=49.75$27π − 40.5 sin 120° = 84.82 − 35.07 = 49.75 cm² to two decimal places.

The sector and the triangle are written here to two decimal places. In the working, keep the full calculator values and round only the answer: rounding the pieces first can change the last digit.

A 120° sector. Its triangle has the same area as the triangle of a 60° sector on the same radius, so all of the extra area of the sector is in the segment.

The usual mistakes

Giving the area of the sector. The sector reaches all the way to the center; the segment stops at the chord, so the triangle must be taken away.

Adding the triangle instead of taking it away. The triangle is already inside the sector, so the sector plus the triangle counts it twice.

Using $\frac{1}{2}{r}^2$½r² for every triangle. That is its area only when the angle at the center is 90°. For any other angle it is $\frac{1}{2}{r}^2\sin \theta$½r² sin θ.

A calculator set to radians. In radians mode, sin 60 gives −0.3048… instead of 0.8660…, so check the mode before finding a sine.