Arc Length and Sector Area

A slice of pizza, the area a windshield wiper sweeps, the path of a swinging pendulum: all of these are parts of circles. This page shows how to find the length of a curved edge (an arc) and the area of a slice (a sector), with the angle measured in degrees or in radians. Then it cuts a sector with a straight line to get a segment.

Key ideas

Arcs, sectors and segments

Two radii of a circle with an angle $\theta$ between them cut off:

• an arc: the curved piece of the circumference, with length $l$;
• a sector: the “pizza slice” bounded by the two radii and the arc;
• a segment: the region between the arc and the straight chord joining its ends.

A sector (left), and the same sector split into a triangle and a segment (right).

With the angle in degrees

A sector with angle $\theta^\circ$ is the fraction $\dfrac{\theta}{360}$ of the whole circle, so it gets that fraction of the circumference and of the area:

$$l = \frac{\theta}{360} \times 2\pi r \qquad\qquad A = \frac{\theta}{360} \times \pi r^2$$

This is the version used in AI SL, where radians aren’t needed.

With the angle in radians

A full turn is $2\pi$ radians (see radian measure), so the fraction is $\dfrac{\theta}{2\pi}$:

$$l = \frac{\theta}{2\pi} \times 2\pi r = r\theta \qquad\qquad A = \frac{\theta}{2\pi} \times \pi r^2 = \frac{1}{2}r^2\theta$$

These short formulas are one big reason radians are used. They’re the version for AA (SL and HL) and AI HL, and they only work when $\theta$ is in radians. Angles can be given as exact multiples of $\pi$ (like $\dfrac{2\pi}{3}$) or as decimals (like $1.2$).

Perimeter of a sector

The boundary of a sector is two radii plus the arc:

$$P = 2r + l$$

A common slip is to give only the arc length when the question asks for the perimeter.

Area of a segment

A segment is a sector with a triangle taken away. The triangle has two sides of length $r$ with the angle $\theta$ between them, so by the triangle area formula its area is $\dfrac{1}{2}r^2\sin\theta$:

$$\text{segment} = \text{sector} - \text{triangle} = \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta = \frac{1}{2}r^2(\theta - \sin\theta) \quad (\theta \text{ in radians})$$

In degrees, use $\dfrac{\theta}{360}\pi r^2 - \dfrac{1}{2}r^2\sin\theta$. Notice the $\sin\theta$ part works in either unit, as long as your calculator is in the matching mode.

On the SAT

Think proportionally: the arc’s fraction of the circumference and the sector’s fraction of the area both equal the angle’s fraction of the full turn ($360^\circ$ or $2\pi$ radians, both given on the SAT reference sheet). For example, a $60^\circ$ sector of a circle with radius $6$ is $\dfrac{1}{6}$ of the circle, so its arc length is $\dfrac{1}{6}(12\pi) = 2\pi$ and its area is $\dfrac{1}{6}(36\pi) = 6\pi$. Answers are usually in terms of $\pi$, so Desmos is only needed for decimal choices; if you evaluate trig there, remember it starts in degrees. See using Desmos on the SAT.

Worked examples

Example 1: Degrees

A sector of a circle has radius $12$ cm and angle $75^\circ$. Find the arc length, the area and the perimeter of the sector.

Solution.

$$l = \frac{75}{360} \times 2\pi(12) = 5\pi \approx 15.7 \text{ cm} \quad (\text{3 s.f.})$$ $$A = \frac{75}{360} \times \pi(12)^2 = 30\pi \approx 94.2 \text{ cm}^2 \quad (\text{3 s.f.})$$ $$P = 2(12) + 5\pi = 24 + 5\pi \approx 39.7 \text{ cm} \quad (\text{3 s.f.})$$

Example 2: Radians

A sector has radius $8$ cm and angle $1.2$ radians. Find the arc length and the area.

Solution. The angle is in radians, so use the short formulas:

$$l = r\theta = 8(1.2) = 9.6 \text{ cm} \qquad\qquad A = \frac{1}{2}r^2\theta = \frac{1}{2}(8)^2(1.2) = 38.4 \text{ cm}^2$$

Example 3: Working backwards

A sector of a circle with radius $9$ cm has a perimeter of $30$ cm. Find the angle of the sector in radians, and its area.

Solution. The perimeter is two radii plus the arc:

$$2(9) + l = 30 \quad\Rightarrow\quad l = 12 \text{ cm}$$ $$l = r\theta \quad\Rightarrow\quad 12 = 9\theta \quad\Rightarrow\quad \theta = \frac{4}{3} \approx 1.33 \text{ radians}$$ $$A = \frac{1}{2}(9)^2\left(\frac{4}{3}\right) = 54 \text{ cm}^2$$

(In degrees, $\dfrac{4}{3}$ radians is about $76.4^\circ$.)

Example 4: A segment

A chord cuts off a segment from a circle of radius $10$ cm. The chord subtends an angle of $\dfrac{\pi}{3}$ (that is, $60^\circ$) at the centre. Find the exact area of the segment, and its value to 3 s.f.

Solution. In radians:

$$\begin{aligned} \text{segment} &= \frac{1}{2}(10)^2\left(\frac{\pi}{3} - \sin\frac{\pi}{3}\right) \\ &= 50\left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right) \\ &= \frac{50\pi}{3} - 25\sqrt{3} \approx 9.06 \text{ cm}^2 \quad (\text{3 s.f.}) \end{aligned}$$

The same in degrees: the sector is $\dfrac{60}{360}\pi(10)^2 = \dfrac{50\pi}{3} \approx 52.36$, the triangle is $\dfrac{1}{2}(10)^2\sin 60^\circ = 25\sqrt{3} \approx 43.30$, and the difference is $9.06 \text{ cm}^2$. (The triangle is equilateral here, since all its sides are $10$.)

Common mistakes

Using $r\theta$ with $\theta$ in degrees. For $\theta = 75^\circ$ and $r = 12$, $r\theta = 900$, which is far longer than the whole circumference. Use $\dfrac{\theta}{360}$ with degrees, or convert to radians first.

Giving the arc length when the perimeter is asked for. The perimeter of a sector includes the two straight radii: $P = 2r + l$.

Forgetting the triangle in a segment. A segment is the sector minus the triangle. If your segment area is bigger than the triangle in a small-angle sector, check that you subtracted.

Mixing modes in the segment formula. In $\frac{1}{2}r^2(\theta - \sin\theta)$, the $\theta$ outside the sine must be in radians, and your calculator must be in radian mode for $\sin\theta$. With degrees, use the degree version of both parts.

Rounding the angle too early. In Example 3, keep $\theta = \dfrac{4}{3}$ exactly. Rounding it to $1.3$ gives an area of $52.65$ instead of $54$.

Practice

1. (Warm-up) A sector has radius $5$ m and angle $144^\circ$. Find its arc length and its area, exactly and to 3 s.f.

Solution

$$l = \frac{144}{360} \times 2\pi(5) = 4\pi \approx 12.6 \text{ m}$$$$A = \frac{144}{360} \times \pi(5)^2 = 10\pi \approx 31.4 \text{ m}^2$$

2. (Warm-up) A sector has radius $6$ cm and angle $\dfrac{5\pi}{6}$. Find its arc length and its area, exactly and to 3 s.f.

Solution

$$l = 6 \times \frac{5\pi}{6} = 5\pi \approx 15.7 \text{ cm}$$$$A = \frac{1}{2}(6)^2 \times \frac{5\pi}{6} = 15\pi \approx 47.1 \text{ cm}^2$$

3. (Warm-up) Find the perimeter of a sector with radius $15$ cm and angle $0.8$ radians.

Solution

$$l = 15(0.8) = 12 \text{ cm} \qquad P = 2(15) + 12 = 42 \text{ cm}$$

4. (Core) An arc of length $14$ cm lies on a circle of radius $8$ cm. Find the angle at the centre in radians and in degrees, and the area of the sector.

Solution

$$\theta = \frac{l}{r} = \frac{14}{8} = 1.75 \text{ radians}$$

In degrees: $1.75 \times \dfrac{180^\circ}{\pi} = 100.26\ldots^\circ \approx 100^\circ$ (3 s.f.).

$$A = \frac{1}{2}(8)^2(1.75) = 56 \text{ cm}^2$$

5. (Core) A car’s windshield wiper blade is fixed to an arm so that the blade cleans the region from $15$ cm to $60$ cm from the pivot. The arm turns through $110^\circ$. Find the area of windshield that the blade cleans.

Solution

The cleaned region is a big sector of radius $60$ cm minus a small sector of radius $15$ cm, both with angle $110^\circ$:

$$A = \frac{110}{360}\pi(60^2 - 15^2) = \frac{110}{360}\pi(3375) = 3239.76\ldots \approx 3240 \text{ cm}^2 \quad (\text{3 s.f.})$$

6. (Core) A sector has area $75 \text{ cm}^2$ and angle $1.5$ radians. Find its radius and its perimeter.

Solution

$$\frac{1}{2}r^2(1.5) = 75 \quad\Rightarrow\quad r^2 = 100 \quad\Rightarrow\quad r = 10 \text{ cm}$$$$l = 10(1.5) = 15 \text{ cm} \qquad P = 2(10) + 15 = 35 \text{ cm}$$

7. (Core) A chord of a circle of radius $20$ cm subtends an angle of $100^\circ$ at the centre. Find the area of the smaller segment.

Solution

$$\text{sector} = \frac{100}{360}\pi(20)^2 = 349.06\ldots \text{ cm}^2$$$$\text{triangle} = \frac{1}{2}(20)^2\sin 100^\circ = 196.96\ldots \text{ cm}^2$$$$\text{segment} = 349.06\ldots - 196.96\ldots = 152.10\ldots \approx 152 \text{ cm}^2 \quad (\text{3 s.f.})$$

8. (Challenge) A chord $AB$ of a circle with centre $O$ subtends an angle of $2$ radians at $O$. The area of the minor segment cut off by $AB$ is $20 \text{ cm}^2$. Find the radius of the circle.

Solution

$$\frac{1}{2}r^2(2 - \sin 2) = 20 \quad\Rightarrow\quad r^2 = \frac{40}{2 - \sin 2} = 36.673\ldots$$$$r = 6.0558\ldots \approx 6.06 \text{ cm} \quad (\text{3 s.f.})$$

(Make sure your calculator is in radian mode for $\sin 2$.)

9. (Challenge) A sector has a perimeter of $20$ cm and radius $r$ cm.

• (a) Show that its area is $A = 10r - r^2$.
• (b) Find the largest possible area, and the angle of the sector (in radians) that gives it.

Solution

(a) The arc is $l = 20 - 2r$. With $\theta$ in radians, $l = r\theta$, so the area is

$$A = \frac{1}{2}r^2\theta = \frac{1}{2}r(r\theta) = \frac{1}{2}r(20 - 2r) = 10r - r^2$$

(b) $A = 10r - r^2 = -(r - 5)^2 + 25$ is a downward parabola with vertex $(5, 25)$, so the largest area is $25 \text{ cm}^2$, when $r = 5$ cm. Then $l = 20 - 10 = 10$ cm and

$$\theta = \frac{l}{r} = \frac{10}{5} = 2 \text{ radians}$$