Precalculus · F.TF.A.1 · G.C.B.5

Arc length and sector area in radians

Once an angle is measured in radians, how long is its arc and how much area does its sector hold?

You need

Radian measurePrecalculus

Where you are 4 in Angles and their measure

Trigonometry7

Angles and their measure4

Angles in standard position and coterminal angles Radian measure Angular and linear speed

Arc length and sector area in radians

When you reach for it

Reach for these formulas whenever a piece of a circle is cut off by a central angle measured in radians: the edge of a folding fan, the paper the fan carries, the strip a pendulum sweeps. The arc length and sector area lessons did the same jobs in degrees, where every answer carried a factor of \displaystyle \frac{\pi}{180} or \displaystyle \frac{1}{360}. With radian measure that factor disappears, and the formulas become short enough to solve backward, combine, and compare.

The procedure

Arc length

The radian measure of a central angle is defined as arc over radius, \displaystyle \theta = \frac{s}{r}. Multiplying both sides by r gives the arc:

s = r\theta

Nothing new has been assumed; this is the definition read the other way. It holds only with \theta in radians, and s comes out in the same unit as r.

Figure 1A sector of a circle. With \theta in radians, its arc is r\theta and its area is \displaystyle \frac{1}{2}r^2\theta.

Sector area

A sector with angle \theta is the fraction \displaystyle \frac{\theta}{2\pi} of a full turn, because a full turn is 2\pi radians. Equal central angles cut equal slices, so the sector holds that same fraction of the circle’s area \pi r^2:

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

Replacing r\theta by s gives a third form, which needs no angle at all:

A = \frac{1}{2}r s

It reads like the area of a triangle with base s and height r, and that is the picture: a thin sector is nearly a tall thin triangle, as in the area of a circle.

Running the formulas backward

Each formula links three quantities, so any two fix the third. Dividing gives

\theta = \frac{s}{r} \qquad \theta = \frac{2A}{r^2} \qquad r = \frac{2A}{s}

The last one is the useful surprise: the arc and the area together fix the radius, with no angle needed.

Perimeter and segment

The perimeter of a sector is its whole boundary: two radii and the arc,

P = 2r + r\theta

A segment is the region between a chord and its arc. The chord splits the sector into the triangle \triangle OAB and the segment, which do not overlap, so

A_{\text{segment}} = \frac{1}{2}r^2\theta - A_{\triangle OAB}

Worked examples

Example 1

Arc and area of a fan

Marianna’s folding fan has ribs 9 cm long and opens to an angle of \displaystyle \frac{7\pi}{12}. Find the length of its curved edge and the area of the sector it covers.

Solution

The angle is in radians, so s = r\theta applies directly:

s = 9 \cdot \frac{7\pi}{12} = \frac{21\pi}{4} \approx 16.49 \text{ cm}

For the area, square the radius first, then take half:

\begin{aligned} A &= \frac{1}{2} \cdot 81 \cdot \frac{7\pi}{12} = \frac{567\pi}{24} \\ &= \frac{189\pi}{8} \approx 74.22 \text{ cm}^2 \end{aligned}

Check. \displaystyle \frac{1}{2}rs = \frac{1}{2} \cdot 9 \cdot \frac{21\pi}{4} = \frac{189\pi}{8}, the same.

Example 2

Finding the angle or the radius

(a) Dinara’s fan has ribs 15 cm long, and its curved edge measures 27 cm when open. Through what angle does it open?

(b) A fan opened to \displaystyle \frac{3\pi}{4} covers 54\pi square centimeters. How long are its ribs?

Solution

(a) The arc and radius are known, so the angle is their ratio:

\theta = \frac{s}{r} = \frac{27}{15} = 1.8

The centimeters cancel, leaving 1.8 radians, which is \displaystyle 1.8 \cdot \frac{180^\circ}{\pi} \approx 103.13^\circ.

(b) Put the known values into \displaystyle A = \frac{1}{2}r^2\theta and solve for r^2:

\begin{aligned} 54\pi &= \frac{1}{2}r^2 \cdot \frac{3\pi}{4} = \frac{3\pi}{8}r^2 \\ r^2 &= 54\pi \cdot \frac{8}{3\pi} = 144 \end{aligned}

A rib is a length, so take the positive root: r = 12 cm.

Example 3

Arc and area together

Ermek measures a sector of fan paper: its curved edge is 40 cm and its area is 360 square centimeters. Find the radius and the angle.

Solution

The form \displaystyle A = \frac{1}{2}rs holds the radius and the arc, both of which matter here:

\begin{aligned} 360 &= \frac{1}{2} \cdot r \cdot 40 = 20r \\ r &= 18 \text{ cm} \end{aligned}

Then the angle is arc over radius:

\theta = \frac{40}{18} = \frac{20}{9} \approx 2.22

about 127.32^\circ. Check. \displaystyle \frac{1}{2} \cdot 18^2 \cdot \frac{20}{9} = \frac{1}{2} \cdot 324 \cdot \frac{20}{9} = 360.

Example 4

A sector from its perimeter and area

A sector has perimeter 40 cm and area 75 square centimeters. Find every possible radius and angle.

Solution

From the perimeter, the arc is what is left after the two radii: r\theta = 40 - 2r. Put that arc into \displaystyle A = \frac{1}{2}r \cdot (r\theta):

\begin{aligned} 75 &= \frac{1}{2}r(40 - 2r) = 20r - r^2 \\ r^2 - 20r + 75 &= 0 \\ (r - 5)(r - 15) &= 0 \end{aligned}

So r = 5 or r = 15. Each must be tested, since an angle is only a sector’s angle if it lies between 0 and 2\pi.

If r = 5: the arc is 40 - 10 = 30, so \displaystyle \theta = \frac{30}{5} = 6. Since 2\pi \approx 6.28, this sector is almost the whole circle, with a thin wedge missing.

If r = 15: the arc is 40 - 30 = 10, so \displaystyle \theta = \frac{10}{15} = \frac{2}{3}, a narrow slice.

Both are real sectors. Check. \displaystyle \frac{1}{2} \cdot 5 \cdot 30 = 75 and \displaystyle \frac{1}{2} \cdot 15 \cdot 10 = 75.

Example 5

The segment under a string

Marianna ties a straight string between the tips of two ribs 14 cm long that are opened to \displaystyle \frac{\pi}{3}. Find the area of the paper between the string and the curved edge, exactly and to two decimal places.

Figure 2The segment in Example 5 is the sector with the triangle \triangle OAB taken away.

Solution

The region is a segment: the sector minus the triangle \triangle OAB. The sector has area

\frac{1}{2} \cdot 14^2 \cdot \frac{\pi}{3} = \frac{98\pi}{3}

The triangle has two sides of 14 cm, so its base angles are equal; with an apex angle of \displaystyle \frac{\pi}{3}, or 60^\circ, the three angles adding to 180^\circ force each base angle to be 60^\circ too. It is equilateral with side 14, and its area is \displaystyle \frac{\sqrt{3}}{4} \cdot 14^2 = 49\sqrt{3}. So

A_{\text{segment}} = \frac{98\pi}{3} - 49\sqrt{3} \approx 17.75 \text{ cm}^2

Round only at the end. Rounding the two pieces first, 102.63 - 84.87, would give 17.76.

Example 6

Comparing two fans

Fan A has ribs of 15 cm and opens to \displaystyle \frac{2\pi}{3}. Fan B has ribs of 20 cm and opens to \displaystyle \frac{\pi}{2}. Compare their curved edges and their areas.

Solution

The arcs:

\begin{aligned} s_A &= 15 \cdot \frac{2\pi}{3} = 10\pi \\ s_B &= 20 \cdot \frac{\pi}{2} = 10\pi \end{aligned}

The edges are equal, even though fan A opens wider. The areas:

\begin{aligned} A_A &= \frac{1}{2} \cdot 225 \cdot \frac{2\pi}{3} = 75\pi \\ A_B &= \frac{1}{2} \cdot 400 \cdot \frac{\pi}{2} = 100\pi \end{aligned}

Figure 3The two fans in Example 6 have the same arc, 10\pi, but the one with the longer ribs holds more area.

Fan B holds a third more paper. The form \displaystyle A = \frac{1}{2}rs explains it at once: with the arc the same, area is proportional to the radius, and \displaystyle \frac{20}{15} = \frac{4}{3}.

Example 7

The most area for a fixed perimeter

Ermek has 48 cm of wire to bend into the outline of a sector. Which radius gives the sector of greatest area, and what angle does it have?

Solution

The arc is the wire left after the two radii, 48 - 2r, so

A = \frac{1}{2}r(48 - 2r) = 24r - r^2

This is a quadratic in r whose graph opens downward, so its greatest value is at the vertex, \displaystyle r = \frac{-24}{2(-1)} = 12. Then A = 24 \cdot 12 - 144 = 144 square centimeters, the arc is 48 - 24 = 24 cm, and

\theta = \frac{24}{12} = 2

The vertex is allowed: the angle 2 lies between 0 and 2\pi. The best sector opens to 2 radians, about 114.59^\circ.

Boundaries

Radians only. s = r\theta and \displaystyle A = \frac{1}{2}r^2\theta are true only with \theta in radians. With \theta = 60 for 60^\circ they give answers \displaystyle \frac{180}{\pi} times too large. Convert first.

One unit for lengths. r and s must be in the same unit; the area then comes out in that unit squared.

A sector’s angle is at most 2\pi, by what a sector is. No law of nature stops an angle from passing 2\pi: a wheel that turns twice turns through 4\pi, and r\theta is still the distance a point on its rim travels. But a sector is a region cut from one disk, and the whole disk is the most it can be, so the word “sector” itself limits the angle to 0 < \theta \le 2\pi. Past 2\pi the path laps the same circle, and \displaystyle \frac{1}{2}r^2\theta counts some of the disk twice instead of measuring a region. A computed angle above 2\pi, as in Exercise 25, means no sector has those measurements.

The segment formula needs the triangle. The triangle’s area is easy for \displaystyle \frac{\pi}{3} (an equilateral triangle) and \displaystyle \frac{\pi}{2} (a right triangle); other angles need its height, which trigonometry supplies later. For a major segment, add the triangle to the major sector instead of subtracting it.

Where this goes next

Dividing s = r\theta by a time turns an angle into a rate: a wheel turning \theta radians each second carries a point on its rim r\theta units each second. That is angular and linear speed. On a circle of radius 1, s = \theta, so radian measure is arc length itself; this is why the unit circle measures angles in radians and why cosine and sine are defined there.

Exercises

Arc and area

Find the arc length and the sector area for a radius of 8 cm and \displaystyle \theta = \frac{5\pi}{6}, exactly and to two decimal places.
Find the arc length and the sector area for a radius of 2.5 m and \theta = 1.6.
A sector has radius 10 in and central angle 135^\circ. Convert the angle to radians, then find the arc and the area, exactly.
A fan with ribs of 21 cm opens to \displaystyle \frac{2\pi}{3}. Find its curved edge and its area, to two decimal places.

Working backward

An arc of 22 cm lies on a circle of radius 8 cm. Find the central angle in radians and in degrees, to two decimal places.
An angle of \displaystyle \frac{5\pi}{9} cuts off an arc of 10\pi cm. Find the radius.
A sector with angle 0.6 has area 30 square centimeters. Find the radius.
A sector of a circle of radius 6 has area 24\pi. Find its angle.
A sector has arc 15 cm and area 60 square centimeters. Find the radius and the angle.

Perimeter

Find the perimeter of a sector with radius 7 cm and angle 1.4.
A sector has radius 15 cm and perimeter 50 cm. Find its angle and its area.
A sector has perimeter 30 cm and area 50 square centimeters. Find every possible radius and angle.

Segments

A circle has radius 10. A chord cuts off an arc of angle \displaystyle \frac{\pi}{3}. Find the area of the minor segment and of the major segment, exactly and to two decimal places.
Find the area of the segment cut off by a chord on a circle of radius 14 when the central angle is \displaystyle \frac{\pi}{2}, exactly and to two decimal places.

Comparing sectors

Sector P has radius 6 and angle 2. Sector Q has radius 4 and angle 4.5. Compare their areas and their arcs.
A sector’s radius is tripled and its angle is halved. By what factor do its arc and its area change?
Which holds more area: a sector with radius 10 and arc 12, or one with radius 8 and angle 2?
A piece of wire 36 cm long is bent into the outline of a sector. Find the radius, angle and area of the largest sector it can enclose.

Fans

Dinara’s fan has ribs of 25 cm, but the paper starts 10 cm from the pivot. Opened to \displaystyle \frac{3\pi}{4}, what area of paper shows, and how long is the outer edge? Give exact values and two decimal places.
Ermek’s fan has ribs of 24 cm and 12 pleats of equal angle. Opened to \displaystyle \frac{5\pi}{6}, how long is each pleat’s piece of the curved edge, and what area does each pleat cover?

Reasoning

Marianna computes the arc of a 60^\circ angle on a circle of radius 5 as 5 \cdot 60 = 300. Explain what went wrong and give the true arc.
Show that \displaystyle A = \frac{1}{2}rs follows from s = r\theta and \displaystyle A = \frac{1}{2}r^2\theta, and explain why it looks like the area of a triangle.
Ermek says that doubling both the radius and the angle of a sector doubles its area. Is he right? What happens to the arc?
Show that for any fixed perimeter P, the sector of greatest area has angle 2 radians.
A sector has perimeter 40 cm. Explain why its radius can be 5 cm but cannot be 4 cm.

Answers

\displaystyle s = \frac{20\pi}{3} \approx 20.94 cm, \displaystyle A = \frac{80\pi}{3} \approx 83.78 \text{cm}^2. \displaystyle s = 8 \cdot \frac{5\pi}{6} and \displaystyle A = \frac{1}{2} \cdot 64 \cdot \frac{5\pi}{6}.
s = 4 m, A = 5 \text{m}^2. 2.5 \cdot 1.6 = 4 and \displaystyle \frac{1}{2} \cdot 6.25 \cdot 1.6 = 5.
\displaystyle 135^\circ = \frac{3\pi}{4}, \displaystyle s = \frac{15\pi}{2} in, \displaystyle A = \frac{75\pi}{2} \text{in}^2. \displaystyle \frac{135\pi}{180} = \frac{3\pi}{4}, then \displaystyle 10 \cdot \frac{3\pi}{4} and \displaystyle \frac{1}{2} \cdot 100 \cdot \frac{3\pi}{4}.
s = 14\pi \approx 43.98 cm, A = 147\pi \approx 461.81 \text{cm}^2. \displaystyle 21 \cdot \frac{2\pi}{3} and \displaystyle \frac{1}{2} \cdot 441 \cdot \frac{2\pi}{3}.
\theta = 2.75, about 157.56^\circ. \displaystyle \frac{22}{8} = \frac{11}{4}, and \displaystyle \frac{11}{4} \cdot \frac{180^\circ}{\pi} = \frac{495^\circ}{\pi}.
r = 18 cm. \displaystyle r = \frac{s}{\theta} = 10\pi \cdot \frac{9}{5\pi}.
r = 10 cm. \displaystyle r^2 = \frac{2A}{\theta} = \frac{60}{0.6} = 100.
\displaystyle \theta = \frac{4\pi}{3}. \displaystyle \theta = \frac{2A}{r^2} = \frac{48\pi}{36}.
r = 8 cm, \displaystyle \theta = \frac{15}{8} = 1.875. \displaystyle r = \frac{2A}{s} = \frac{120}{15}, then \displaystyle \theta = \frac{s}{r}.
P = 23.8 cm. 2 \cdot 7 + 7 \cdot 1.4 = 14 + 9.8.
\displaystyle \theta = \frac{4}{3}, A = 150 \text{cm}^2. The arc is 50 - 30 = 20, so \displaystyle \theta = \frac{20}{15} and \displaystyle A = \frac{1}{2} \cdot 15 \cdot 20.
r = 5 with \theta = 4, or r = 10 with \theta = 1. \displaystyle \frac{1}{2}r(30 - 2r) = 50 gives r^2 - 15r + 50 = 0, so (r - 5)(r - 10) = 0. The arcs are 20 and 10; both angles are below 2\pi.
Minor: \displaystyle \frac{50\pi}{3} - 25\sqrt{3} \approx 9.06. Major: \displaystyle \frac{250\pi}{3} + 25\sqrt{3} \approx 305.10. The sector is \displaystyle \frac{1}{2} \cdot 100 \cdot \frac{\pi}{3} and the triangle is equilateral, area \displaystyle \frac{\sqrt{3}}{4} \cdot 100. The major segment is the circle, 100\pi, minus the minor one.
49\pi - 98 \approx 55.94. The sector is \displaystyle \frac{1}{2} \cdot 196 \cdot \frac{\pi}{2} = 49\pi; the triangle has the two radii as legs, area \displaystyle \frac{1}{2} \cdot 14 \cdot 14 = 98.
Equal areas, 36 each; Q’s arc is longer, 18 against 12. \displaystyle \frac{1}{2} \cdot 36 \cdot 2 and \displaystyle \frac{1}{2} \cdot 16 \cdot 4.5; 6 \cdot 2 and 4 \cdot 4.5.
The arc is multiplied by \displaystyle \frac{3}{2} and the area by \displaystyle \frac{9}{2}. s = r\theta gains \displaystyle 3 \cdot \frac{1}{2}; \displaystyle A = \frac{1}{2}r^2\theta gains \displaystyle 3^2 \cdot \frac{1}{2}.
The second, 64 against 60. \displaystyle \frac{1}{2} \cdot 10 \cdot 12 = 60 and \displaystyle \frac{1}{2} \cdot 64 \cdot 2 = 64.
r = 9 cm, \theta = 2, A = 81 \text{cm}^2. \displaystyle A = \frac{1}{2}r(36 - 2r) = 18r - r^2, greatest at r = 9; the arc is 18.
Area \displaystyle \frac{1575\pi}{8} \approx 618.50 \text{cm}^2, outer edge \displaystyle \frac{75\pi}{4} \approx 58.90 cm. The paper is the large sector minus the small one: \displaystyle \frac{1}{2}(625 - 100) \cdot \frac{3\pi}{4}.
Each pleat has angle \displaystyle \frac{5\pi}{72}, edge \displaystyle \frac{5\pi}{3} \approx 5.24 cm and area 20\pi \approx 62.83 \text{cm}^2. \displaystyle 24 \cdot \frac{5\pi}{72} and \displaystyle \frac{1}{2} \cdot 576 \cdot \frac{5\pi}{72}.
s = r\theta needs radians, and 60 is a degree measure. \displaystyle 60^\circ = \frac{\pi}{3}, so the arc is \displaystyle \frac{5\pi}{3} \approx 5.24; her answer is \displaystyle \frac{180}{\pi} times too big.
\displaystyle A = \frac{1}{2}r^2\theta = \frac{1}{2}r(r\theta) = \frac{1}{2}rs. Cut the sector into many thin slices: each is nearly a triangle of height r, and their bases add to s.
No. \displaystyle A = \frac{1}{2}r^2\theta becomes \displaystyle \frac{1}{2}(2r)^2(2\theta) = 8 \cdot \frac{1}{2}r^2\theta, so the area is multiplied by 8. The arc r\theta is multiplied by 4.
The arc is P - 2r, so \displaystyle A = \frac{1}{2}r(P - 2r) = \frac{P}{2}r - r^2, a downward quadratic with its vertex at \displaystyle r = \frac{P}{4}. There the arc is \displaystyle P - \frac{P}{2} = \frac{P}{2} and \displaystyle \theta = \frac{P/2}{P/4} = 2, whatever P is.
With r = 5 the arc is 30 and \theta = 6 < 2\pi, a real sector. With r = 4 the arc is 32 and \theta = 8 > 2\pi \approx 6.28: the arc would be longer than the whole circumference 8\pi \approx 25.13, so no sector fits.