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Arc length. Area of a sector and an annulus

Lesson 5.3 · 20 min read · Updated

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A car's windscreen wiper turns about a pivot OO. The rubber blade is fixed on the arm from the point 1010 cm from OO to the point 5050 cm from OO. On each sweep the arm turns through 120∘120^\circ. How many square centimetres of glass does one sweep clean?

Arc length

Explore: split the circle into 360360 parts

A circle of radius RR has circumference C=2πRC=2\pi R. The whole circle corresponds to 360∘360^\circ, so an arc of 1∘1^\circ has length 2πR360=πR180\dfrac{2\pi R}{360}=\dfrac{\pi R}{180}. An arc of n∘n^\circ is nn times as long as an arc of 1∘1^\circ.

From the formula, when we know ll and RR we can find nn.

Area of a sector

A sector is the part of a disc bounded by an arc and the two radii through its end points. A disc of area πR2\pi R^2 corresponds to 360∘360^\circ, so a sector of n∘n^\circ has area n360\dfrac{n}{360} of the area of the disc.

Rn°OAB
The sector OABOAB of radius RR with an arc of n∘n^\circ

Area of an annulus

An annulus is the region between two concentric circles. Its area is the area of the large disc minus the area of the small disc.

RrOAB
An annulus between two concentric circles

The glass cleaned in the opening problem is the part of an annulus inside an angle of 120∘120^\circ.

Key terms

Summary

  • Circumference C=2πRC=2\pi R; area of a disc πR2\pi R^2.
  • Arc of n∘n^\circ: l=πRn180l=\dfrac{\pi Rn}{180}.
  • Sector of n∘n^\circ: S=πR2n360=lR2S=\dfrac{\pi R^2n}{360}=\dfrac{lR}{2}.
  • Annulus: S=π(R2−r2)S=\pi\left(R^2-r^2\right).

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

Find the length of an arc of 45∘45^\circ on a circle of radius 1212 cm (give your answer to 2 decimal places).

What you need and how to start

What you need: l=πRn180l=\dfrac{\pi Rn}{180}.

Steps:

  1. Substitute R=12R=12 and n=45n=45.
Show solution

We have l=π⋅12⋅45180=3π≈9.42l=\dfrac{\pi\cdot12\cdot45}{180}=3\pi\approx9.42 (cm).

So the arc is about 9.429.42 cm long.

Answer. l=3π≈9.42l=3\pi\approx9.42 cm.

Exercise 2

Find the area of a sector of radius 66 cm with an arc of 150∘150^\circ (give your answer to 2 decimal places).

What you need and how to start

What you need: S=πR2n360S=\dfrac{\pi R^2n}{360}.

Steps:

  1. Substitute R=6R=6 and n=150n=150.
Show solution

We have S=π⋅36⋅150360=15π≈47.12S=\dfrac{\pi\cdot36\cdot150}{360}=15\pi\approx47.12 (cm²).

So the area of the sector is about 47.1247.12 cm².

Answer. S=15π≈47.12S=15\pi\approx47.12 cm².

Exercise 3

Find the area of the annulus between two concentric circles of radii 55 cm and 33 cm (give your answer to 2 decimal places).

What you need and how to start

What you need: S=π(R2−r2)S=\pi\left(R^2-r^2\right).

Steps:

  1. Compute R2−r2R^2-r^2 and multiply by π\pi.
Show solution

We have S=π(52−32)=16π≈50.27S=\pi\left(5^2-3^2\right)=16\pi\approx50.27 (cm²).

So the area of the annulus is about 50.2750.27 cm².

Answer. S=16π≈50.27S=16\pi\approx50.27 cm².

Exercise 4

An arc of 90∘90^\circ has length 4π4\pi cm. Find the radius of the circle.

What you need and how to start

What you need: l=πRn180l=\dfrac{\pi Rn}{180}, so R=180lπnR=\dfrac{180l}{\pi n}.

Steps:

  1. Substitute l=4πl=4\pi and n=90n=90.
Show solution

We have R=180⋅4π90π=8R=\dfrac{180\cdot4\pi}{90\pi}=8 (cm).

So the radius is 88 cm.

Answer. R=8R=8 cm.

Exercise 5

Nam's bicycle wheel has diameter 7070 cm. How many metres does the bicycle travel when the wheel turns 100100 times (give your answer to 1 decimal place)?

What you need and how to start

What you need: in one turn, the bicycle travels the circumference C=πdC=\pi d.

Steps:

  1. Find the circumference.
  2. Multiply by 100100 and convert to metres.
Show solution

In one turn the bicycle travels C=70πC=70\pi (cm). Hence 100100 turns give 7000π≈219917000\pi\approx21991 (cm) ≈219.9\approx219.9 (m).

So the bicycle travels about 219.9219.9 m.

Answer. About 219.9219.9 m.

Exercise 6

An annulus has area 40π40\pi cm², and the difference of its two radii is 22 cm. Find the two radii. (Revise solving simultaneous linear equations.)

What you need and how to start

What you need: R2−r2=(R−r)(R+r)R^2-r^2=(R-r)(R+r).

Steps:

  1. From the area, (R−r)(R+r)=40(R-r)(R+r)=40, so R+r=20R+r=20.
  2. Solve the simultaneous equations R+r=20R+r=20 and R−r=2R-r=2.
Show solution

From the question, π(R2−r2)=40π\pi\left(R^2-r^2\right)=40\pi, so (R−r)(R+r)=40(R-r)(R+r)=40. Substituting R−r=2R-r=2, we get R+r=20R+r=20. Solving

{R+r=20R−r=2⇒R=11, r=9.\begin{cases}R+r=20\\ R-r=2\end{cases}\Rightarrow R=11,\ r=9.

So the two radii are 1111 cm and 99 cm.

Answer. R=11R=11 cm, r=9r=9 cm.

Exercise 7

(1.5 marks) A clock has a minute hand 1010 cm long and an hour hand 66 cm long.

a) In 2020 minutes, how far does the tip of the minute hand travel (give your answer to 2 decimal places)?

b) Find the area of the sector swept by the minute hand in those 2020 minutes (give your answer to 2 decimal places).

c) In those 2020 minutes, through how many degrees does the hour hand turn, and how far does its tip travel (give your answer to 2 decimal places)?

What you need and how to start

What you need: the minute hand turns 6∘6^\circ per minute and the hour hand 0.5∘0.5^\circ per minute.

Steps:

  1. Find the measure of the arc, then use l=πRn180l=\dfrac{\pi Rn}{180}.
  2. Use S=πR2n360S=\dfrac{\pi R^2n}{360}.
  3. Do the same for the hour hand.
Show solution

a) The minute hand turns 20⋅6∘=120∘20\cdot6^\circ=120^\circ, so l=π⋅10⋅120180=20π3≈20.94l=\dfrac{\pi\cdot10\cdot120}{180}=\dfrac{20\pi}{3}\approx20.94 (cm).

b) We have S=π⋅102⋅120360=100π3≈104.72S=\dfrac{\pi\cdot10^2\cdot120}{360}=\dfrac{100\pi}{3}\approx104.72 (cm²).

c) The hour hand turns 20⋅0.5∘=10∘20\cdot0.5^\circ=10^\circ, so l=π⋅6⋅10180=π3≈1.05l=\dfrac{\pi\cdot6\cdot10}{180}=\dfrac{\pi}{3}\approx1.05 (cm).

So a) about 20.9420.94 cm; b) about 104.72104.72 cm²; c) 10∘10^\circ and about 1.051.05 cm.

Answer. a) 20π3≈20.94\dfrac{20\pi}{3}\approx20.94 cm; b) 100π3≈104.72\dfrac{100\pi}{3}\approx104.72 cm²; c) 10∘10^\circ and π3≈1.05\dfrac{\pi}{3}\approx1.05 cm.

Exercise 8 Challenge

Consider the circle (O,6 cm)(O, 6\text{ cm}) and a chord AB=6AB=6 cm. Find the area of the segment bounded by the chord ABAB and the minor arc ABAB (give your answer to 2 decimal places).

What you need and how to start

What you need: the area of the segment is the area of the sector OABOAB minus the area of the triangle OABOAB.

Steps:

  1. Show that the triangle OABOAB is equilateral, so the minor arc ABAB measures 60∘60^\circ.
  2. Find the area of the sector and the area of the equilateral triangle.
Show solution

We have OA=OB=AB=6OA=OB=AB=6 cm, so the triangle OABOAB is equilateral, hence ∠AOB=60∘\angle AOB=60^\circ. The sector has area π⋅36⋅60360=6π\dfrac{\pi\cdot36\cdot60}{360}=6\pi (cm²). The equilateral triangle with side 66 cm has height 333\sqrt3 cm, so its area is 12⋅6⋅33=93\dfrac12\cdot6\cdot3\sqrt3=9\sqrt3 (cm²). Hence

S=6π−93≈3.26 (cm2).S=6\pi-9\sqrt3\approx3.26\ \text{(cm}^2\text{)}.

So the area of the segment is about 3.263.26 cm².

Answer. S=6π−93≈3.26S=6\pi-9\sqrt3\approx3.26 cm².

Mini test

Mini test

Marked like the Vietnamese graduation exam: Part I 0.25 points per question; Part II up to 1 point per question (1, 2, 3, 4 correct statements give 0.1, 0.25, 0.5, 1 point); Part III 0.5 points per question.

Part I. Multiple choice

Question 1. The length of an arc of 40∘40^\circ on a circle of radius 99 cm is
Question 2. The area of a sector of radius 44 cm with an arc of 90∘90^\circ is
Question 3. The area of the annulus between two concentric circles of radii 77 cm and 44 cm is
Question 4. A wheel has diameter 6060 cm. The circumference of the wheel is

Part II. True or false

Question 1. Consider a sector of radius 66 cm with an arc of 120∘120^\circ. Decide whether each statement is true or false.
a) The arc of the sector has length 4π4\pi cm.
b) The area of the sector is 12π12\pi cm².
c) The area of the sector equals the arc length times the radius.
d) The perimeter of the sector is (4π+12)(4\pi+12) cm.

Part III. Short answer

Question 1. Find the length (in cm) of an arc of 36∘36^\circ on a circle of radius 1010 cm (give your answer to 2 decimal places).

Question 2. Find the area (in cm²) of the annulus between two concentric circles of radii 66 cm and 44 cm (give your answer to the nearest whole number).

Show answers and explanations
  1. Part I, question 1: B. l=π⋅9⋅40180=2πl=\dfrac{\pi\cdot9\cdot40}{180}=2\pi. A uses 2πR2\pi R instead of πR\pi R, C uses the sector-area formula, and D divides by 360360 instead of 180180.
  2. Part I, question 2: D. S=π⋅16⋅90360=4πS=\dfrac{\pi\cdot16\cdot90}{360}=4\pi. A is the arc length, B is the whole disc, and C divides by 180180 instead of 360360.
  3. Part I, question 3: A. S=π(49−16)=33πS=\pi\left(49-16\right)=33\pi. B computes π(R−r)2\pi(R-r)^2, and C adds the two squares.
  4. Part I, question 4: B. C=πd=60πC=\pi d=60\pi cm. C takes 6060 as the radius, and D computes the area.
  5. Part II, question 1: a) True; b) True; c) False; d) True. a) True: l=π⋅6⋅120180=4πl=\dfrac{\pi\cdot6\cdot120}{180}=4\pi. b) True: S=π⋅36⋅120360=12πS=\dfrac{\pi\cdot36\cdot120}{360}=12\pi. c) False: S=lR2S=\dfrac{lR}{2}. d) True: the perimeter is the arc plus two radii.
  6. Part III, question 1: 6.28. l=π⋅10⋅36180=2π≈6.28l=\dfrac{\pi\cdot10\cdot36}{180}=2\pi\approx6.28 cm.
  7. Part III, question 2: 63. S=π(36−16)=20π≈62.83≈63S=\pi\left(36-16\right)=20\pi\approx62.83\approx63 cm².