Sec 3 G2 Maths Arc length and sector area worksheet with answers

A free Sec 3 G2 Maths worksheet on arc length and sector area: 7 questions, 30 marks, short questions first and then structured ones, in the school exam style. The answers and full worked solutions are on the last page.

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Sec 3 G2 Maths worksheet: Arc length and sector area

Total: 30 marksSuggested time: 50 minutes

Name: ______________________________Class: ____________ Date: ____________

Answer all questions. Show all your working. Give non-exact answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless the question says otherwise.

Section A

A sector of a circle has radius

5

cm and angle

0.6

radians. Find the length of its arc.

Answercm[2]

(a)Express

\dfrac{\pi}{5}

radians in degrees.

Answer (a)°[1]

(b)Express

\dfrac{7\pi}{10}

radians in degrees.

Answer (b)°[1]

A sector of a circle has radius

5

cm and angle

\dfrac{2\pi}{5}

radians. Find its area, correct to 3 significant figures.

Answercm²[2]

AB

is a chord of a circle with centre

O

and radius

4

cm, and angle

AOB = \dfrac{2\pi}{5}

radians. Find the perimeter of the minor segment, correct to 3 significant figures.

Answercm[3]

Section B

OAB

is a sector of a circle with centre

O

and radius

6

cm. Angle

AOB = 108^\circ

. The chord

AB

is drawn.

(a)Express angle

AOB

in radians, in terms of

\pi

Answer (a)[1]

(b)Find the perimeter of the sector

OAB

, correct to 3 significant figures.

Answer (b)cm[2]

(c)Find the area of the shaded segment, correct to 3 significant figures.

Answer (c)cm²[3]

OAB

is a sector of a circle with centre

O

and radius

6

cm. Angle

AOB = 150^\circ

. The chord

AB

is drawn.

(a)Express angle

AOB

in radians, in terms of

\pi

Answer (a)[1]

(b)Find the perimeter of the sector

OAB

, correct to 3 significant figures.

Answer (b)cm[2]

(c)Find the area of the shaded segment, correct to 3 significant figures.

Answer (c)cm²[3]

The diagram shows a circle with centre

O

and radius

12

cm.

AB

is a chord of length

13

cm. The minor segment cut off by

AB

is shaded.

(a)Show that angle

AOB = 1.145

radians, correct to 4 significant figures.

[2]

(b)Find the length of the major arc

AB

Answer (b)cm[2]

(c)Find the area of the shaded segment.

Answer (c)cm²[3]

(d)Find the perimeter of the shaded segment.

Answer (d)cm[2]

END OF PAPER

Answers and worked solutions

Question 1

Answer:

s = 3

cm[2]

s = r\theta

\mathstrut s = 5 \times 0.6

s = 3

Question 2

(a)Answer:

\dfrac{\pi}{5} = 36^\circ

°[1]

\mathstrut \dfrac{\pi}{5} = \dfrac{180^\circ}{5}

\dfrac{\pi}{5} = 36^\circ

(b)Answer:

\dfrac{7\pi}{10} = 126^\circ

°[1]

\mathstrut \dfrac{7\pi}{10} = \dfrac{7 \times 180^\circ}{10}

\dfrac{7\pi}{10} = 126^\circ

Question 3

Answer:

\phantom{A} = 15.7

cm² (3 s.f.)[2]

A = \dfrac{1}{2}r^2\theta

\mathstrut A = \dfrac{1}{2} \times 5^2 \times \dfrac{2\pi}{5}

A = 15.7079\ldots

\phantom{A} = 15.7

cm² (3 s.f.)

Question 4

Answer:

\phantom{\text{Perimeter}} = 9.73

cm (3 s.f.)[3]

\mathstrut s = 4 \times \dfrac{2\pi}{5}

s = 5.0265\ldots

\text{Cosine rule}

AB^2 = 4^2 + 4^2 - 2(4)(4) \cos \dfrac{2\pi}{5}

AB^2 = 22.1114\ldots

AB = 4.7022\ldots

\text{Perimeter} = 9.7288\ldots

\phantom{\text{Perimeter}} = 9.73

cm (3 s.f.)

Question 5

(a)Answer:

\dfrac{3\pi}{5}

[1]

108^\circ = 108 \times \dfrac{\pi}{180}

\phantom{108^\circ} = \dfrac{3\pi}{5}

(b)Answer:

23.3

cm[2]

s = r\theta

\phantom{s} = 6 \times \dfrac{3\pi}{5}

\phantom{s} = 11.3097\ldots

\text{Perimeter} = 6 + 6 + 11.3097\ldots

\phantom{\text{Perimeter}} = 23.3

(c)Answer:

16.8

cm²[3]

\text{Sector} = \dfrac{1}{2} \times 6^2 \times \dfrac{3\pi}{5}

\phantom{\text{Sector}} = 33.9292\ldots

\text{Triangle} = \dfrac{1}{2} \times 6^2 \times \sin 108^\circ

\phantom{\text{Triangle}} = 17.1190\ldots

\text{Segment} = 33.9292\ldots - 17.1190\ldots

\phantom{\text{Segment}} = 16.8

cm²

Question 6

(a)Answer:

\dfrac{5\pi}{6}

[1]

150^\circ = 150 \times \dfrac{\pi}{180}

\phantom{150^\circ} = \dfrac{5\pi}{6}

(b)Answer:

27.7

cm[2]

s = r\theta

\phantom{s} = 6 \times \dfrac{5\pi}{6}

\phantom{s} = 15.7079\ldots

\text{Perimeter} = 6 + 6 + 15.7079\ldots

\phantom{\text{Perimeter}} = 27.7

(c)Answer:

38.1

cm²[3]

\text{Sector} = \dfrac{1}{2} \times 6^2 \times \dfrac{5\pi}{6}

\phantom{\text{Sector}} = 47.1238\ldots

\text{Triangle} = \dfrac{1}{2} \times 6^2 \times \sin 150^\circ

\phantom{\text{Triangle}} = 8.9999\ldots

\text{Segment} = 47.1238\ldots - 8.9999\ldots

\phantom{\text{Segment}} = 38.1

cm²

Question 7

(a)[2]

Let

M

be the midpoint of

AB

OM \perp AB

(⊥ from centre bisects chord)

\sin \angle AOM = \dfrac{AM}{OA}

\phantom{\sin \angle AOM} = \dfrac{6.5}{12}

\angle AOM = 0.5724\ldots

rad

\angle AOB = 2 \times 0.5724\ldots

\phantom{\angle AOB} = 1.1448\ldots

\angle AOB = 1.145

rad (to 4 s.f.) (shown)

(b)Answer:

61.7

cm[2]

\text{Reflex } \angle AOB = 2\pi - 1.1448\ldots

\phantom{\text{Reflex } \angle AOB} = 5.1383\ldots

rad

\text{Major arc } AB = 12 \times 5.1383\ldots

\phantom{\text{Major arc } AB} = 61.6601\ldots

\text{Major arc } AB = 61.7

(c)Answer:

16.9

cm²[3]

\text{Area of sector } OAB = \dfrac{1}{2} \times 12^2 \times 1.1448\ldots

\phantom{\text{Area of sector } OAB} = 82.4282\ldots

\text{Area of } \triangle OAB = \dfrac{1}{2} \times 12^2 \times \sin 1.1448\ldots

\phantom{\text{Area of } \triangle OAB} = 65.5662\ldots

\text{Area of segment} = 82.4282\ldots - 65.5662\ldots

\phantom{\text{Area of segment}} = 16.8619\ldots

\text{Area of segment} = 16.9

cm²

(d)Answer:

26.7

cm[2]

\text{Minor arc } AB = 12 \times 1.1448\ldots

\phantom{\text{Minor arc } AB} = 13.7380\ldots

\text{Perimeter} = 13.7380\ldots + 13

\phantom{\text{Perimeter}} = 26.7380\ldots

\text{Perimeter} = 26.7

Sec 3 G2 Maths Arc length and sector area worksheet with answers

In the print window, choose Save as PDF to download it.

Need the formulas first? Read the Arc length and sector area notes. Or practise it online with step-by-step help when you get one wrong.

Section B

OAB

is a sector of a circle with centre

O

and radius

6

cm. Angle

AOB = 108^\circ

. The chord

AB

is drawn.

(a)Express angle

AOB

in radians, in terms of

\pi

Answer (a)[1]

(b)Find the perimeter of the sector

OAB

, correct to 3 significant figures.

Answer (b)cm[2]

(c)Find the area of the shaded segment, correct to 3 significant figures.

Answer (c)cm²[3]

Answers and worked solutions

Question 1

Answer:

s = 3

cm[2]

s = r\theta

\mathstrut s = 5 \times 0.6

s = 3

Question 2

(a)Answer:

\dfrac{\pi}{5} = 36^\circ

°[1]

\mathstrut \dfrac{\pi}{5} = \dfrac{180^\circ}{5}

\dfrac{\pi}{5} = 36^\circ

(b)Answer:

\dfrac{7\pi}{10} = 126^\circ

°[1]

\mathstrut \dfrac{7\pi}{10} = \dfrac{7 \times 180^\circ}{10}

\dfrac{7\pi}{10} = 126^\circ

Question 3

Answer:

\phantom{A} = 15.7

cm² (3 s.f.)[2]

A = \dfrac{1}{2}r^2\theta

\mathstrut A = \dfrac{1}{2} \times 5^2 \times \dfrac{2\pi}{5}

A = 15.7079\ldots

\phantom{A} = 15.7

cm² (3 s.f.)

Question 4

Answer:

\phantom{\text{Perimeter}} = 9.73

cm (3 s.f.)[3]

\mathstrut s = 4 \times \dfrac{2\pi}{5}

s = 5.0265\ldots

\text{Cosine rule}

AB^2 = 4^2 + 4^2 - 2(4)(4) \cos \dfrac{2\pi}{5}

AB^2 = 22.1114\ldots

AB = 4.7022\ldots

\text{Perimeter} = 9.7288\ldots

\phantom{\text{Perimeter}} = 9.73

cm (3 s.f.)

Question 5

(a)Answer:

\dfrac{3\pi}{5}

[1]

108^\circ = 108 \times \dfrac{\pi}{180}

\phantom{108^\circ} = \dfrac{3\pi}{5}

(b)Answer:

23.3

cm[2]

s = r\theta

\phantom{s} = 6 \times \dfrac{3\pi}{5}

\phantom{s} = 11.3097\ldots

\text{Perimeter} = 6 + 6 + 11.3097\ldots

\phantom{\text{Perimeter}} = 23.3

(c)Answer:

16.8

cm²[3]

\text{Sector} = \dfrac{1}{2} \times 6^2 \times \dfrac{3\pi}{5}

\phantom{\text{Sector}} = 33.9292\ldots

\text{Triangle} = \dfrac{1}{2} \times 6^2 \times \sin 108^\circ

\phantom{\text{Triangle}} = 17.1190\ldots

\text{Segment} = 33.9292\ldots - 17.1190\ldots

\phantom{\text{Segment}} = 16.8

cm²

Question 6

(a)Answer:

\dfrac{5\pi}{6}

[1]

150^\circ = 150 \times \dfrac{\pi}{180}

\phantom{150^\circ} = \dfrac{5\pi}{6}

(b)Answer:

27.7

cm[2]

s = r\theta

\phantom{s} = 6 \times \dfrac{5\pi}{6}

\phantom{s} = 15.7079\ldots

\text{Perimeter} = 6 + 6 + 15.7079\ldots

\phantom{\text{Perimeter}} = 27.7

(c)Answer:

38.1

cm²[3]

\text{Sector} = \dfrac{1}{2} \times 6^2 \times \dfrac{5\pi}{6}

\phantom{\text{Sector}} = 47.1238\ldots

\text{Triangle} = \dfrac{1}{2} \times 6^2 \times \sin 150^\circ

\phantom{\text{Triangle}} = 8.9999\ldots

\text{Segment} = 47.1238\ldots - 8.9999\ldots

\phantom{\text{Segment}} = 38.1

cm²

Question 7

(a)[2]

Let

M

be the midpoint of

AB

OM \perp AB

(⊥ from centre bisects chord)

\sin \angle AOM = \dfrac{AM}{OA}

\phantom{\sin \angle AOM} = \dfrac{6.5}{12}

\angle AOM = 0.5724\ldots

rad

\angle AOB = 2 \times 0.5724\ldots

\phantom{\angle AOB} = 1.1448\ldots

\angle AOB = 1.145

rad (to 4 s.f.) (shown)

(b)Answer:

61.7

cm[2]

\text{Reflex } \angle AOB = 2\pi - 1.1448\ldots

\phantom{\text{Reflex } \angle AOB} = 5.1383\ldots

rad

\text{Major arc } AB = 12 \times 5.1383\ldots

\phantom{\text{Major arc } AB} = 61.6601\ldots

\text{Major arc } AB = 61.7

(c)Answer:

16.9

cm²[3]

\text{Area of sector } OAB = \dfrac{1}{2} \times 12^2 \times 1.1448\ldots

\phantom{\text{Area of sector } OAB} = 82.4282\ldots

\text{Area of } \triangle OAB = \dfrac{1}{2} \times 12^2 \times \sin 1.1448\ldots

\phantom{\text{Area of } \triangle OAB} = 65.5662\ldots

\text{Area of segment} = 82.4282\ldots - 65.5662\ldots

\phantom{\text{Area of segment}} = 16.8619\ldots

\text{Area of segment} = 16.9

cm²

(d)Answer:

26.7

cm[2]

\text{Minor arc } AB = 12 \times 1.1448\ldots

\phantom{\text{Minor arc } AB} = 13.7380\ldots

\text{Perimeter} = 13.7380\ldots + 13

\phantom{\text{Perimeter}} = 26.7380\ldots

\text{Perimeter} = 26.7