Sec 3 E-Math Arc length and sector area worksheet with answers

A free Sec 3 E-Math 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 E-Math worksheet: Arc length and sector area

Total: 30 marksSuggested time: 45 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

Express

263^\circ

in radians, correct to 3 significant figures.

Answerrad[2]

A sector of a circle of radius

10

cm has an arc of length

10

cm. Find the area of the sector.

Answercm²[2]

A sector of a circle has radius

4

cm and angle

2.4

radians. Find the length of its arc.

Answercm[2]

AB

is a chord of a circle with centre

O

and radius

8

cm, and angle

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

radians. Find the area of the shaded segment, correct to 3 significant figures.

Answercm²[3]

AB

is a chord of a circle with centre

O

and radius

7

cm, and angle

AOB = 135^\circ

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

Answercm²[3]

Section B

In the diagram,

OAB

is a sector of a circle with centre

O

and radius

7

cm, and angle

AOB = 1

radians. The tangent to the circle at

A

meets

OB

produced at

T

. The region bounded by

AT

BT

and the arc

AB

is shaded.

(a)Find the length of

AT

, correct to 3 significant figures.

Answer (a)cm[2]

(b)Find the length of

BT

Answer (b)cm[2]

(c)Find the perimeter of the shaded region.

Answer (c)cm[2]

(d)Find the area of the shaded region.

Answer (d)cm²[3]

The diagram shows a circle with centre

O

and radius

12

cm.

AB

is a chord of length

19

cm. The minor segment cut off by

AB

is shaded.

(a)Show that angle

AOB = 1.827

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:

263^\circ = 4.59

rad (3 s.f.)[2]

To change degrees to radians, multiply by

\dfrac{\pi}{180}

\mathstrut 263^\circ = 263 \times \dfrac{\pi}{180}

263^\circ = 4.5902\ldots

263^\circ = 4.59

rad (3 s.f.)

Question 2

Answer:

A = 50

cm²[2]

\mathstrut \theta = \dfrac{10}{10}

\theta = 1

\mathstrut A = \dfrac{1}{2} \times 10^2 \times 1

A = 50

cm²

Question 3

Answer:

s = 9.6

cm[2]

s = r\theta

\mathstrut s = 4 \times 2.4

s = 9.6

Question 4

Answer:

\phantom{\text{Segment}} = 9.78

cm² (3 s.f.)[3]

\mathstrut \text{Sector} = \dfrac{1}{2} \times 8^2 \times \dfrac{2\pi}{5}

\text{Sector} = 40.2123\ldots

\text{Triangle} = \dfrac{1}{2} \times 8 \times 8 \times \sin \dfrac{2\pi}{5}

\text{Triangle} = 30.4338\ldots

\mathstrut \text{Segment} = \text{Sector} - \text{Triangle}

\text{Segment} = 9.7785\ldots

\phantom{\text{Segment}} = 9.78

cm² (3 s.f.)

Question 5

Answer:

\phantom{\text{Segment}} = 40.4

cm² (3 s.f.)[3]

\mathstrut \text{Sector} = \dfrac{135}{360} \times \pi(7)^2

\text{Sector} = 57.7267\ldots

\text{Triangle} = \dfrac{1}{2} \times 7 \times 7 \times \sin 135^\circ

\text{Triangle} = 17.3241\ldots

\mathstrut \text{Segment} = \text{Sector} - \text{Triangle}

\text{Segment} = 40.4026\ldots

\phantom{\text{Segment}} = 40.4

cm² (3 s.f.)

Question 6

(a)Answer:

10.9

cm[2]

\angle OAT = 90^\circ

(tangent ⊥ radius)

\tan 1 = \dfrac{AT}{7}

AT = 7 \tan 1

\phantom{AT} = 10.9018\ldots

AT = 10.9

(b)Answer:

5.96

cm[2]

\cos 1 = \dfrac{7}{OT}

OT = \dfrac{7}{\cos 1}

\phantom{OT} = 12.9557\ldots

BT = 12.9557\ldots - 7

\phantom{BT} = 5.9557\ldots

BT = 5.96

(c)Answer:

23.9

cm[2]

\text{Arc } AB = 7 \times 1

\phantom{\text{Arc } AB} = 7

\text{Perimeter} = 10.9018\ldots + 5.9557\ldots + 7

\phantom{\text{Perimeter}} = 23.8575\ldots

\text{Perimeter} = 23.9

(d)Answer:

13.7

cm²[3]

\text{Area of } \triangle OAT = \dfrac{1}{2} \times 7 \times 10.9018\ldots

\phantom{\text{Area of } \triangle OAT} = 38.1564\ldots

\text{Area of sector } OAB = \dfrac{1}{2} \times 7^2 \times 1

\phantom{\text{Area of sector } OAB} = 24.5

\text{Shaded area} = 38.1564\ldots - 24.5

\phantom{\text{Shaded area}} = 13.6564\ldots

\text{Shaded area} = 13.7

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{9.5}{12}

\angle AOM = 0.9135\ldots

rad

\angle AOB = 2 \times 0.9135\ldots

\phantom{\angle AOB} = 1.8270\ldots

\angle AOB = 1.827

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

(b)Answer:

53.5

cm[2]

\text{Reflex } \angle AOB = 2\pi - 1.8270\ldots

\phantom{\text{Reflex } \angle AOB} = 4.4561\ldots

rad

\text{Major arc } AB = 12 \times 4.4561\ldots

\phantom{\text{Major arc } AB} = 53.4734\ldots

\text{Major arc } AB = 53.5

(c)Answer:

61.9

cm²[3]

\text{Area of sector } OAB = \dfrac{1}{2} \times 12^2 \times 1.8270\ldots

\phantom{\text{Area of sector } OAB} = 131.5486\ldots

\text{Area of } \triangle OAB = \dfrac{1}{2} \times 12^2 \times \sin 1.8270\ldots

\phantom{\text{Area of } \triangle OAB} = 69.6486\ldots

\text{Area of segment} = 131.5486\ldots - 69.6486\ldots

\phantom{\text{Area of segment}} = 61.8999\ldots

\text{Area of segment} = 61.9

cm²

(d)Answer:

40.9

cm[2]

\text{Minor arc } AB = 12 \times 1.8270\ldots

\phantom{\text{Minor arc } AB} = 21.9247\ldots

\text{Perimeter} = 21.9247\ldots + 19

\phantom{\text{Perimeter}} = 40.9247\ldots

\text{Perimeter} = 40.9

Practice questions by skill

Sec 3 E-Math 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

In the diagram,

OAB

is a sector of a circle with centre

O

and radius

7

cm, and angle

AOB = 1

radians. The tangent to the circle at

A

meets

OB

produced at

T

. The region bounded by

AT

BT

and the arc

AB

is shaded.

(a)Find the length of

AT

, correct to 3 significant figures.

Answer (a)cm[2]

(b)Find the length of

BT

Answer (b)cm[2]

(c)Find the perimeter of the shaded region.

Answer (c)cm[2]

(d)Find the area of the shaded region.

Answer (d)cm²[3]

Answers and worked solutions

Question 1

Answer:

263^\circ = 4.59

rad (3 s.f.)[2]

To change degrees to radians, multiply by

\dfrac{\pi}{180}

\mathstrut 263^\circ = 263 \times \dfrac{\pi}{180}

263^\circ = 4.5902\ldots

263^\circ = 4.59

rad (3 s.f.)

Question 2

Answer:

A = 50

cm²[2]

\mathstrut \theta = \dfrac{10}{10}

\theta = 1

\mathstrut A = \dfrac{1}{2} \times 10^2 \times 1

A = 50

cm²

Question 3

Answer:

s = 9.6

cm[2]

s = r\theta

\mathstrut s = 4 \times 2.4

s = 9.6

Question 4

Answer:

\phantom{\text{Segment}} = 9.78

cm² (3 s.f.)[3]

\mathstrut \text{Sector} = \dfrac{1}{2} \times 8^2 \times \dfrac{2\pi}{5}

\text{Sector} = 40.2123\ldots

\text{Triangle} = \dfrac{1}{2} \times 8 \times 8 \times \sin \dfrac{2\pi}{5}

\text{Triangle} = 30.4338\ldots

\mathstrut \text{Segment} = \text{Sector} - \text{Triangle}

\text{Segment} = 9.7785\ldots

\phantom{\text{Segment}} = 9.78

cm² (3 s.f.)

Question 5

Answer:

\phantom{\text{Segment}} = 40.4

cm² (3 s.f.)[3]

\mathstrut \text{Sector} = \dfrac{135}{360} \times \pi(7)^2

\text{Sector} = 57.7267\ldots

\text{Triangle} = \dfrac{1}{2} \times 7 \times 7 \times \sin 135^\circ

\text{Triangle} = 17.3241\ldots

\mathstrut \text{Segment} = \text{Sector} - \text{Triangle}

\text{Segment} = 40.4026\ldots

\phantom{\text{Segment}} = 40.4

cm² (3 s.f.)

Question 6

(a)Answer:

10.9

cm[2]

\angle OAT = 90^\circ

(tangent ⊥ radius)

\tan 1 = \dfrac{AT}{7}

AT = 7 \tan 1

\phantom{AT} = 10.9018\ldots

AT = 10.9

(b)Answer:

5.96

cm[2]

\cos 1 = \dfrac{7}{OT}

OT = \dfrac{7}{\cos 1}

\phantom{OT} = 12.9557\ldots

BT = 12.9557\ldots - 7

\phantom{BT} = 5.9557\ldots

BT = 5.96

(c)Answer:

23.9

cm[2]

\text{Arc } AB = 7 \times 1

\phantom{\text{Arc } AB} = 7

\text{Perimeter} = 10.9018\ldots + 5.9557\ldots + 7

\phantom{\text{Perimeter}} = 23.8575\ldots

\text{Perimeter} = 23.9

(d)Answer:

13.7

cm²[3]

\text{Area of } \triangle OAT = \dfrac{1}{2} \times 7 \times 10.9018\ldots

\phantom{\text{Area of } \triangle OAT} = 38.1564\ldots

\text{Area of sector } OAB = \dfrac{1}{2} \times 7^2 \times 1

\phantom{\text{Area of sector } OAB} = 24.5

\text{Shaded area} = 38.1564\ldots - 24.5

\phantom{\text{Shaded area}} = 13.6564\ldots

\text{Shaded area} = 13.7

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{9.5}{12}

\angle AOM = 0.9135\ldots

rad

\angle AOB = 2 \times 0.9135\ldots

\phantom{\angle AOB} = 1.8270\ldots

\angle AOB = 1.827

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

(b)Answer:

53.5

cm[2]

\text{Reflex } \angle AOB = 2\pi - 1.8270\ldots

\phantom{\text{Reflex } \angle AOB} = 4.4561\ldots

rad

\text{Major arc } AB = 12 \times 4.4561\ldots

\phantom{\text{Major arc } AB} = 53.4734\ldots

\text{Major arc } AB = 53.5

(c)Answer:

61.9

cm²[3]

\text{Area of sector } OAB = \dfrac{1}{2} \times 12^2 \times 1.8270\ldots

\phantom{\text{Area of sector } OAB} = 131.5486\ldots

\text{Area of } \triangle OAB = \dfrac{1}{2} \times 12^2 \times \sin 1.8270\ldots

\phantom{\text{Area of } \triangle OAB} = 69.6486\ldots

\text{Area of segment} = 131.5486\ldots - 69.6486\ldots

\phantom{\text{Area of segment}} = 61.8999\ldots

\text{Area of segment} = 61.9

cm²

(d)Answer:

40.9

cm[2]

\text{Minor arc } AB = 12 \times 1.8270\ldots

\phantom{\text{Minor arc } AB} = 21.9247\ldots

\text{Perimeter} = 21.9247\ldots + 19

\phantom{\text{Perimeter}} = 40.9247\ldots

\text{Perimeter} = 40.9