Circular Measure: Additional Mathematics Tingkatan 5 questions

Section A: objective questions

9 marks

Choose the correct answer. Each question carries 1 mark.

What is the size of the angle subtended at the centre of a circle by an arc whose length is equal to the radius?

1°
π rad
1 rad
2 rad

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This is the definition of one radian. An arc of length πr subtends π rad and an arc of length 2r subtends 2 rad.

For the teacher: Remembering, low difficulty, standard 1.1.1

Textbook p. 3

Convert 225° into radians, in terms of π.

3π4
π4
3π2
5π4

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225° × π180° = 5π4 rad. 3π4 is 135° and 3π2 is 270°.

For the teacher: Applying, low difficulty, standard 1.1.1

Textbook p. 4

Convert 2.4 rad into degrees. [Use π = 3.142]

137.5°
13.75°
7.5°
0.042°

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2.4 × 180°3.142 = 137.5°. 13.75° and 0.042° come from dividing by the wrong number.

For the teacher: Applying, medium difficulty, standard 1.1.1

Textbook p. 3

A sector has a radius of 7 cm and subtends an angle of 0.8 rad at the centre. Its arc length is

8.75 cm
0.114 cm
5.6 cm
2.8 cm

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s = rθ = 7 × 0.8 = 5.6 cm. 8.75 comes from dividing r by θ.

For the teacher: Applying, low difficulty, standard 1.2.1

Textbook p. 6

An arc of length 15 cm is drawn in a circle of radius 6 cm. Find the angle subtended at the centre.

2.5 rad
90 rad
0.4 rad
9 rad

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θ = s ÷ r = 15 ÷ 6 = 2.5 rad. 0.4 is r ÷ s.

For the teacher: Applying, medium difficulty, standard 1.2.1

Textbook p. 7

Find the area of a sector with radius 10 cm and angle 1.2 rad.

12 cm²
120 cm²
6 cm²
60 cm²

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A = 12r²θ = 12 × 100 × 1.2 = 60 cm². 120 forgets the 12.

For the teacher: Applying, medium difficulty, standard 1.3.1

Textbook p. 13

A sector has radius 5 cm and angle 2 rad. What is the perimeter of the sector?

10 cm
15 cm
20 cm
25 cm

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Arc = 5 × 2 = 10 cm. Perimeter = arc + two radii = 10 + 5 + 5 = 20 cm. 10 cm is only the arc.

For the teacher: Understanding, medium difficulty, standard 1.2.1

Textbook p. 7

A sector has an area of 24 cm² and a radius of 6 cm. Its angle at the centre is

23 rad
43 rad
83 rad
4 rad

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12 × 36 × θ = 24, so 18θ = 24 and θ = 43 rad. 83 forgets the 12.

For the teacher: Analysing, medium difficulty, standard 1.3.1

Textbook p. 14

Sector AOB has radius 10 cm and ∠AOB = 1.2 rad. Find the area of the segment cut off by chord AB. (sin 1.2 = 0.932)

60.0 cm²
46.6 cm²
106.6 cm²
13.4 cm²

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Sector = 60 cm². Triangle = 12 × 100 × 0.932 = 46.6 cm². Segment = 60 − 46.6 = 13.4 cm². 60 is the sector alone; 106.6 adds instead of subtracting.

For the teacher: Applying, high difficulty, standard 1.3.2

Textbook p. 15

Section B: structured questions

36 marks

Answer all questions. Show your working. Use π = 3.142 unless told otherwise.

Convert each angle as asked.

[1 mark] (a)3π4 rad into degrees

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3π4 × 180°π = 135° [1]1

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[1 mark] (b)0.5 rad into degrees, correct to 2 decimal places

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0.5 × 180°3.142 = 28.64° [1]1

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[1 mark] (c)72° into radians, in terms of π

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72° × π180° = 2π5 rad [1]1

72 ÷ 180 simplifies to 25.

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For the teacher: Applying, low difficulty, standard 1.1.1

Textbook p. 4

Diagram 1 shows sector AOB with centre O.

Sector AOB with centre O, radius 8 cm and angle AOB of 1.5 radians — Diagram 1

[1 mark] (a)Find the arc length AB, in cm.

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s = 8 × 1.5 = 12 cm [1]1

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[1 mark] (b)Find the perimeter of the sector, in cm.

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8 + 8 + 12 = 28 cm [1]1

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[1 mark] (c)Find the area of the sector, in cm².

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12 × 8² × 1.5 = 48 cm² [1]1

Check: 12rs = 12 × 8 × 12 = 48.

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For the teacher: Applying, low difficulty, standard 1.2.1

Textbook p. 6

In a circle with centre O, the minor arc AB has length 9.1 cm and subtends ∠AOB = 1.4 rad at O.

[1 mark] (a)Find the radius of the circle, in cm.

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r = 9.1 ÷ 1.4 = 6.5 cm [1]1

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[1 mark] (b)Point C lies on the circle such that ∠BOC = 0.8 rad. Find the arc length BC, in cm.

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6.5 × 0.8 = 5.2 cm [1]1

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[2 marks] (c)Find the length of the major arc AB, in cm.

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Reflex angle = 2π − 1.4 = 4.884 rad [1]1
Major arc = 6.5 × 4.884 = 31.75 cm [1]1

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For the teacher: Applying, medium difficulty, standard 1.2.1

Textbook p. 7

Diagram 2 shows a circle with centre O and radius 12 cm. The chord AB subtends 80° at the centre.

Circle with centre O and radius 12 cm; chord AB cuts off a shaded segment and angle AOB is 80 degrees — Diagram 2

[1 mark] (a)Convert 80° into radians, correct to 3 decimal places.

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80° × 3.142180° = 1.396 rad [1]1

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[1 mark] (b)Find the length of the arc AB, in cm.

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12 × 1.396 = 16.75 cm [1]1

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[2 marks] (c)Find the length of the chord AB, in cm.

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AB = 2 × 12 × sin 40° or AB² = 12² + 12² − 2(12)(12) cos 80° [1]1
AB = 15.43 cm [1]1

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[1 mark] (d)Find the perimeter of the shaded segment, in cm.

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16.75 + 15.43 = 32.18 cm [1]1

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For the teacher: Applying, medium difficulty, standard 1.2.2

Textbook p. 8

Diagram 3 shows two sectors AOB and POQ with the same centre O. OA = 5 cm, AP = 4 cm and ∠AOB = 1.2 rad.

Two sectors with centre O and angle 1.2 radians; the inner has radius 5 cm and the shaded ring part adds 4 cm — Diagram 3

[1 mark] (a)Find the area of sector POQ, in cm².

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OP = 9 cm, A = 12 × 81 × 1.2 = 48.6 cm² [1]1

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[1 mark] (b)Find the area of sector AOB, in cm².

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12 × 25 × 1.2 = 15 cm² [1]1

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[1 mark] (c)Find the area of the shaded region, in cm².

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48.6 − 15 = 33.6 cm² [1]1

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For the teacher: Applying, medium difficulty, standard 1.3.3

Textbook p. 17

A sector of a circle of radius r cm subtends an angle θ rad at the centre. Its perimeter is 16 cm and its area is 16 cm².

[2 marks] (a)Write down two equations in r and θ.

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2r + rθ = 16 [1]1
12r²θ = 16 [1]1

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[2 marks] (b)Hence find the value of r.

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rθ = 16 − 2r, so 12r(16 − 2r) = 16 and r² − 8r + 16 = 0 [1]1
(r − 4)² = 0, r = 4 cm [1]1

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[1 mark] (c)Find the value of θ.

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θ = (16 − 8) ÷ 4 = 2 rad [1]1

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For the teacher: Analysing, high difficulty, standard 1.3.1

Textbook p. 14

Diagram 5 shows the cross-section of a tunnel. ABC is a major segment of a circle with centre O and radius 5 m, and ∠AOC = 2 rad. [sin 2 = 0.9093, sin 1 = 0.8415]

Major segment ABC of a circle with centre O and radius 5 m; angle AOC at the bottom is 2 radians — Diagram 5

[2 marks] (a)Calculate the length of the chord AC, in m.

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AC = 2 × 5 × sin 1 or AC² = 5² + 5² − 2(5)(5) cos 2 [1]1
AC = 8.415 m [1]1

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[1 mark] (b)Find the length of the major arc ABC, in m.

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5 × (2π − 2) = 5 × 4.284 = 21.42 m [1]1

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[3 marks] (c)Find the area of the cross-section of the tunnel, in m².

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Major sector = 12 × 25 × 4.284 = 53.55 m² [1]1
Triangle AOC = 12 × 25 × sin 2 = 11.37 m² [1]1
Total = 53.55 + 11.37 = 64.92 m² [1]1

O is inside the major segment, so the triangle is added.

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For the teacher: Analysing, high difficulty, standard 1.4.1

Textbook p. 20

Aiman wrote: "A sector has r = 6 cm and θ = 60°, so A = 12r²θ = 12 × 36 × 60 = 1 080 cm²."

[1 mark] (a)State the mistake in Aiman's working.

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A = 12r²θ needs θ in radians, but he used 60° [1]1

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[2 marks] (b)Find the correct area, in cm².

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60° = 60° × 3.142180° = 1.047 rad [1]1
A = 12 × 36 × 1.047 = 18.85 cm² [1]1

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For the teacher: Evaluating, medium difficulty, standard 1.3.1

Textbook p. 12

A school plans a sector-shaped flower bed with radius 6 m and area 18 m².

[1 mark] (a)Find the angle of the sector, in radians.

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12 × 36 × θ = 18, θ = 1 rad [1]1

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[1 mark] (b)Find the length of fencing needed to go all round the bed, in m.

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6 + 6 + 6 × 1 = 18 m [1]1

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[2 marks] (c)The school has 18 m of fencing and decides on a semicircular bed instead. Find the radius that uses all the fencing, correct to 2 decimal places.

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Perimeter = 2r + πr = r(2 + 3.142) = 18 [1]1
r = 18 ÷ 5.142 = 3.50 m [1]1

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For the teacher: Creating, medium difficulty, standard 1.4.1

Textbook p. 19

Section A: objective questions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Section B: structured questions

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18