Lesson 2, Topic 2
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# Area of Sector

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Recall, the area of a circle is given as $$\scriptsize \pi r^2$$

In general, the area of a sector of a circle is proportional to the angle of the sector as shown in the diagram above i.e. the area of the sector XOY is $$\frac{\theta}{360^o}$$ of the whole circle

i.e. Area of Sector XOY = $$\frac {θ}{360} \scriptsize \: \times \: \pi r^2$$

Example 1

A pie chart is divided into four sectors as shown in the diagram below. Each sector represents a percentage of the whole. The two larger sectors are equal and each represents X%. What is the angle subtended by one of those larger sectors? (WAEC)

Solution:

Note: x% + x% + 21 + 9 = 100%

i.e. 2x% = 100 – 30

i.e. x% = $$\frac{70}{2}$$ = 35%

x% = 35%

by proportions 35% ≡ x°

100% = 360o

x° = $$\scriptsize 36^o \: \times \: \normalsize \frac{35}{100} \\ \scriptsize 18^o \: \times \: 7^o \\ \scriptsize 126^o$$

Example 2

In the diagram below, ABCD is a rhombus with dimensions as shown BXD is a circular arc with centre A. Calculate the area of the shaded section to the nearest cm2.

Solution

Area of shaded section = Area of rhombus – Area of sector

Area of rhombus = absinθ

= 9 x 9 x sin 70° = 81 x 0.9397

= 76.1157cm2

Area of sector = $$\frac {70}{360} \: \times \: \frac{22}{7} \: \times \; 9 \\ = \frac{99}{2} \\ \scriptsize 49.5 cm^2$$

Area of shaded section = 76.1157cm2 – 49.5cm2

= 26.6157

= 27cm2

Example 3

The diagram below shows the cross-section of a tunnel. It is in the shape of a major segment of a circle of radius 1m on a chord of length 1.6m.

Calculate

(a) the angle subtends at the centre of the circle by the major arc correct to the nearest 0.1o

(b) the area of the cross-section of the tunnel, correct to 2decimal places.

Solution

(a) To get angle of the minor segment

Sin θ = $$\frac{0.8}{1}$$

θ = sin-1 (0.8)

θ = 53.1o

Angle of minor segment = 2θ

2θ = 2 x 53.1

Angle of minor segment = 106.2°

∴ Angle of major segment = 360 – 106.2 = 253.8°

(b) Area of tunnel = Area of major Arc + Area of Triangle

Area of major Arc = $$\frac{253.8}{360} \: \times \: \frac{22}{7} \: \times \: \scriptsize 1^2 \\ \scriptsize = 2.21571429$$

Area of Triangle = $$\frac{1}{2} \: \times \: \scriptsize 1.6 \: \times \ 1 \: \times \: sin C \\ = \frac{1}{2} \scriptsize \: \times \: 1.6 \: \times \ 1 \: \times \: sin 36.9$$

(C = 90 – 53.1 = 36.9)

= 0.8 x 0.60042022

= 0.48033

Area of tunnel = 2.21571429 + 0.48033

= 2.6960504702

= 2.69m2

Example 4

Calculate the area of the shaded segment in the diagram above. Leave your answer in terms of  π

Solution:

$$\scriptsize \hat{BOA} = 120^o$$

$$\scriptsize \hat {OAB} = 30^o, \: \hat{OBA} = 30^o$$ base angles of isosceles triangle

Sin 30 = $$\frac{opp}{hyp} = \frac{x}{3}$$

x = sin 30° x 3

= 0.5 x 3 = 1.5

x = 1.5

Cos 30 = $$\frac{adj}{hyp} = \frac{y}{3}$$

y = 3 x Cos 30°

y = $$\scriptsize 3 \: \times \: \normalsize \frac{\sqrt{3}}{2}$$

y = $$\frac{3\sqrt{3}}{2}$$

or

y = 3 x 0.866

y = 2.59807621

AB = 2y

AB = $$\frac{2 \: \times \: 3 \sqrt{3}}{2}$$

AB = $$\scriptsize 3 \sqrt{3}$$

or

AB = 2 x 2.59807621

AB = 5.1961524

Area of $$\scriptsize \Delta AOB = \frac{1}{2} \scriptsize \: \times \: AB \: \times \: x$$

= $$\frac{1}{2} \scriptsize \: \times \: 3 \sqrt{3} \: \times \: 1.5$$

= $$\frac{4.5 \sqrt{3}}{2} \\ = \frac{9\sqrt{3}}{4}$$

or Area = $$\frac{1}{2} \scriptsize \: \times \: AB \: \times \: x \\ = \frac{1}{2} \scriptsize \: \times \: 5.1961\: \times \: 1.5 \\ \scriptsize = 3.897$$

Area of sector AOB =$$\frac{120}{360} \: \times \: \frac{22}{7} \: \times \: \scriptsize 3^2 \\ = \frac{3 \: \times \: 22}{7} \\ = \frac{66}{7}$$

Area = 9.42857143 = 3π

Area of shaded portion = Area of sector – Area of Triangle

Area of shaded portion = 9.42857143 – 3.897

= 5.53157143

In terms of π

Area = $$\scriptsize 3 \pi \: – \: \normalsize \frac{9 \sqrt{3}}{4}$$

Area = $$\scriptsize 3 \left [ \pi \: – \: \normalsize \frac {3 \sqrt{3}}{4} \right] \scriptsize units^2$$

Example 5

The area of circle PQR with centre O is 72cm2. What is the area of sector POQ if POQ = 40? (WAEC)

Solution

Recall, area of a sector is proportional to the angle it subtends at the centre.

i.e. let 360 $$\scriptsize \equiv$$ 72cm2

if x is the area of sector POQ

40o $$\scriptsize \equiv$$ xcm2

x = $$\frac{40}{360}\scriptsize \; \times \; 72 = 8cm^2$$

i.e x = 8cm2

Example 6

Calculate the area of the shaded part in the diagram below. All dimensions are in cm and all arcs are circular.

Solution:

Area shaded portion = 2[Area of sector – Area of Triangle]

= $$\scriptsize 2 \left [ \normalsize \frac{90}{360}\scriptsize \: \times \: \normalsize \frac{22}{7}\scriptsize \: \times \: 14^2 \: – \: \normalsize \frac{1}{2} \scriptsize \: \times \: 14 \; \times \: 14 \right ]$$

= $$\scriptsize 2 \left [ \normalsize \frac{1}{2}\scriptsize \: \times \: \normalsize \frac{22}{7}\scriptsize \: \times \: 14 \: \times \: 14 \: – \: 7 \: \times \: 14 \right]$$

= 2[11 x 14 – 98]

= 2[154 – 98]

= 2

= 112cm2

Or using the concept of Venn diagram

Area square = Area square – sector + shaded portion + Area square – sector

$$\scriptsize 14^2 = \left (14^2 \: – \: 154 \right) \: + \: shaded \: portion \: + \left (14^2 \: – \: 154 \right)$$

$$\scriptsize 196 = \left (196 \: – \: 154 \right) \: + \: shaded \: portion \: + \: \left (196 \: – \: 154 \right)$$

196 = 42 + shaded portion + 42

i.e. 196 = shaded portion + 84

∴ shaded portion = 196 – 84 = 112cm2 error: