Lesson Progress
0% Complete
area of sector 2

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)

Screen Shot 2021 11 26 at 4.52.27 AM

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.

Screen Shot 2021 11 26 at 5.45.27 AM

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.

Screen Shot 2021 11 26 at 5.55.47 AM

Solution

(a) To get angle of the minor segment

Screen Shot 2021 11 26 at 6.05.51 AM

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

Screen Shot 2021 11 26 at 6.28.10 AM

Solution: 

area of sector
\( \scriptsize \hat{BOA} = 120^o \)

\( \scriptsize \hat {OAB} = 30^o, \: \hat{OBA} = 30^o \) 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)

Screen Shot 2021 11 26 at 7.44.03 AM

 

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.

Screen Shot 2021 11 28 at 6.09.34 PM

 

Solution:

Screen Shot 2021 02 08 at 1.41.31 PM

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[56]

= 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

back-to-top
error: