## Area of a part of a circle

Lesson objective

#### By the end of the lesson you should be able to

:Find the area of a common region between two circles.

**INTRODUCTION**

Trigonometric ratio:

Given a right angled triangle, ABC.

OppositeHypotenus
A

Adjacent

B
C

__Diagram 9__

The longest side is called the Hypotenuse (AC) (Blink AC as it is read)

Side BC is Adjacent to angle ? and side AB is opposite to angle ?.
(Blink AB and BC as they read)

Trigonometric ratios are stated with reference to angle ?

Sin of ? =

Cosine of ?=

Tangent of ? =

The ratios are abbreviated as Cos ?, Sino ? and Tan ?.

Area of sector

The following is a diagram showing a sector.

The arc subtend an angle of ?o at the centre of circle
of radius r

The area of sector AOB is

Note that the area of a sector is a fraction of the area of a circle
which is
r2.
The size of the fraction is determined by the angle of the sector eg;

- If the angle is ?; the fraction is
- If the angle is 60o, the fraction is

**Example 1**

Find the area of the following sector.

Solution:

Area of the sector =

__Area of the segment__**.**

** **

** **

A sector is made up of a triangle (OAB) and a segment
(the shaded part).

The area of the segment:

- Find the area of the sector.
- Find the area of triangle OAB.
- Subtract the area of the triangle OAB from the area of the sector.

Area of the sector =

Area of the triangle = ' x r2 x sin?

Area of the segment =

=
r2sin?.

**Example 2**

Find the area of the shaded part in the figure below. (? r = 3.142).

Diagram 13

Solution:

The area of the sector is

The area of the triangle OPQ is

x
5 x 5 x sin60 =
x
25cm x 0.866

=10.83cm2

Area of the segment (shaded part) is

13.09cm2 ' 10.83cm2

=2.26cm2.

**Example 3**

** **

**Diagram 14**

Find the area of the segment in the following diagram given that XY is
12cm and radius is 10 cm.

**Solution:**

** **

**Diagram 15**

Is perpendicular OT bisector of the chord XY?

XT=6cm and TY=6cm

Diagram 15

Using the right angled triangle OTY, find the height OT by Pythagoras
theory.

Diagram 16

Height (H) = DY2 ' TY2

H = 102 ' 62

H = 100-36 = 64

H=8cm.

Then:

Area of the triangle XOY

=
x
base x height

Base = 12cm.

Height = 8cm

Area =
x
12cm x 8cm =48cm2

To find the area of the sector, determine the value of ? which the
sector angle.

Notice that OT bisects the angle ?.

Using the right angled TOY which is

We use the trigonometric ratios,

Therefore cosine

To find the value of
,
find the cosine invade of 0.8, Cos-1 0.8 =

Read off the value from the cosine tables

Therefore the sector angle

? = 36.87 x 2 = 73.74

The area of the sector is

The area of the segment then = Area of the sector '
Area of triangle OXY

=64.36cm2 ' 48cm2

=16.36cm2

Area of a common regions, between two circles.

When two circles interact they form a common region.

Diagram 18

(i) the following are two circles with centre A and B.

Diagram 19

(ii) When the two circles intersect they form a common region.

Diagram 20

- When the circles are separated, we note that the common region is made of the two segments.

(Animate the two circles by letting them intersect to show the common region and then separate them to show the two segments).

The common area that is shaded is calculated by adding the areas of the two segments.

**EXAMPLE 1**

Find the area of the common shaded in diagram below where the circles
have an equal reading of 7cm and is

PAQ=PBQ=600

Solution

Separate the circles to show the segments.

Diagram 22

To find the area of the common region

Find the area for the segment in the circle centre A

Area of the sector =

=25.67cm2

Area of the triangle APQ

=
x
7 x 7x sin60

=21.22cm2

Area of the segment = 25.67-21.22

=4.45cm2

Note that the two segments are equal.

Therefore the area of the common region is

4.45cm x 2 = 8.9cm2

**EXAMPLE 2**

Find the area of the shaded region in the following diagram.

**Solution**

- To find the area of the sector in the circle centre P draw sector PUV.

Diagram 24

Find the area of triangle PUV

To get
find
the sin inverse 0.5= 30o.

Therefore PM = 10 Cos 30

=10 x 0.866 = 8.66cm

Area of the triangle PVU =
x
10 x8.66

=43.3 cm2

To find the area of the sector PUV, find the sector angle.

Sector angle is 2 x 30 =60o

= 37cm2

Area of the segment

= 52.37 ' 43.3

= 9.07 cm2

- To find the area of the section in circle centre Q. Draw sector

Find the area of the triangle

Sin-1 =0.7143=45.590

=
45.59o

To find QM, use

Cos 45.59 =

QM = 4.899cm

Area of triangle UVQ

=
x
10 x 4.899

=24.5cm2

To find the area of sector QUV, find the sector angle.

Sector angle = 45.59 x 2

=91.81o

=

=39cm2

Area of the segment

=39 ' 24.5

=14.49

=14.5 cm2

Area of the common region therefore is the sum of the, area of the
segment in circle centre P = 9.07 cm

And

Area of the segment in circle centre

Q = 14.5cm

Which is =9.07 + 14.5

=23.57cm2

(ii) A sector: This an area bounded by two radii and an arc.

Area of a part of a circle

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