# Area of Part of a Circle - Mathematics Form 2 Notes

## Sector

A sector is an area bounded by two radii and an arc .A minor sector has a smaller area compared to a major sector.

The orange part is the major sector while the yellow part is the minor sector.

### The Area of a Sector

The area of a sector subtending an angle θ at the Centre of the circle is given by; A= θ/360 X πr2

Example

Find the area of a sector of radius 3 cm, if the angle subtended at the Centre is given as 1400 take π as 22/7

Solution

Area A of a sector is given by;
A
=  θ  × πr2
360
Area= 140 x 22 x 32
360    7
= 11 cm2

Example

The area of the sector of a circle is 38.5 cm. Find the radius of the circle if the angle subtended at the Centre is 900.

Solution

From A = θ  × πr2, we get
360
90/360 x22/7 x r2 = 38.5
r238.5 x 360 x 7
90 x 22
r2=
49
R = 7 cm

Example

The area of a sector of radius 63 cm is 4158 cm .Calculate the angle subtended at the Centre of the circle.

Solution

4158 = θ/360 x 22/x 63 x 63
θ = 4158 x 360 x 7
22 x 63 x 63
= 120
0

## Area of a Segment of a Circle

A segment is a region of a circle bounded by a chord and an arc.

Example

In the figure above the shaded region is a segment of the circle with Centre O and radius r. AB=8 cm, ON = 3 cm, ANGLE AOB =106.30. Find the area of the shaded part.

Solution

Area of the segment = area of the sector OAPB – area of triangle OAB
=
[106.3/360 × 3.142 × 52] - [1/2 × 8 × 3 ]
= 23.19 – 12
= 11.19
cm2

## Area of a Common Region Between Two Intersecting Circles.

Example

Find the area of the intersecting circles above. If the common chord AB is 9 cm.

Solution

From ∆AO1M;
O1M = √(82 - 4.52)
= √43.75
=
6.614 cm

From ∆AO2 M;
O2M = √(62 - 4.52)
= 15.75
=
3.969 cm
The area between the intersecting circles is the sum of the areas of segments
AP1B and AP2B.
Area of segment
AP1B = area of sector O2AP1B - area of ∆O2AB
Using trigonometry, sin < AO2M = AM = 4.5= 0.75
AO2    6
Find the sine inverse of 0.75 to get 48.590 hence < AO2M = 48.590

∠AO2B = 2 X ∠ AO2M
= 2 X 48.590 = 97.180
Area of segment AP1B = 97.18/360 × 3.12 × 62 1/2 × 9 × 3.969
= 30.53 − 17.86
= 12.67 cm
2
Area of segment AP2B = area of sector O1AP2B − area of ∆O1AB
Using trigonometry, sin ∠ AO1M = AM = 4.5 = 0.5625
AO1     8
Find the sine inverse of 0.5625 to get 34.230 hence ∠ AO1M = 34.230
∠AO1B = 2X∠ AO1M
= 2 X 34.230
= 68.460

Area of segment AP2B = 68.46/360 × 3.12 × 821/2 × 9 × 6.614
= 38.24 − 29.76
= 8.48 cm
2
Therefore the area of the region between the intersecting circles is given by;
Area of segmnet AP1B + area of segment AP2B
= 12.67 + 8.48
= 21.15cm2

## Past KCSE Questions on the Topic.

1. The figure below shows a circle of radius 9cm and centre O. Chord AB is 7cm long. Calculate the area of the shaded region. (4mks)

2. The figure below shows two intersecting circles with centres P and Q of radius 8cm and 10cm respectively. Length AB = 12cm

Calculate:
1. APB (2mks)
2. AQB (2mks)
3. Area of the shaded region (6mks)
3.

The diagram above represents a circle centre o of radius 5cm. The minor arc AB subtends an angle of 1200 at the centre. Find the area of the shaded part. (3mks)
4. The figure below shows a regular pentagon inscribed in a circle of radius 12cm, centre O.

Calculate the area of the shaded part. (3mks)
5. Two circles of radii 13cm and 16cm intersect such that they share a common chord of length 20cm. Calculate the area of the shaded part. (π=22/7) (10mks)

6. Find the perimeter of the figure below, given AB, BC and AC are diameters. (4mks)

7. The figure below shows two intersecting circles. The radius of a circle A is 12cm and that of circle B is 8 cm.

If the angle MBN = 72o, calculate
1. The size of the angle MAN
2. The length of MN
3. The area of the shaded region.
8.

In the diagram above, two circles, centres A and C and radii 7cm and 24cm respectively intersect at B and D. AC = 25cm.
1. Show that angel ABC = 900
2. Calculate
1. the size of obtuse angel BAD
2. the area of the shaded part (10 mks)
9. The ends of the roof of a workshop are segments of a circle of radius 1 0m. The roof is 20m long. The angle at the centre of the circle is 120o as shown in the figure below:

1. Calculate :-
1. The area of one end of the roof
2. The area of the curved surface of the roof
2. What would be the cost to the nearest shilling of covering the two ends and the curved surface with galvanized iron sheets costing shs. 310 per square metre
10. The diagram below, not drawn to scale, is a regular pengtagon circumscribed in a circle of radius 10cm at centre O

Find;
1. The side of the pentagon
2. The area of the shaded region
11. Triangle PQR is inscribed in he circle PQ= 7.8cm, PR = 6.6cm and QR = 5.9cm. Find:

1. The radius of the circle, correct to one decimal place
2. The angles of the triangle
3. The area of shaded region

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