The document defines key terms and formulas related to circles, including circumference, diameter, radius, area, arcs, sectors, segments, chords, and semicircles. It provides formulas for calculating the circumference, area, arc length, area of sectors and segments. Examples are included to demonstrate how to apply the formulas to calculate values for various circle properties and parts.

1. CIRCLE
Prepared By:
SHARDA CHAUHAN
TGT MATHEMATICS

2. CIRCLE
A circle is a plain figure enclosed by a curved line,
every point on which is equidistant from a point
within, called the centre.

3. Circumference - The circumference of a
circle is the perimeter
Diameter - The diameter of a circle is longest
distance across a circle.
Radius - The radius of a circle is the distance
from the center of the circle to the outside
edge.

4. CIRCUMFERENCE
C = 2πr
C = πd
* Where π = 3.142

5. EXAMPLE (CIRCUMFERENCE)
C = πd
= 3.142 x 6 cm
= 18.85 cm
C = 2πr
= 2 x 3.142 x 4 cm
= 25.14 cm

6. AREA OF CIRCLE
A = πr2
* Where π = 3.142

7. EXAMPLE 1 (AREA OF CIRCLE)
A = πr2
= 3.142 x 62
= 3.142 x 36
= 113.11 cm2

8. EXAMPLE 2 (AREA OF CIRCLE)
A = πr2
= 3.142 x 42
= 3.142 x 16
= 50.27 cm2
𝑑
r =
2
8
=
2
= 4cm

9. ARC
A portion of the circumference of a circle.

10. ARC LENGTH (DEGREE)
𝑎
𝑙 =
𝑛
360𝑜
2𝜋r
* A circle is 360𝑜

11. EXAMPLE 1 (ARC LENGTH)
𝑎
𝑙 =
𝑛
360𝑜
2𝜋r
=
45𝑜
360𝑜
x 2 x 3.142 x 12
=
1
8
x 75.41
= 9.43 cm

12. RADIAN
The angle made by taking the radius and
wrapping it along the edge of the circle.

13. FROM RADIAN TO DEGREE
Degree =
180𝑜
π
x Radians
Radians =
π
180𝑜
x Degree
FROM DEGREE TO RADIAN

14. EXAMPLE (FROM DEGREE TO RADIAN)
1. 30𝑜 =
π
180𝑜 6
x 30𝑜 = π
rad
3. 270𝑜 =
π
180𝑜 2
x 270𝑜 = 3π
rad
2. 150𝑜 =
π
180𝑜 6
x 150𝑜 = 5π
rad

15. EXAMPLE (FROM RADIAN TO DEGREE)
3
1. π
rad =
180𝑜
π
x
π
3
rad = 60𝑜
3
2. 2π
rad =
180𝑜
π
x
2π
3
rad = 120𝑜
4
3. 5π
rad =
180𝑜
π
x
5π
4
rad = 225𝑜

16. ARC LENGTH (RADIAN)
𝑙𝑎 = r θ
* Where θ is radians

17. EXAMPLE 2 (ARC LENGTH)
𝑙𝑎 = r θ
= 4.16 cm x 2.5 rad
= 10.4 cm

18. EXAMPLE 3 (ARC LENGTH)
𝑙𝑎 = r θ
= 10 cm x
= 7.86 cm
π
4
rad
𝑙𝑎 = r θ
= 25 cm x 0.8 rad
= 20 cm

19. SECTOR
A sector is the part of a circle enclosed by two
radii of a circle and their intercepted arc.

20. AREA OF SECTOR (DEGREE)
By propotion,
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑆𝑒𝑐𝑡𝑜𝑟
= 𝐶𝑒𝑛𝑡𝑟𝑎𝑙 𝐴𝑛𝑔𝑙𝑒
=
𝐴𝑟𝑒𝑎 𝑜𝑓 𝐶𝑖𝑟𝑐𝑙𝑒 360𝑜
𝐴 𝑛
A =
𝜋𝑟2 360𝑜
𝑛
360𝑜
𝜋𝑟2

21. EXAMPLE 1 (AREA OF SECTOR)
Area =
𝑛
𝜋𝑟2
x 3.142 x 62
=
360𝑜
45𝑜
360𝑜
=
1
8
x 3.142 x 36
= 14.14 cm2

22. AREA OF SECTOR (RADIAN)
By propotion,
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑆𝑒𝑐𝑡𝑜𝑟
= 𝐶𝑒𝑛𝑡𝑟𝑎𝑙 𝐴𝑛𝑔𝑙𝑒
𝐴𝑟𝑒𝑎 𝑜𝑓 𝐶𝑖𝑟𝑐𝑙𝑒 2𝜋
𝐴
𝜋𝑟2 =
2𝜋
θ
A =
1
2
2
𝑟 θ

23. EXAMPLE 2 (AREA OF SECTOR)
Area
5 cm
O
1.4 rad

1
r2

2
1
 17.5cm2
2
5 1.4
2


24. SEGMENT
The segment of a circle is the region bounded by
a chord and the arc subtended by the chord.

25. AREA OF SEGMENT
2
 1
r2

2
1
r2
sin
2

1
r2
( sin
)

1
2 1800
2
r 
 -sin

or

26. EXAMPLE (AREA OF SEGMENT)
Solution:
(i) 𝑙𝑎 = 8 cm
𝑙𝑎 = r θ
8 = r θ
8 = 6 θ
θ = 1.333 radians
Ð AOB = 1.333 radians
The above diagram shows a sector of a
circle, with centre O and a radius 6 cm.
The length of the arc AB is 8 cm. Find
(i) Ð AOB
(ii) the area of the shaded segment.
(ii) the area of the shaded segment
2
1
𝑟2(θ - sin θ)
𝑜
= 1
(6)2(1.333 - sin (1.333 x 180
))
2 ϴ
= 1
(36)(1.333 – sin 76.38𝑜)
2
= 6.501 cm2

27. CHORD
Chord of a circle is a line segment whose ends
lie on the circle.

28. GIVEN THE RADIUS AND CENTRAL ANGLE
Chord length = 2r sin
θ
2

29. EXAMPLE 1
Chord length = 2r sin
= 2(6) sin
θ
2
90
2
= 12 x sin 45
= 8.49 cm

30. GIVEN THE RADIUS AND DISTANCE TO CENTER
This is a simple application of Pythagoras'
Theorem.
Chord length = 2 𝑟2 − 𝑑2

31. EXAMPLE 2
Find the chord of the circle where the radius
measurement is about 8 cm that is 6 units from the
middle.
Solution:
Chord length = 2
= 2
= 2
= 2
𝑟2 − 𝑑2
82 − 62
64 − 36
28
= 10.58 cm

32. SEMICIRCLE

33. PERIMETER OF A SEMICIRCLE
 Remember that the perimeter is the distance
round the outside. A semicircle has two edges.
One is half of a circumference and the other is
a diameter
 So, the formula for the perimeter of a semicircle
is:
Perimeter = πr + 2r

34. EXAMPLE (PERIMETER)
Perimeter = πr + 2r
= (3.142)
8
2
+ 8
= 20.56 cm

35. AREA OF A SEMICIRCLE
 A semicircle is just half of a circle. To find the
area of a semicircle we just take half of the
area of a circle.
 So, the formula for the area of a semicircle is:
Area =
1
2
π𝑟2

36. EXAMPLE (AREA)
2
1
Area = π𝑟2
1
= x 3.142 x 42
2
= 25.14 cm2

37. SUMMARY

38. THANK YOU !
Any Questions!