enables better understandings about circles chapter . carrying all basics about circles . according to class 10

1. POWERPOINT PRESETATION OF
MATHEMATICS
SUBMITTED TO:-
INDU BATRA
SUBMITTED BY:-
MOHINI KUMARI
CLASS- X A
ROLL NO.- 42

2. -
CHAPTER- 12
AREAS RELATED TO CIRCLES

3. CIRCLE

4. INTRODUCTION

5. CONTINUE….

6. CONTINUE……
• Circumference of a circle- 2πr
• Area of a circle – πr2
• Length of an Arc of a sector of a circle with radius r and angle with degree
measure θ is θ/360x2πr
• Area of a circle with radius r and angle with degrees measure θ is θ/360 x πr2
• Area of a segment of a circle = Area of the corresponding sector – Area of the
corresponding triangle

7. PERIMETER AND AREA OF
A CIRCLE

8. Perimeter/
Circumference = 2πr
Example :- find circumference of the circle ?
Solution :- circumference of a circle = 2πr
c = 2 x 22/7 x 7
c = 44 cm
7 cm

9. Area of a circle = πr2
Example :- Find the area of circle?
solution A = πr2
A= π x 52
A = 3.14 x 5 x 5
A = 3.14 x 25
A = 78.5 cm2
5cm

10. AREAS OF SECTOR AND SEGMENT OF A
CIRCLE

11. Area of the sector of angle θ = θ / 360 x πr2
Example :- Find the area of the sector of a circle with radius 4 cm and of
angle 30 degree ?
Solution :- Area of the sector = θ / 360 x πr2
= 30 / 360 x 3.14 x 4 x4 cm2
= 12.56 / 3 cm2
= 4.19 cm2 (approx.)
30

12. Length of an arc of a sector of angle θ = θ / 360 x 2πr
Example :- find the length of the major arc given in the
fig..?
Solution :- arc length = θ/ 360 x 2 πr
= 240 / 360 x 2π x 6
= 8π
= 25.2
240
120
6

13. Area of segment of a circle = area of the corresponding sector –area of
the corresponding triangle
Example:- find the area of the segment of a circle in which area of sector is
16.75 m and area of triangle is 6.93m ?
Solution :- area of segment = area of the corresponding sector – area of the
corresponding triangle.
= 16.75 – 6.93
= 9.82 m2

14. AREAS OF COMBINATIONS OF PLAIN FIGURES
Example :- Find the area of the shaded region in fig. given, where ABCD is a
square of side 14 cm.
Solution :- Area of square ABCD = 14 x 14 cm2 = 196 cm2
Diameter of each circle = 14/ 2 cm= 7 cm
Radius of each circle = 7/2 cm
so, area of one circle = πr2 = 22/7 x 7/2 x 7/2 cm2
= 154/ 4 cm
= 77/2 cm2
Therefore, area of the four circles = 4 x 77 / 2
= 154 cm 2
Hence, area of the shaded region = (196 – 154 ) cm2
= 42 cm2
A B
C D

15. MULTIPLE
CHOICE
QUESTIONS

16. 1. If TP and TQ are two tangents to a circle with centre O so that angle
POQ = 110 degree , then angle PTQ is equal to
a) 60 b) 70 c) 80 d) 90
2. PQ is a tangent to a circle with centre O at the point P. if triangle OPQ is
an isosceles triangle than angle OQP =
a) 30 b) 45 c) 60 d) 90
3. Two equal circles touch each other externally at C and AB is a common
tangent to the circles . Then , angle ACB =
a)60 b) 45 c) 30 d) 90
4. ABC is a right angled triangle, right angled at B such that BC = 6 cm and
AB = 8cm . A circle with centre O is inscribed in triangle ABC the radius
of the circle is
a) 1cm b) 2 cm c) 3cm d) 4cm
5. PQ is a tangent drawn from a point P to a circle 2with centre O and
QOR is a diameter of the circle such that angle POR = 120 degree ,
then angle OPQ is
a) 60 b) 45 c) 30 d) 90
ANSWERS- 1. b) 70 2. b) 45 3. d) 90 4. b) 2cm 5. c) 30

17. THANK YOU