Topics
Number Systems
Algebra
Geometry
Trigonometry
Statistics and Probability
Coordinate Geometry
Mensuration
Internal Assessment
Real Numbers
Pair of Linear Equations in Two Variables
 Linear Equation in Two Variables
 Graphical Method of Solution of a Pair of Linear Equations
 Substitution Method
 Elimination Method
 Cross  Multiplication Method
 Equations Reducible to a Pair of Linear Equations in Two Variables
 Consistency of Pair of Linear Equations
 Inconsistency of Pair of Linear Equations
 Algebraic Conditions for Number of Solutions
 Simple Situational Problems
 Pair of Linear Equations in Two Variables
 Relation Between Coefficient
Arithmetic Progressions
Quadratic Equations
 Quadratic Equations
 Solutions of Quadratic Equations by Factorization
 Solutions of Quadratic Equations by Completing the Square
 Nature of Roots of a Quadratic Equation
 Relationship Between Discriminant and Nature of Roots
 Situational Problems Based on Quadratic Equations Related to Day to Day Activities to Be Incorporated
 Application of Quadratic Equation
Polynomials
Circles
 Concept of Circle  Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles
 Tangent to a Circle
 Number of Tangents from a Point on a Circle
 Concept of Circle  Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles
Triangles
 Similar Figures
 Similarity of Triangles
 Basic Proportionality Theorem (Thales Theorem)
 Criteria for Similarity of Triangles
 Areas of Similar Triangles
 Rightangled Triangles and Pythagoras Property
 Similarity of Triangles
 Application of Pythagoras Theorem in Acute Angle and Obtuse Angle
 Triangles Examples and Solutions
 Angle Bisector
 Similarity of Triangles
 Ratio of Sides of Triangle
Constructions
Heights and Distances
Trigonometric Identities
Introduction to Trigonometry
Probability
Statistics
Lines (In Twodimensions)
Areas Related to Circles
Surface Areas and Volumes
 Concept of Surface Area, Volume, and Capacity
 Surface Area of a Combination of Solids
 Volume of a Combination of Solids
 Conversion of Solid from One Shape to Another
 Frustum of a Cone
 Concept of Surface Area, Volume, and Capacity
 Surface Area and Volume of Different Combination of Solid Figures
 Surface Area and Volume of Three Dimensional Figures
definition
Area of a circle: The area of a circle is the region occupied by the circle in a twodimensional plane.
formula
Area of the circle = πr^{2}
notes
Area of a circle:

The area of a circle is the region occupied by the circle in a twodimensional plane.

Draw a circle and cut them into a number of sectors and rearranging them as shown in Figure.
Arrange the separate pieces as shown, which is roughly a parallelogram.
The more sectors we have, the nearer we reach an appropriate parallelogram.
As done above if we divide the circle into 64 sectors, and arrange these sectors. It gives nearly a rectangle
The breadth of this rectangle is the radius of the circle, i.e., ‘r’.
As the whole circle is divided into 64 sectors and on each side we have 32 sectors, the length of the rectangle is the length of the 32 sectors, which is half of the circumference.
Area of the circle = Area of rectangle thus formed = l × b
= (Half of the circumference) × radius
= (`1/2` × 2πr) × r
= πr^{2}
So, the area of the circle = πr^{2}
Example
Find the area of a circle of radius 30 cm (use π = 3.14).
Radius, r = 30 cm
Area of the circle = πr^{2}
= 3.14 × 302
= 2,826 cm^{2}
Example
Diameter of a circular garden is 9.8 m. Find its area.
Diameter, d = 9.8 m.
Therefore, radius r = 9.8 ÷ 2 = 4.9 m
Area of the circle = πr^{2}
= `22/7` × (49.2)^{2} m^{2}
= `22/7` × 4.9 × 4.9 m^{2}
= 75.46 m^{2}
Example
The following figure shows two circles with the same center. The radius of the larger circle is 10 cm and the radius of the smaller circle is 4 cm.
Find:
(a) the area of the larger circle
(b) the area of the smaller circle
(c) the shaded area between the two circles. (π = 3.14)
(a) Radius of the larger circle = 10 cm
So, area of the larger circle= πr^{2 }= 3.14 × 10 × 10 = 314 cm^{2}
(b) Radius of the smaller circle = 4 cm
Area of the smaller circle = πr^{2 }= 3.14 × 4 × 4 = 50.24 cm^{2}
(c) Area of the shaded region = (314 – 50.24) cm^{2} = 263.76 cm^{2}
Video Tutorials
Shaalaa.com  Areas Related to Circles Ex 12 1 , Q1
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