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 Co-efficient
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
- Right-angled 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 Two-dimensions)
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 two-dimensional plane.
formula
Area of the circle = πr2
notes
Area of a circle:
-
The area of a circle is the region occupied by the circle in a two-dimensional 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
= πr2
So, the area of the circle = πr2
Example
Find the area of a circle of radius 30 cm (use π = 3.14).
Radius, r = 30 cm
Area of the circle = πr2
= 3.14 × 302
= 2,826 cm2
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 = πr2
= `22/7` × (49.2)2 m2
= `22/7` × 4.9 × 4.9 m2
= 75.46 m2
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= πr2 = 3.14 × 10 × 10 = 314 cm2
(b) Radius of the smaller circle = 4 cm
Area of the smaller circle = πr2 = 3.14 × 4 × 4 = 50.24 cm2
(c) Area of the shaded region = (314 – 50.24) cm2 = 263.76 cm2
Video Tutorials
Shaalaa.com | Areas Related to Circles Ex 12 1 , Q1
Series:
00:03:15 undefined
00:03:04 undefined
00:06:26 undefined
00:02:07 undefined
00:01:33 undefined
00:03:10 undefined
00:04:49 undefined
00:06:23 undefined
00:06:50 undefined
00:05:58 undefined
00:06:05 undefined
00:05:24 undefined
00:04:51 undefined
00:03:27 undefined
00:04:10 undefined
00:03:34 undefined
00:06:37 undefined
00:02:05 undefined
00:07:57 undefined
00:04:15 undefined
00:03:42 undefined
00:05:01 undefined
00:05:47 undefined
00:08:27 undefined
00:04:57 undefined
00:05:35 undefined
00:06:19 undefined
00:03:34 undefined
00:04:25 undefined
00:05:10 undefined
00:05:18 undefined
