Topics
Number Systems
Real Numbers
Algebra
Polynomials
Pair of Linear Equations in Two Variables
- Introduction to linear equations in two variables
- Graphical Method
- 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
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
Arithmetic Progressions
Coordinate Geometry
Lines (In Two-dimensions)
Constructions
- Division of a Line Segment
- Construction of Tangents to a Circle
- Constructions Examples and Solutions
Geometry
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
- Concept of Angle Bisector
- Similarity of Triangles
- Ratio of Sides of Triangle
Circles
Trigonometry
Introduction to Trigonometry
- Trigonometry
- Trigonometry
- Trigonometric Ratios
- Trigonometric Ratios and Its Reciprocal
- Trigonometric Ratios of Some Special Angles
- Trigonometric Ratios of Complementary Angles
- Trigonometric Identities
- Proof of Existence
- Relationships Between the Ratios
Trigonometric Identities
Some Applications of Trigonometry
Mensuration
Areas Related to Circles
- Perimeter and Area of a Circle - A Review
- Areas of Sector and Segment of a Circle
- Areas of Combinations of Plane Figures
- Circumference of a Circle
- Area of Circle
Surface Areas and Volumes
- 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
Statistics and Probability
Statistics
Probability
Internal Assessment
Circumference of a Circle
The circumference, or the perimeter, of a circle, refers to the measurement of the border across any 2D circular shape, including the circle. 
Formula: `"Circumference"/"diameter"` = π
circumference = π × diameter
= π × 2r (where r is the radius of the circle)
= 2πr
Circumference of a circle: 2πr
Example
What is the circumference of a circle of diameter 10 cm (Take π = 3.14)?
Diameter of the circle (d) = 10 cm
Circumference of circle = πd
= 3.14 × 10 cm
= 31.4 cm
So, the circumference of the circle of diameter 10 cm is 31.4 cm.
Example
Find the perimeter of the given shape (Take π = `22/7`).
The outer boundary, of this figure, is made up of semicircles.
Diameter of each semicircle is 14 cm.
We know that:
Circumference of the circle = πd
Circumference of the semicircle = `1/2`πd
= `1/2 xx 22/7` × 14 cm
= 22 cm
Circumference of each of the semicircles is 22 cm
Therefore, perimeter of the given figure = 4 × 22 cm = 88 cm.
Example
Sudhanshu divides a circular disc of radius 7 cm in two equal parts. What is the perimeter of each semicircular shape disc? (Use π = `22/7`)
Given that radius (r) = 7 cm.
We know that the circumference of circle = 2πr
So, the circumference of the semicircle = `1/2` × 2πr = πr
= `22/7` × 7 cm
= 22 cm
So, the diameter of the circle =2r = 2 × 7 cm = 14 cm
Thus, perimeter of each semicircular disc = 22 cm + 14 cm = 36 cm.
Example
The radius of a circular plot is 7.7 metres. How much will it cost to fence the plot with 3 rounds of wire at the rate of 50 rupees per metre?
Circumference of circular plot = 2πr = `2 xx 22/7 xx 7.7` = 48.4
Length of wire required for one round of fencing = 48.4 m.
Cost of one round of fence = length of wire × cost per metre.
= 48.4 × 50
= 2420 rupees.
Cost of 3 rounds of fencing = 3 × 2420 = 7260 rupees
Example
The radius of the wheel of a bus is 0.7 m. How many rotations will a wheel complete while traveling a distance of 22 km?
Circumference of circle = πd = `22/7` × 1.4 = 4.4 m
When the wheel completes one rotation it crosses a distance of 4.4 m., (1 rotation = 1 circumference)
Total number of rotations = `"distance"/"circumference"`
= `(22000)/(4.4)`
= `(220000)/(44)`
= 5000
A wheel completes 10,000 rotations to cover a distance of 22 km.
Video Tutorials
Shaalaa.com | How to solve problems based on circumference and area of the circle Part 1
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