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
- Significance of the Fundamental Theorem of Arithmetic
Notes
Around 300 BC, a philosopher known as Euclid of Alexandria understood that all numbers could be split into these two distinct categories. He realized that any number can be divided down over and over until you reach a group of the smallest equal numbers. And by definition, these smallest numbers are always prime numbers. All numbers are built out of smaller primes.
Take any composite number, ex: 125, and break it down, it will always be left with prime numbers i.e. 5 × 5 × 5. Euclid knew that every number could be expressed using a group of smaller primes. No matter what number you choose, it can always be built with the addition of smaller primes. This is the root of his discovery, known as the fundamental theorem of arithmetic. Every possible number has one and only one prime factorization.
Every composite number can be written as the product of the powers of primes. For example, 4 is a composite number that can be split into 2 × 2. Similarly, 9, 18, 21, 33, and 55 could be split into 3 × 3, 3 × 3 × 2, 7 × 3, 11 × 3, and 11 × 5, respectively.
Let's consider an example 4n, where n is a natural number. Check whether there is any value for which 4n ends with the digit zero. If 4n has to end with the digit zero, then its factors must be 5 × 2. Therefore, 4n = k × 5 where k is a constant.
4n = (2 × 2)n i.e. 2 × 2 × 2 × 2 ×......×n
This does not have 5 in the prime factors. Therefore, 4n doesn't end with 0.
Let us learn how to find LCM and HCF by the prime factorisation method. For example, what are the LCF and HCF of 96 and 404 by prime factorisation?
96 = 2 × 2 × 2 ×2 × 3 Therefore 96 = 25 × 3
404 = 2 × 2 × 101 Hence 404 = 22 × 101
While finding HCF, we will take the lowest common prime factor, whereas to find LCM, the highest prime factors are considered.
HCF = 22 = 4
LCM= 25 × 3 × 101 = 9696