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1.Number System
- Numbers and its Classification, Rational Number and its Representation on the Number Line
- Decimal Representation of Rational Number and Conversion of Decimal Number into Rational Number
- Representation of Decimal Number on Number Line, Finding Irrational Numbers Between Given Two Numbers
- Representing Irrational Number on the Number Line
- Rationalisation
- Introduction to Exponents, Laws of Exponents, BODMAS Rule
- Integral Exponents of Real Numbers
- Integral Exponents of Real Numbers and Solving Equation for x
- Rational Exponents of Real Numbers
- Examples Based on Rational Exponents of Real Numbers
- Chapter Notes – Number System
- NCERT Solutions – Number System Exercise 1.1 – 1.6
- R D Sharma Solutions Number System
- Revision Notes – Number System
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2.Polynomials
- Introduction to Polynomials and Some Basics Questions
- Value of Polynomial, Zeroes of Polynomial and Relationship between Zeroes and Coefficient of Linear Equation
- Division of Polynomials and Division Algorithm and Remainder Theorem
- Factorising by Spliting Method
- Factorisation By Different Methods and Factor Theorem
- Factorisation of Polynomial by Using Algebric Identities
- Chapter Notes – Polynomials
- NCERT Solutions – Polynomials Exercise 2.1 – 2.5
- R S Aggarwal Solutions Polynomials
- Revision Notes Polynomials
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3.Coordinate Geometry
- Representation of Position on Number Line, Finding the location by Using the Axis
- Quadrant, Coordinates of Point on Axis, Distance from Axis
- Finding the Area of Triangle and Rectangle and Mirror Image of Point On Axis
- Chapter Notes – Coordinate Geometry
- NCERT Solutions – Coordinate Geometry Exercise 3.1 – 3.3
- R D Sharma Solutions Coordinate Geometry
- R S Aggarwal Solutions Coordinate Geometry
- Revision Notes Coordinate Geometry
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4.Linear Equations
- Introduction of Linear Equation in Two Variables and Graphical Representation
- Example to find the value of Constant and Graph of Linear Equation in Two Variables
- Equations of lines Parallel to the x-axis and y-axis
- Chapter Notes – Linear Equations
- NCERT Solutions – Linear Equations Exercise 4.1 – 4.4
- R D Sharma Solutions Linear Equations
- R S Aggarwal Solutions Linear Equations
- Revision Notes Linear Equations
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5.Euclid's Geometry
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6.Lines and Angles
- Introduction to Lines and Angles, Types of Angle and Axioms
- Vertically opposite Angles and Angle bisector
- Angle made by a transversal with two Lines and Parallel Lines
- Lines Parallel to same Line; Exterior angle of a triangle; Angle Sum property of a Triangle
- Angle Sum Property of Triangle Including Interior and Exterior Angles
- Chapter Notes – Lines and Angles
- NCERT Solutions – Lines and Angles Exercise 6.1 – 6.3
- R D Sharma Solutions Lines and Angles
- R S Aggarwal Solutions Lines and Angles
- Revision Notes Lines and Angles
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7.Triangles
- Introduction, Types of Triangle, Altitude & Median of Triangle, Congruences, SAS criterion of Congruence
- Questions Based on SAS Criterion
- ASA Criterion of Congruence and Its based Questions
- Questions Based on AAS and SSS Criterion
- RHS Criterion of Congruence and Its based Questions, Questions Based on Multiple Criterion
- Inequalities of Triangle
- Inequalities of Triangle cont., Distance Between a line and The Point, Miscellaneous Questions
- Chapter Notes – Triangles
- NCERT Solutions – Triangles Exercise 7.1 – 7.5
- R D Sharma Solutions Triangles
- Revision Notes Triangles
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8.Quadrilaterals
- Congruence of Triangles- SAS Congruence Criteria, ASA Congruence Criteria, SSS Congruence Criteria, RHS Congruence Criteria; Types of Quadrilateral- Parallelogram, Rectangle, Rhombus, Square, Trapezium, Kite
- Theorem 1; Properties of parallelogram- Theorem 2, Theorem 3;
- Theorem 4; Theorem 5;
- Conditions for a Quadrilateral to be a parallelogram- Theorem 6
- Theorem 8
- Theorem 9
- Theorem 9(Mid-point Theorem);
- Theorem 10
- Chapter Notes – Quadrilaterals
- NCERT Solutions – Quadrilaterals Exercise 8.1 8.2
- R D Sharma Solutions Quadrilaterals
- R S Aggarwal Solutions Quadrilaterals
- Revision Notes Quadrilaterals
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9.Area of Parallelogram
- Parallelograms on the Same Base and between the Same Parallels; A Diagonal of a parallelogram divides it into two triangles of Equal Area
- Questions based on parallelogram and Triangle are on same base & parallel lines and Area of Triangle
- Triangles on the same Base and between the same Parallels
- Miscellaneous Questions Based on Triangles on the same Base and between the same Parallels
- Sums Based on Median of a Triangle
- Two Triangles on the same Base and equal Area lie b/w the same parallel
- Summary of All Concepts and Miscellaneous Questions
- Chapter Notes – Area of Parallelogram
- NCERT Solutions – Area of Parallelogram Exercise 9.1 – 9.4
- R D Sharma Solutions Area of Parallelogram
- Revision Notes Area of Parallelogram
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10.Constructions
- Construction of Perpendicular Bisector of a Line Segment, Different Angles and Angle bisector
- Construction of Different types of Triangle
- Chapter Notes – Constructions
- NCERT Solutions – Constructions Exercise
- R D Sharma Solutions Constructions
- R S Aggarwal Solutions Constructions
- Revision Notes Constructions
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11.Circles
- Introduction, Important Terms related to circle, Theorem-1, 2, 3 and 4
- Problems Solving
- Problems Based on Parallel Chords, Common Chord of Circles
- Problems Based on Concentric Circles, Triangle in Circle
- Equal Chords and Their Distance, Theorem-5, 6, 7, 8
- Circle Through Points, Theorem-9, Arc of a Circle, Theorem-10
- Chapter Notes – Circles
- NCERT Solutions – Circles Exercise
- R D Sharma Solutions Circles
- R S Aggarwal Solutions Circles
- Revision Notes Circles
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12.Heron's Formula
- Introduction of Heron’s Formula; Area of Different Triangles
- Area of Triangles Using Heron’s Formula (Type-1 level-1 and 2)
- Area of Quadrilaterals Using Heron’s Formula (Type-2 level-1)
- Area of Quadrilaterals Using Heron’s Formula (Type-2 level-2)
- Chapter Notes – Heron’s Formula
- NCERT Solutions – Heron’s Formula Exercise 12.1 12.2
- R D Sharma Solutions Heron’s Formula
- Revision Notes Heron’s Formula
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13.Surface Area and Volume
- Introduction, Surface Area of Cube, Cuboid, Cylinder, Hollow cylinder, Cone, Sphere
- Type-1: Surface Area and Volume of Cube and Cuboid
- Level-2, Question based on Cube and Cuboid, Level-1 Questions based on Cylinder
- Level-2 Questions based on Cylinder, Type-3 Surface Area and Volume of Right Circular Cone (Level-1)
- Level-2 Questions based on Right Circular Cone, Level-1& 2 Questions Based on Sphere & Hemisphere
- Type-5 Short Answer Types Questions,
- Chapter Notes – Surface Area and Volume of a Cuboid and Cube
- Chapter Notes – Surface Area and Volume of A Right Circular Cylinder
- Chapter Notes – Surface Area ad Volume of A Right Circular Cone
- Chapter Notes – Surface Area ad Volume of A Sphere
- NCERT Solutions – Surface Area and Volume Exercise 13.1 – 13.9
- R D Sharma Solutions Surface Areas And Volumes of A cuboid and A cube
- R D Sharma Solutions Surface Areas And Volumes of A Right Circular Cylinder
- R D Sharma Solutions Surface Areas And Volumes of A Right Circular Cone
- R D Sharma Solutions Surface Areas And Volumes of A Sphere
- Revision Notes Surface Areas And Volumes
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14.Statistics
- Data; Presentation of Data; Understanding Basic terms;
- Questions;
- Bar Graph, Histogram, Frequency Polygon;
- Histogram for Grouped Frequency Distribution (Class interval in Inclusive form), Frequency Polygon, Measures of Central Tendency,
- Examples Based on Mean of Ungrouped Data
- Direct Method of Finding Mean of Ungrouped Data, Median of Cont. Freq. Distribution (Grouped Data)
- Mean; Mean for an ungrouped freq. Dist. Table
- Chapter Notes – Statistics – Tabular Representation of Statistical Data
- Chapter Notes – Statistics – Graphical Representation of Statistical Data
- Chapter Notes – Statistics – Measures of Central Tendency
- NCERT Solutions – Statistics Exercise 14.1 – 14.4
- R D Sharma Solutions Tabular Representation of Statistical Data
- R D Sharma Solutions Graphical Representation of Statistical Data
- R S Aggarwal Solutions Statistics
- Revision Notes Statistics
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15.Probability
- Activities; Understanding term Probability; Experiments;
- Event; Probability; Empirical Probability- Tossing a coin;
- Rolling a dice; Miscellaneous Example;
- Chapter Notes – Probability
- NCERT Solutions – Probability Exercise 15.1
- R D Sharma Solutions Probability
- R S Aggarwal Solutions Probability
- Revision Notes Probability
Chapter Notes – Heron’s Formula
(1) Area of triangle: Generally, the area of triangle is given by the following formula:
Area of a triangle = ½ x base x height
(2) Area of a triangle by Heron’s Formula: The formula given by Heron about the area of triangle is given below:
Area of a triangle =
Where a, b and c are the sides of the triangle and s is semi-perimeter i.e. half the perimeter of the triangle = (a + b + c)/2.
For Example: Find the area of triangle having three sides as 4 cm, 6 cm and 8 cm using Heron’s formula.
Let a = 4 cm, b = 6 cm and c = 8 cm.
Therefore, s = (a + b + c)/2 = (4 + 6 + 8)/2 = 9 cm.
Using Heron’s formula, we have,
Area of triangle =
=
=
= cm2 = 11.62 cm2
For Example: An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Let the given perimeter p = 30 cm and the two equal sides a = b = 12 cm.
Let the third side be denoted by c.
Now, we know that, perimeter of triangle = sum of all sides
So, p = a + b + c
30 = 12 + 12 + c
Hence, c = 6 cm.
Now, s = Perimeter of triangle/2 = 30/2 = 15 cm.
Now, using Heron’s formula,
Area of triangle =
=
=
= cm2 = 11.62 cm2
(3) Application of Heron’s Formula to Find Area of Quadrilaterals:
Let us take examples to understand this:
For Example: Kavita has a piece of land which is in the shape of a rhombus. She wants her one daughter and one son to work on the land and produce different crops. She divided the land in two equal parts. If the perimeter of the land is 400 m and one of the diagonals is 160 m, how much area each of them will get for their crops?
Let the piece of land be denoted by ABCD as shown in figure.
Given, perimeter of land is 400 m and diagonal is of 160 m.
Since, given land is in shape of rhombus, therefore each side will be 400/4 = 100 m.
Let diagonal BD = 160 m.
Now, considering ∆ ABD, its semi-perimeter will be
s = (100 + 100 + 160)/2 = 180 m.
Now, area of ∆ ABD =
=
= 4800 m2.
Hence, each one of them will get 4800 m2 of area.
For Example: A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
Now, for the given figure draw line BE parallel to AD and a perpendicular BF on CD as shown in the figure below:
It can be seen from the figure that ABED is a parallelogram, where,
BE = AD = 13 m; ED = AB = 10 m and EC = 25 – ED = 25 – 10 = 15 m.
Now, for ∆BEC, semi – perimeter s = (13 + 14 +15)/2 = 21 m.
Therefore, area of ∆BEC =
=
=
= 84 m2
We know that, area of ∆BEC = ½ x base x height
= ½ x CE x BF
84 = ½ x 15 x BF
On solving,
BF = 11.2 m
Now, area of quadrilateral ABED = base x height
= DE x BF
= 10 x 11.2 = 112 m2.
Now, total area of field = area of ∆BEC + area of quadrilateral ABED
= 84 + 112 = 196 m2.
Thus, area of field is 196 m2.