NCERT Solutions Class 9 Maths

NCERT Solutions for Class 9 Maths – Free PDF Updated for 2021-22 Session

NCERT Solutions for Class 9 Maths includes solutions to all the questions given in the NCERT textbook for Class 9. The students can download PDF of chapter-wise solutions to these problems, from the links provided further below in this page. These NCERT Solutions for Class 9 cover all the topics included in the NCERT textbook-like Number System, Coordinate Geometry, Polynomials, Euclid’s Geometry, Quadrilaterals, Triangles, Circles, Constructions, Surface Areas and Volumes, Statistics, Probability, etc.

With the help of these Solutions of NCERT Books for Class 9 Maths, students can practise all types of questions from the chapters. The CBSE Class 9 Maths Solutions have been designed by our experts in a well-structured format to provide several possible methods of answering the problems and ensure a proper understanding of concepts. The students are suggested to practise all these solutions thoroughly for their exams. It will also help them in building a foundation for higher-level classes.

Along with NCERT Solutions, students are also provided with other online learning materials, available in BYJU’S, such as notes, books, question papers, exemplar problems, worksheets etc. These materials are designed with respect to the CBSE syllabus and NCERT curriculum. Students are also advised to practise CBSE Class 9 Sample Papers to get an idea of the question pattern in the final exam.

15 chapters are present in the NCERT Solutions for Class 9 Maths. These chapters given in the NCERT Class 9 Maths lay a foundation for the chapters present in Class 10. The PDFs given at BYJU’S is accessible to everyone and can be downloaded by the students for free.

NCERT Solutions of Class 9th Maths Book Chapter brief:

Students having trouble in solving tough Math problems can refer to these CBSE Maths Class 9 Solutions of NCERT for better guidance and for quick review. Solving these exercises in each chapter will assure positive results.

NCERT Solutions Class 9 Maths Chapter 1 Number System – Term I

This chapter discusses different topics, including rational numbers and irrational numbers. Students will also be learning the extended version of the number line and how to represent numbers (integers, rational and irrational) on it. A total of 6 exercises are present in this chapter that contains the problems based on all the topics asked in the chapter. This chapter also teaches students the representation of terminating/non-terminating recurring decimals (and successive magnification method) as well as the presentation of square roots of 2, 3 and other non-rational numbers on the number line. The chapter also deals with the laws of integral powers and rational exponents with positive real bases in Number System.

Topics Covered in Class 9 Maths Chapter 1 Number System for First Term:

Review of representation of natural numbers, integers, rational numbers on the number line. Rational numbers as recurring/ terminating decimals. Operations on real numbers.

1. Examples of non-recurring/non-terminating decimals. Existence of non-rational numbers (irrational numbers) such as, √2, √3 and their representation on the number
2. Rationalization (with precise meaning) of real numbers of the type [latex]\frac{1}{a+b\sqrt{x}}\: and\: \frac{1}{\sqrt{x}+\sqrt{\sqrt{y}}}[/latex] (and their combinations) where x and y are natural number and a and b are integers.
3. Recall of laws of exponents with integral powers. Rational exponents with positive real bases (to be done by particular cases, allowing learner to arrive at the general laws.)

Important Formulas – 

Operations on Real Numbers

√(a/b) = √a/√b

√ab = √a √b

(√a + √b) (√a – √b) = a – b

(a + √b) (a – √b) = a² – b

(√a + √b) (√c + √d) = √ac + √ad + √bc + √bd

(√a + √b)² = a + 2√ab + b

Laws of Exponents for Real Numbers

a^m . aⁿ = a^{m + n}

(a^m)ⁿ = a^mn

a^m/aⁿ = a^{m – n}, m > n

a^mb^m = (ab)^m

Also access the following resources for Class 9 Chapter 1 Number System at BYJU’S:

• NCERT Number System Class 9 Notes
• Number System Class 9 Important Questions
• NCERT Exemplar Solutions Class 9 Number Systems
• RD Sharma Solutions Number Systems Class 9

NCERT Solutions Class 9 Maths Chapter 2 Polynomials – Term II

This chapter discusses a particular type of algebraic expression called polynomial and terminology related to it. Polynomial is an expression that consists of variables and coefficients, involving the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. The chapter also deals with Remainder Theorem and the Factor Theorem with the uses of these theorems in the factorisation of polynomials. Students will be taught several examples as well as the definition of different terms like polynomial, degrees, coefficient, zeros and terms of a polynomial. A total of 5 exercises are present in this chapter which includes problems related to all the topics mentioned in the chapter.

Topics Covered in Class 9 Maths Chapter 2 Polynomials for Second Term:

Definition of a polynomial in one variable, with examples and counter examples. Coefficients of a polynomial, terms of a polynomial and zero polynomial. Degree of a polynomial. Constant, linear, quadratic and cubic polynomials. Monomials, binomials, trinomials. Factors and multiples. Zeros of a polynomial. Factorization of ax² + bx + c, a ≠ 0 where a, b and c are real numbers, and of cubic polynomials using the Factor Theorem.

Recall of algebraic expressions and identities. Verification of identities

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

(x ± y)³ = x³ ± y³ ± 3xy (x ± y)

x³ ± y³ = (x ± y) (x² ± xy + y²)

and their use in factorization of polynomials.

Important Formulas –

(x + y)² = x² + 2xy + y²

(x – y)² = x² – 2xy + y²

x² – y² = (x + y) (x – y)

(x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx

(x + y)³ = x³ + y³ + 3xy(x + y)

(x – y)³ = x³ – y³ – 3xy(x – y)

x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – zx)

Dividend = (Divisor × Quotient) + Remainder

Remainder theorem

Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number. If p(x) is divided by the linear polynomial x – a, then the remainder is p(a).

Factor theorem

If p(x) is a polynomial of degree n > 1 and a is any real number, then (i) x – a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x – a is a factor of p(x).

Also access the following resources for Class 9 Chapter 2 Polynomials at BYJU’S:

• NCERT Polynomials Class 9 Notes 
• Polynomials Class 9 Important Questions 
• NCERT Exemplar Solutions Class 9 Polynomials
• RD Sharma Solutions factorization of Polynomials Class 9

NCERT Solutions Class 9 Maths Chapter 3 Coordinate Geometry – Term I

The chapter Coordinate Geometry includes the concepts of the cartesian plane, coordinates of a point in xy – plane, terms, notations associated with the coordinate plane, including the x-axis, y-axis, x- coordinate, y-coordinate, origin, quadrants and more. Students, in this chapter, will also be studying the concepts of Abscissa and ordinates of a point as well as plotting and naming a point in xy – plane. There are 3 exercises in this chapter that contain questions revolving around the topics mentioned in the chapter, helping the students get thorough with the concepts.

Topics Covered in Class 9 Maths Chapter 3 Coordinate Geometry for First Term:

The Cartesian plane, coordinates of a point, names and terms associated with the coordinate plane, notations, plotting points in the plane.

Important Points – 

1. To locate the position of an object or a point in a plane, we require two perpendicular lines. One of them is horizontal, and the other is vertical.
2. The plane is called the Cartesian, or coordinate plane and the lines are called the coordinate axes.
3. The horizontal line is called the x -axis, and the vertical line is called the y – axis.
4. The coordinate axes divide the plane into four parts called quadrants.
5. The point of intersection of the axes is called the origin.
6. The distance of a point from the y – axis is called its x-coordinate, or abscissa, and the distance of the point from the x-axis is called its y-coordinate, or ordinate.

Also access the following resources for Class 9 Chapter 3 Coordinate Geometry at BYJU’S:

• NCERT Coordinate Geometry Class 9 Notes
• Coordinate Geometry Class 9 Important Questions
• NCERT Exemplar Solutions Class 9 Coordinate Geometry
• RD Sharma Solutions Coordinate Geometry Class 9

NCERT Solutions Class 9 Maths Chapter 4 Linear Equations in Two Variables – Term I

Along with recalling the knowledge of linear equations in one variable, this chapter will introduce the students to the linear equation in two variables, i.e., ax + by + c = 0. Students will also learn to plot the graph of a linear equation in two variables. There are 4 exercises in this chapter that consist of questions related to finding the solutions of a linear equation, plotting a linear equation on the graph and other topics discussed in the chapter.

Topics Covered in Class 9 Maths Chapter 4 Linear Equations in Two Variables for First Term:

Recall of linear equations in one variable. Introduction to the equation in two variables. Focus on linear equations of the type ax+by+c=0. Explain that a linear equation in two variables has infinitely many solutions and justify their being written as ordered pairs of real numbers, plotting them and showing that they lie on a line. Graph of linear equations in two variables. Examples, problems from real life with algebraic and graphical solutions being done simultaneously.

Important Points –

1. An equation of the form ax + by + c = 0, where a, b and c are real numbers, such that a and b are not both zero, is called a linear equation in two variables.
2. A linear equation in two variables has infinitely many solutions.
3. The graph of every linear equation in two variables is a straight line.
4. x = 0 is the equation of the y-axis and y = 0 is the equation of the x-axis.
5. The graph of x = a is a straight line parallel to the y-axis.
6. The graph of y = a is a straight line parallel to the x-axis.
7. An equation of the type y = mx represents a line passing through the origin.

Also access the following resources for Class 9 Chapter 4 Linear Equations in Two Variables at BYJU’S:

• NCERT Linear Equations in Two Variables Class 9 Notes
• Linear Equations in Two Variables Class 9 Important Questions
• NCERT Exemplar Solutions Class 9 Linear Equations in Two Variables
• RD Sharma Solutions Linear Equations in Two Variables Class 9

NCERT Solutions Class 9 Maths Chapter 5 Introduction to Euclids Geometry

This chapter discusses Euclid’s approach to geometry and tries to link it with the present-day geometry. Introduction to Euclid’s Geometry provides the students with a method of defining common geometrical shapes and terms. Students will be taken deeper into the topic of axioms, postulates and theorems with the two exercises present in the chapter.

Topics Covered in Class 9 Maths Chapter 5 Introduction to Euclids Geometry:

History – Geometry in India and Euclid’s geometry. Euclid’s method of formalizing observed phenomenon into rigorous Mathematics with definitions, common/obvious notions, axioms/postulates and theorems. The five postulates of Euclid. Equivalent versions of the fifth postulate. Showing the relationship between axiom and theorem, for example:
(Axiom) 1. Given two distinct points, there exists one and only one line through them.
(Theorem) 2. (Prove) Two distinct lines cannot have more than one point in common.

Important Axioms and Postulates – 

Some of Euclid’s axioms are:

(1) Things which are equal to the same thing are equal to one another.

(2) If equals are added to equals, the wholes are equal.

(3) If equals are subtracted from equals, the remainders are equal.

(4) Things which coincide with one another are equal to one another.

(5) The whole is greater than the part.

(6) Things which are double of the same things are equal to one another.

(7) Things which are halves of the same things are equal to one another.

Euclid’s Five Postulates

Postulate 1 – A straight line may be drawn from any one point to any other point.

Postulate 2 – A terminated line can be produced indefinitely.

Postulate 3 – A circle can be drawn with any centre and any radius.

Postulate 4 – All right angles are equal to one another.

Postulate 5 – If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles.

Also access the following resources for Class 9 Chapter 5 Introduction to Euclids Geometry at BYJU’S:

• NCERT Introduction to Euclids Geometry Class 9 Notes
• Introduction to Euclids Geometry Class 9 Important Questions
• NCERT Exemplar Solutions Class 9 Introduction to Euclids Geometry
• RD Sharma Solutions Introduction to Euclids Geometry Class 9

NCERT Solutions Class 9 Maths Chapter 6 Lines and Angles – Term I

This chapter revolves around the theorems present in the topics of Lines and Angles. Students might often be asked to prove the statements given in the questions. There are 3 exercises in the chapter, solving which students would be able to understand the concepts covered in the chapter thoroughly. There are four axioms and eight theorems covered in the chapter.

Topics Covered in Class 9 Maths Chapter 6 Lines and Angles for First Term:

1. (Motivate) If a ray stands on a line, then the sum of the two adjacent angles so formed is 180˚ and the converse.
2. (Prove) If two lines intersect, vertically opposite angles are equal.
3. (Motivate) Results on corresponding angles, alternate angles, interior angles when a transversal intersects two parallel lines.
4. (Motivate) Lines which are parallel to a given line are parallel.
5. (Prove) The sum of the angles of a triangle is 180˚.
6. (Motivate) If a side of a triangle is produced, the exterior angle so formed is equal to the sum of the two interior opposite angles.

Important Points –

1. If a ray stands on a line, then the sum of the two adjacent angles so formed is 180° and viceversa. This property is called as the Linear pair axiom.
2. If two lines intersect each other, then the vertically opposite angles are equal.
3. If a transversal intersects two parallel lines, then

(i) each pair of corresponding angles is equal,

(ii) each pair of alternate interior angles is equal,

(iii) each pair of interior angles on the same side of the transversal is supplementary.

4. If a transversal intersects two lines such that, either

(i) any one pair of corresponding angles is equal, or

(ii) any one pair of alternate interior angles is equal, or

(iii) any one pair of interior angles on the same side of the transversal is supplementary, then the lines are parallel.

Also access the following resources for Class 9 Chapter 6 Lines and Angles at BYJU’S:

• NCERT Lines and Angles Class 9 Notes
• Lines and Angles Class 9 Important Questions
• NCERT Exemplar Solutions Class 9 Lines and Angles
• RD Sharma Solutions Lines and Angles Class 9

NCERT Solutions Class 9 Maths Chapter 7 Triangles – Term I

In this chapter, students will study in detail about the congruence of triangles, rules of congruence, some properties of triangles and the inequalities in triangles. The chapter has a total of 5 exercises, in which the students are asked “to-prove” as well as application-level problems. With this chapter, students will also learn to prove the properties that they learnt in earlier classes. The chapter also teaches the students to apply the various congruence rules while solving the problems. There are about eight theorems covered in this chapter.

Topics Covered in Class 9 Maths Chapter 7 Triangles for First Term:

1. (Motivate) Two triangles are congruent if any two sides and the included angle of one triangle is equal to any two sides and the included angle of the other triangle (SAS Congruence).
2. (Motivate) Two triangles are congruent if any two angles and the included side of one triangle is equal to any two angles and the included side of the other triangle (ASA Congruence).
3. (Motivate) Two triangles are congruent if the three sides of one triangle are equal to three sides of the other triangle (SSS Congruence).
4. (Motivate) Two right triangles are congruent if the hypotenuse and a side of one triangle are equal (respectively) to the hypotenuse and a side of the other triangle. (RHS Congruence)
5. (Prove) The angles opposite to equal sides of a triangle are equal.
6. (Motivate) The sides opposite to equal angles of a triangle are equal.
7. (Motivate) The sides opposite to equal angles of a triangle are equal.

Important Axioms and Theorems –

Axiom 7.1 (SAS congruence rule) – Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.

Theorem 7.1 (ASA congruence rule) – Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle.

Theorem 7.2 – Angles opposite to equal sides of an isosceles triangle are equal.

Theorem 7.3 – The sides opposite to equal angles of a triangle are equal.

Theorem 7.4 (SSS congruence rule) – If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

Theorem 7.5 (RHS congruence rule) – If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.

Theorem 7.6 – If two sides of a triangle are unequal, the angle opposite to the longer side is larger (or greater).

Theorem 7.7 – In any triangle, the side opposite to the larger (greater) angle is longer

Theorem 7.8 – The sum of any two sides of a triangle is greater than the third side.

Also access the following resources for Class 9 Chapter 7 Triangles at BYJU’S:

• NCERT Triangles Class 9 Notes
• Triangles Class 9 Important Questions
• NCERT Exemplar Solutions Class 9 Triangles
• RD Sharma Solutions Triangles Class 9

NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals – Term II

A figure obtained by joining four points in order is called a quadrilateral. This chapter takes the students to the depth of the topics of Quadrilaterals. The chapter contains 2 exercises that contain only one theorem to prove. However, there are a total of nine theorems that can be used to solve the application or conceptual level questions asked. Angle sum property of a Quadrilateral, types of quadrilaterals, properties of a parallelogram, and the mid-point theorem are taught explained in this chapter to help the students in learning the concepts thoroughly.

Topics Covered in Class 9 Maths Chapter 8 Quadrilaterals for Second Term:

1. (Prove) The diagonal divides a parallelogram into two congruent triangles.
2. (Motivate) In a parallelogram opposite sides are equal, and conversely.
3. (Motivate) In a parallelogram opposite angles are equal, and conversely.
4. (Motivate) A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and equal.
5. (Motivate) In a parallelogram, the diagonals bisect each other and conversely.
6. (Motivate) In a triangle, the line segment joining the mid points of any two sides is parallel to the third side and in half of it and (motivate) its converse.

Important Theorems –

A quadrilateral has four sides, four angles and four vertices.

Angle Sum Property of a Quadrilateral – The sum of the angles of a quadrilateral is 360^o.

Theorem 8.1 – A diagonal of a parallelogram divides it into two congruent triangles

Theorem 8.2 – In a parallelogram, opposite sides are equal.

Theorem 8.3 – If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Theorem 8.4 – In a parallelogram, opposite angles are equal.

Theorem 8.5 – If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.

Theorem 8.6 – The diagonals of a parallelogram bisect each other.

Theorem 8.7 – If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Theorem 8.8 – A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.

Theorem 8.9 – The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

Theorem 8.10 – The line drawn through the mid-point of one side of a triangle, parallel to another side bisects the third side.

Also access the following resources for Class 9 Chapter 8 Quadrilaterals at BYJU’S:

• NCERT Quadrilaterals Class 9 Notes
• Quadrilaterals Class 9 Important Questions
• NCERT Exemplar Solutions Class 9 Quadrilaterals
• RD Sharma Solutions Quadrilaterals Class 9

NCERT Solutions Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles

In this chapter, an attempt has been made to consolidate the knowledge about the formulae to find the areas of different figures, by studying relationships between the areas of geometric figures, provided they lie on the same base and between the same parallels. This study will also be useful in the understanding of some results on ‘similarity of triangles’. The chapter contains 4 exercises of which, most of the questions ask the students to prove the statements given.

Topics Covered in Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles:

1. (Prove) Parallelograms on the same base and between the same parallels have equal area.
2. (Motivate) Triangles on the same base (or equal bases) and between the same parallels are
equal in area

Important Theorems –

Theorem 9.1 – Parallelograms on the same base and between the same parallels are equal in area.

Theorem 9.2 – Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

Theorem 9.3 – Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.

Also access the following resources for Class 9 Chapter 9 Areas of Parallelograms and Triangles at BYJU’S:

• NCERT Areas of Parallelograms and Triangles Class 9 Notes
• Areas of Parallelograms and Triangles Class 9 Important Questions
• NCERT Exemplar Solutions Class 9 Areas of Parallelograms and Triangles
• RD Sharma Solutions Areas of Parallelograms and Triangles Class 9

NCERT Solutions Class 9 Maths Chapter 10 Circles – Term II

A circle can be defined as a collection of all the points in a plane, at a fixed distance from a fixed point in the plane. Topics like Angle Subtended by a Chord at a Point, Equal Chords and their respective distances from the Centre, the Angle Subtended by an Arc of a Circle, Cyclic Quadrilaterals and other terms related to circles are covered in this chapter. A total of twelve theorems are present in this chapter, learning which the students will get a clearer idea of the concepts taught. There are 6 exercises in this chapter which consist of questions from all the concepts present in the chapter.

Topics Covered in Class 9 Maths Chapter 10 Circles for Second Term:

Through examples, arrive at definition of circle and related concepts-radius, circumference, diameter, chord, arc, secant, sector, segment, subtended angle.
1. (Prove) Equal chords of a circle subtend equal angles at the centre and (motivate) its converse.
2. (Motivate) The perpendicular from the centre of a circle to a chord bisects the chord and conversely, the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.
3. (Motivate) Equal chords of a circle (or of congruent circles) are equidistant from the centre (or their respective centres) and conversely.
4. (Motivate) The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
5. (Motivate) Angles in the same segment of a circle are equal.
6. (Motivate) The sum of either of the pair of the opposite angles of a cyclic quadrilateral is 180° and its converse.

Important Theorems –

Theorem 10.1 – Equal chords of a circle subtend equal angles at the centre.

Theorem 10.2 – If the angles subtended by the chords of a circle at the centre are equal, then the chords are equal.

Theorem 10.3 – The perpendicular from the centre of a circle to a chord bisects the chord.

Theorem 10.4 – The line drawn through the centre of a circle to bisect a chord is perpendicular to the chord.

Theorem 10.5 – There is one and only one circle passing through three given non-collinear points.

Theorem 10.6 – Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres).

Theorem 10.7 – Chords equidistant from the centre of a circle are equal in length.

Theorem 10.8 – The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Theorem 10.9 – Angles in the same segment of a circle are equal.

Theorem 10.10 – If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle (i.e. they are concyclic).

Theorem 10.11 – The sum of either pair of opposite angles of a cyclic quadrilateral is 180º.

Theorem 10.12 – If the sum of a pair of opposite angles of a quadrilateral is 180º, the quadrilateral is cyclic.

Also access the following resources for Class 9 Chapter 10 Circles at BYJU’S:

• NCERT Circles Class 9 Notes
• Circles Class 9 Important Questions
• NCERT Exemplar Solutions Class 9 Circles
• RD Sharma Solutions Circles Class 9

NCERT Solutions Class 9 Maths Chapter 11 Constructions – Term II

In this chapter, students will learn some basic constructions. The method learnt will then be used to construct certain kinds of triangles. There are 2 exercises present in this chapter, of which the first exercise deals with the construction of a certain angle or the bisector of a given angle. On the other hand, the second exercise deals with the constructions of triangles, when different parameters are given.

Topics Covered in Class 9 Maths Chapter 11 Constructions for Second Term:

1. Construction of bisectors of line segments and angles of measure 60˚, 90˚, 45˚ etc., equilateral triangles.
2. Construction of a triangle given its base, sum/difference of the other two sides and one base angle.

Important Points –

Construction 11.1 – To construct the bisector of a given angle.

Construction 11.2 – To construct the perpendicular bisector of a given line segment.

Construction 11.3 – To construct an angle of 600 at the initial point of a given ray.

Construction 11.4 – To construct a triangle, given its base, a base angle and sum of other two sides.

Construction 11.5 – To construct a triangle given its base, a base angle and the difference of the other two sides.

Construction 11.6 – To construct a triangle, given its perimeter and its two base angles.

Also access the following resources for Class 9 Chapter 11 Constructions at BYJU’S:

• NCERT Constructions Class 9 Notes
• Constructions Class 9 Important Questions
• NCERT Exemplar Solutions Class 9 Constructions
• RD Sharma Solutions Constructions Class 9

NCERT Solutions Class 9 Maths Chapter 12 Heron’s Formula – Term I

The chapter discusses Heron’s formula, which can be used to calculate the area of a triangle when the length of all three sides is given. In this method, there is no need to calculate the angles or other distances in the triangle. This formula can be used not only to find the area of triangles but also to find the areas of quadrilaterals and other polygons by dividing them into triangles. There are 2 exercises in this chapter which help the student in understanding the method of solving the problems based on Heron’s formula.

Topics Covered in Class 9 Maths Chapter 12 Heron’s Formula for First Term:

Area of a triangle using Heron’s formula (without proof).

Important Formulas –

Area of a triangle = 1/2 × base × height

Area of a triangle by Heron’s Formula = [latex]\sqrt{s(s-a)(s-b)(s-c)}[/latex]

Where a, b and c are the sides of a triangle

s is the semi perimeter i.e, half of the perimeter of a triangle = (a + b + c)/2

Also access the following resources for Class 9 Chapter 12 Heron’s Formula at BYJU’S:

• NCERT Heron’s Formula Class 9 Notes
• Heron’s Formula Class 9 Important Questions
• NCERT Exemplar Solutions Class 9 Heron’s Formula
• RD Sharma Solutions Heron’s Formula Class 9

NCERT Solutions Class 9 Maths Chapter 13 Surface Areas and Volumes – Term II

In this chapter, students shall learn to find the surface areas and volumes of cuboids and cylinders in detail and broaden the study to some other solids, such as cones and spheres. This chapter is just an extended version of the chapter mensuration, in which the students learnt about the surface areas and volumes in earlier classes. There are 8 exercises in this chapter, and these exercises contain problems that are based on surface areas and volumes of different solids such as cubes, cuboids, spheres, cylinders, cones, and hemispheres.

Topics Covered in Class 9 Maths Chapter 13 Surface Areas and Volumes for Second Term:

Surface areas and volumes of cubes, cuboids, spheres (including hemispheres) and right circular cylinders/cones.

Important Formulas –

Surface Area of a Cuboid = 2(lb + bh + hl)

where l, b and h are respectively the three edges of the cuboid

Surface Area of a Cube = 6a²

where a is the edge of the cube.

Curved Surface Area of a Cylinder = 2πrh

where r is the radius of the base of the cylinder and h is the height of the cylinder

Total Surface Area of a Cylinder = 2πr(r + h)

where h is the height of the cylinder and r its radius

Curved Surface Area of a Cone = 1/2 × l × 2πr = πrl

where r is its base radius and l its slant height

Total Surface Area of a Cone = πrl + πr² = πr(l + r)

Surface Area of a Sphere = 4 π r²

where r is the radius of the sphere.

Curved Surface Area of a Hemisphere = 2πr²

where r is the radius of the sphere of which the hemisphere is a part.

Total Surface Area of a Hemisphere = 3πr²

Volume of a Cuboid = base area × height = length × breadth × height

Volume of a Cube = edge × edge × edge = a³

Volume of a Cylinder = πr²h

where r is the base radius and h is the height of the cylinder.

Volume of a Cone = 1/3 πr²h

where r is the base radius and h is the height of the cone.

Volume of a Sphere = 4/3 πr³

where r is the radius of the sphere.

Volume of a Hemisphere = 2/3 πr³

where r is the radius of the hemisphere.

Also access the following resources for Class 9 Chapter 13 Surface Areas and Volumes at BYJU’S:

• NCERT Surface Areas and Volumes Class 9 Notes
• Surface Areas and Volumes Class 9 Important Questions
• NCERT Exemplar Solutions Class 9 Surface Areas and Volumes
• RD Sharma Solutions Surface Areas and Volumes Class 9

NCERT Solutions Class 9 Maths Chapter 14 Statistics – Term I

The branch of Mathematics in which the extraction of meaningful information is studied is known as Statistics. It can also be defined as the collection of data on different aspects of the life of people, useful to the State. The chapter teaches about the different presentation of the data, including the frequency distribution as well. The chapter also helps the students learn the graphical representation of data, using different graphs such as Bar graphs, Histograms, Frequency polygons, etc. The chapter also lets the students learn the measure of central tendency mean, median and mode of the raw data. A total of 4 exercises are present in the chapter that includes problems related to all these concepts.

Topics Covered in Class 9 Maths Chapter 14 Statistics for First Term:

Introduction to Statistics: Collection of data, presentation of data — tabular form, ungrouped / grouped, bar graphs, histograms.

Important Formulas –

Statistics deals with the collection, presentation, analysis of data as well as drawing of meaningful conclusions from the data.

Mean – It is found by adding all the values of the observations and dividing it by the total number of observations. It is denoted by [latex]\overline{x}[/latex].

So, [latex]\overline{x}=\frac{\sum_{i=1}^{n}x_{i}}{n}[/latex]

For an ungrouped frequency distribution, it is [latex]\overline{x}=\frac{\sum_{i=1}^{n}f_{i}x_{i}}{\sum_{i=1}^{n}f_{i}}[/latex]

Median – It is the value of the middle-most observation (s).

If n is an odd number, the median = value of the [(n + 1)/2]^th observation

If n is an even number, median = Mean of the values of the (n/2)^th and [(n/2) + 1]^th observations.

Mode – The mode is the most frequently occurring observation.

Also access the following resources for Class 9 Chapter 14 Statistics at BYJU’S:

• NCERT Statistics Class 9 Notes
• Statistics Class 9 Important Questions
• NCERT Exemplar Solutions Class 9 Statistics
• RD Sharma Solutions Statistics Class 9

NCERT Solutions Class 9 Maths Chapter 15 Probability – Term II

The collection of some outcomes of an experiment is known as an event of an experiment. The chances of occurrence of an event are known as probability. In this chapter, students will learn to measure the chance of occurrence of a particular outcome in an experiment. This chapter contains only 1 exercise. The problems covered in this exercise are based on real-life incidents, enhancing the interest of the students in solving the questions.

Topics Covered in Class 9 Maths Chapter 15 Probability for Second Term:

History, Repeated experiments and observed frequency approach to probability. Focus is on empirical probability. (A large amount of time to be devoted to group and to individual activities to motivate the concept; the experiments to be drawn from real – life situations, and from examples used in the chapter on statistics).

Important Formulas –

The empirical probability P(E) of an event E happening, is given by

P (E) = Number of trials in which the event happened/ The total number of trials

Also access the following resources for Class 9 Chapter 15 Probability at BYJU’S:

• NCERT Probability Class 9 Notes
• Probability Class 9 Important Questions
• NCERT Exemplar Solutions Class 9 Probability
• RD Sharma Solutions Probability Class 9

Benefits of BYJU’S NCERT Maths Solutions for Class 9

• The solutions are given in easy steps so that students can easily understand them.
• Students can access these solutions for free and clear any doubts instantly.
• Exercise-wise PDFs are also provided for all chapters of CBSE 9th Class Maths.
• The detailed explanations also help students to develop problem-solving skills and help them to learn concepts more effectively.
• The questions are solved using the easiest method.
• Diagrams are also given to help students visualize the questions and solutions.

CBSE Class 9 Maths Unit wise weightage for First Term

Internal Assessment Class 9 Term I

CBSE Class 9 Maths Unit wise weightage for Second Term

Internal Assessment Class 9 Term II

During the exam preparation, students should first complete the units carrying more weightage and then move on to the lower weightage. This helps students to first cover the important concepts and improves their confidence level to perform well in the annual exam.

The students are advised to analyze all the concepts covered in the Syllabus of Class 9 Maths and prepare a proper preparation strategy. It’s merely impossible to score well in Maths by simply reading and memorizing the concepts. Students should always try to understand the concept and logic given in any particular topic. The students are suggested to practise these NCERT Solutions Class 9 materials on a regular basis to develop a strong understanding of basic mathematics concepts.

Along with the NCERT Solutions, we have also provided a brief summary of important formulas and different methods of solving similar problems to help students in understanding the concepts thoroughly. Keep visiting BYJU’S to get more updated learning materials and download the BYJU’S app for a better and personalized learning experience, with engaging video lessons.

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1. 

   Vinayak Dubey

   January 21, 2022 at 8:25 pm

   Thank you Byjus for providing solutions for free?

   Reply