Class 9th Math Ncert Solution

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NCERT Solutions for Class 9 Maths Chapter 4 Linear Equation in one variable 

In earlier classes, you have studied linear equations in one variable. Can you write down a linear equation in one variable? You may say that x + 1 = 0, x + 2 = 0 and 2 y + 3 = 0 are examples of linear equations in one variable. You also know that such equations have a unique (i.e., one and only one) solution. You may also remember how to represent the solution on a number line. In this chapter, the knowledge of linear equations in one variable shall be recalled and extended to that of two variables. You will be considering questions like: Does a linear equation in two variables have a solution? If yes, is it unique? What does the solution look like on the Cartesian plane? You shall also use the concepts you studied in Chapter 3 to answer these questions.

NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclids Geometry

In Chapter 5, you have studied that a minimum of two points are required to draw a line. You have also studied some axioms and, with the help of these axioms, you proved some other statements. In this chapter, you will study the properties of the angles formed when two lines intersect each other, and also the properties of the angles formed when a line intersects two or more parallel lines at distinct points. Further you will use these properties to prove some statements using deductive reasoning (see Appendix 1). You have already verified these statements through some activities in the earlier classes. In your daily life, you see different types of angles formed between the edges of plane surfaces. For making a similar kind of model using the plane surfaces, you need to have a thorough knowledge of angles. For instance, suppose you want to make a model of a hut to keep in the school exhibition using bamboo sticks. Imagine how you would make it? You would keep some of the sticks parallel to each other, and some sticks would be kept slanted. Whenever an architect has to draw a plan for a multistoried building, she has to draw intersecting lines and parallel lines at different angles. Without the knowledge of the properties of these lines and angles, do you think she can draw the layout of the building? In science, you study the properties of light by drawing the ray diagrams. For example, to study the refraction property of light when it enters from one medium to the other medium, you use the properties of intersecting lines and parallel lines. When two or more forces act on a body, you draw the diagram in which forces are represented by directed line segments to study the net effect of the forces on the body. At that time, you need to know the relation between the angles when the rays (or line segments) are parallel to or intersect each other. To find the height of a tower or to find the distance of a ship from the light house, one needs to know the angle formed between the horizontal and the line of sight. Plenty of other examples can be given where lines and angles are used. In the subsequent chapters of geometry, you will be using these properties of lines and angles to deduce more and more useful properties

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.

You have studied about triangles and their various properties in your earlier classes. You know that a closed figure formed by three intersecting lines is called a triangle. (‘Tri’ means ‘three’). A triangle has three sides, three angles and three vertices. For example, in triangle ABC, denoted as ∆ ABC (see Fig. 7.1); AB, BC, CA are the three sides, ∠ A, ∠ B, ∠ C are the three angles and A, B, C are three vertices. In Chapter 6, you have also studied some properties of triangles. In this chapter, you will study in details about the congruence of triangles, rules of congruence, some more properties of triangles and inequalities in a triangle. You have already verified most of these properties in earlier classes. We will now prove some of them

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.

NCERT Solutions for 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:

Review concept of area, recall area of a rectangle.

1. (Prove) Parallelograms on the same base and between the same parallels have the same area.

2. (Motivate) Triangles on the same base and between the same parallels are equal in area and its converse.

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

Class 9 Maths NCERT Solutions Chapter 9 Exercises
Exercise 9.1 – 1 Question (1 short answer)
Exercise 9.2 – 6 Questions (5 short answers, 1 long answer)
Exercise 9.3 – 16 Questions (12 short answers, 4 long answer)
Exercise 9.4 – 8 Questions (4 short answer, 1 long answer, 3 very long answers)

NCERT Solutions for Class 9 Maths Chapter 10 Circles

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:

Through examples, arrive at definitions of circle related concepts, radius, circumference, diameter, chord, arc, 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) There is one and only one circle passing through three given non-collinear points.

4. (Motivate) Equal chords of a circle (or of congruent circles) are equidistant from the centre(s) and conversely.

5. (Prove) 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.

6. (Motivate) Angles in the same segment of a circle are equal.

7. (Motivate) If a line segment joining two points subtends equal angle at two other points lying on the same side of the line containing the segment, the four points lie on a circle.

8. (Motivate) The sum of the either pair of the opposite angles of a cyclic quadrilateral is 1800 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.

Class 9 Maths NCERT Solutions Chapter 10 Exercises
Exercise 10.1 – 2 Questions (2 short answers)
Exercise 10.2 – 2 Questions (2 long answers)
Exercise 10.3 – 3 Questions (3 long answers)
Exercise 10.4 – 6 Questions (6 long answers)
Exercise 10.5 – 12 Questions (12 long answers)
Exercise 10.6 – 10 Questions (10 long answers)

NCERT Solutions for Class 9 Maths Chapter 11 Constructions

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 :

1. Construction of bisectors of a line segment and angle, 60°, 90°, 45° angles etc, equilateral triangles.

2. Construction of a triangle given its base, sum/difference of the other two sides and one base angle.

3. Construction of a triangle of given perimeter and base angles.

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.

Class 9 Maths NCERT Solutions Chapter 11 Exercises
Exercise 11.1 – 5 Questions (2 short answers, 2 long answers, 1 very long answer)
Exercise 11.2 – 5 Questions (5 very long answer)

NCERT Solutions for Class 9 Maths Chapter 12  Heron’s Formula

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 :

Area of a triangle using Heron’s formula (without proof) and its application in finding the area of a quadrilateral.

Important Formulas –

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

Area of a triangle by Heron’s Formula =

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

Class 9 Maths NCERT Solutions Chapter 12 Exercises
Exercise 12.1 – 6 Questions (2 short answers, 2 long answers, 2 very long answers)
Exercise 12.2 – 9 Questions (4 long answers, 5 very long answers)

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes

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 :

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.

Class 9 Maths NCERT Solutions Chapter 13 Exercises
Exercise 13.1 – 8 questions
Exercise 13.2 – 11 questions
Exercise 13.3 – 8 questions
Exercise 13.4 – 9 questions
Exercise 13.5 – 9 questions
Exercise 13.6 – 8 questions
Exercise 13.7 – 9 questions
Exercise 13.8 – 10 questions
Exercise 13.9 – 3 questions

NCERT Solutions for Class 9 Maths Chapter 14 Statistics

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 :

Introduction to Statistics: Collection of data, presentation of data – tabular form, ungrouped/ grouped, bar graphs, histograms (with varying base lengths), frequency polygons, qualitative analysis of data to choose the correct form of presentation for the collected data. Mean, median, mode of ungrouped data.

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

so,

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.

Class 9 Maths NCERT Solutions Chapter 14 Exercises
Exercise 14.1 – 2 questions (2 short answers)
Exercise 14.2 – 2 questions (2 short answers)
Exercise 14.3 – 9 questions (6 short answers, 3 long answers)
Exercise 14.4 – 6 questions (2 short answers, 4 long answers)

NCERT Solutions for Class 9 Maths Chapter 15 Probability

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 :

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

Class 9 Maths NCERT Solutions Chapter 15 Exercises
Exercise 15.1 – 10 Question ( 4 short answers, 3 long answers, 3 very long answers)