# NCERT Solutions for Class 9 Maths

NCERT Solutions for Class 9 Maths for UP Board (High School) and CBSE Board in PDF format free in Hindi Medium and English Medium updated for new academic session 2020-21 following the latest CBSE Syllabus 2020-21 for new session.

NCERT solutions for session 2020-2021 is now available to download in PDF form. Download NCERT Solutions Apps for class 9 Hindi & English Medium Free. These Apps works well without internet also.## NCERT Solutions for Class 9 Maths

Class: | 9 |

Subject: | Maths – गणित |

Contents: | NCERT Solutions |

### Class 9 Maths Solutions in Hindi & English

NCERT Solutions for class 9 Maths all chapters in PDF form are given below for session 2020-21. A separate PDF of all exercises are given to download. Download Apps for class 9 Maths in Hindi and 9 Maths in English Free. NCERT Books 2020-21 as well as Exemplar problems books both are equally important for the CBSE exams. Chapter wise assignments Test Papers, Previous year Question Papers issued by CBSE and other schools, Chapter wise tests, Syllabus for the academic year 2020–2021 and other online study material.

### NCERT Solutions for Class 9 Maths in English

- Chapter 01
- Chapter 02
- Chapter 03
- Chapter 04
- Chapter 05
- Chapter 06
- Chapter 07
- Chapter 08
- Chapter 09
- Chapter 10
- Chapter 11
- Chapter 12
- Chapter 13
- Chapter 14
- Chapter 15

### Class 9 Maths Important Points related to all Chapters.

#### Chapter 1: Number Systems

Class 9 Maths Chapter 1 Consists of the concepts of Natural number, Whole numbers, Integers, etc. and plotting on number lines. Find some rational or Irrational Numbers between two numbers are generally asked in CBSE Exams. For example, converting an irrational number into p/q is asked in most of CBSE Final Term exams 2019-20. Plotting a square root number on number line is also important as examination point of view.

#### Chapter 2: Polynomials

In 9tt Maths Chapter 2 important Questions are based on remainder theorem and factor theorem. For polynomial in one variable, terms, factors and zeroes are to be evaluated. We also have to identify monomials, binomials or trinomials in the given polynomial expression. The concepts of Degree of constant polynomial and others should must be known to all students for periodic test or final exams. Classification of linear, quadratic or cubic polynomial are frequently asked in terminal exams. Factorisation by middle term splitting or using identities are generally asked in all exams.

#### Chapter 3: Coordinate Geometry

Class 9 Maths Chapter 3 Coordinate Geometry help us to find the location of any object by using two independent information (x-axis and y-axis). This chapter explains the plotting a point in Cartesian Plane as well as to find the coordinate to a point, using two dimensional coordinate system. Concepts about Abscissa, Ordinate, Origin, x-axis, y-axis, coordinate plane, quadrants, etc., are describe properly in the NCERT Solutions provided by Tiwari Academy.

#### Chapter 4: Linear Equations in Two Variables

Linear equations in two variables of Class 9 Maths provides the techniques to find the multiple solutions of an equations, so that a line on graph can be drawn. Introduction of standard form of linear equations help the students to understand the terms related to a linear equation, like coefficient of x or y or constant term. The application of linear equations in day to day life are also included in the contents provided by Tiwari Academy NCERT Solutions.

#### Chapter 5: Introduction to Euclid’s Geometry

This chapter explains that what is geometry, its origin and use, its initial fundamental provided by Euclid. Postulates and Axioms are main point to be studied in the chapter 5 Introduction to Euclid’s Geometry. The other version of Euclid’s Fifth postulate is frequently asked in unit test and terminal exams of class 9. Different definitions of geometrical terms like Point, Line, Straight Line, Surface, Curves, etc. are also important part of chapter 5 of class 9 Mathematics.

#### Chapter 6: Lines and Angles

Class 9 Maths Chapter 6 – Lines and Angles tells us about the concepts of line segments, rays, angles, vertex and its arms, collinear and non-collinear points. It involves the applications of acute, right, obtuse, straight angles also including adjacent angle, reflex angle, complementary & Supplementary angles and vertically opposite angles. Theorem on Vertically Opposite angles is also important as per terminal exams. Exercise 6.1 is based on the application of vertically opposite angles.

#### Chapter 7: Triangles

Class 9 Maths Chapter 7 Triangles is based on mainly the criteria of congruency of triangles. There are two theorems which may be asked for proof but axioms based on theorems are also important for problem solving. Exercise 7.1 is totally based on SAS and ASA criteria of Congruence Triangles. Exercise 7.2 and 7.3 includes the questions based on RHS and SSS including sum applications of other theorems also. Exercise 7.4 is based on inequalities of sides and angles.

#### Chapter 8: Quadrilaterals

Chapter 8 Quadrilaterals of Class 9 Maths is based on angle sum property of quadrilateral, types of quadrilateral (Rectangles, Trapezium, Square, Rhombus, Parallelograms, etc.) and its properties. Applications based on parallelogram including theorems and axioms are given to prove. Application of Mid-Point Theorem and its converse is one of the important segment of Chapter 8. Mid-Point Theorem is generally asked to prove in terminal exams as well as unit tests of most of the CBSE School Examinations.

#### Chapter 9: Areas of Parallelograms and Triangles

In 9 Maths Chapter 9 the area of parallelograms and triangles is to be determined using theorems and corollaries. The correlation between a triangle and a parallelogram, on the same base and between same parallel line, is also explained with multiple practice questions. The result of examples also can be used as a corollary to prove some questions. Median divides triangle into two triangles in equal area. This result helps a lot in solving the questions of Exercise 9.3 onwards.

#### Chapter 10: Circles

In Class 9 Maths Chapter 10 Circles and its components – like segment, arc, semicircle, chords, diameter, minor and major sectors are to be discussed. The theorem, Perpendicular from the centre bisect the chord and Equal chords are equidistant from the centre, are widely used for solving questions in Exercise 10.2, 10.3 and 10.4. These are also useful in other exercises of Chapter 10 of class 9 Maths. Exercise 10.5 contains most of the important questions as per examination point of view.

#### Chapter 11: Constructions

Constructions chapter of Class 9 Maths is generally based on making angles and triangles only using ruler and compass. Exercise 11.1 illustrate the questions related to making various angles like 15, 30, 60, 120, 150, 45, 90, etc., with the help of ruler and compass only i.e. without using protector. Exercise 11.2 provide practice of constructing a triangle when sum of two side or difference of two side and a base angle is given or perimeter of triangle and two based angle is given.

#### Chapter 12: Heron’s Formula

Class 9 Maths Chapter 12 Heron’s Formula help us to find out the area of a triangle or quadrilateral (splitting it into two triangles). Heron’s Formula is derived for area of all type of triangle when its three sides are known. Area of other polygons also can be evaluated by divided it into suitable triangles. In this chapter, various application based Heron’s Formula are given which are related to day to day life.

#### Chapter 13: Surface Areas and Volumes

Class 9 Maths Chapter 13 explains the concepts of mensuration. Exercise 13.1, 13.2, 13.3 and 13.4 describes the surface areas of solid figures like Cube, Cuboid, Cone, Cylinder, Sphere and Hemisphere, etc., whereas the Exercise 13.5, 13.6, 13.7 and 13.8 illustrates the concepts of finding volumes of these solid figures. Exercise 13.9 is optional exercise contains very good set of questions for practice. To do this chapter properly, students must derive all the formulae and then learn them.

#### Chapter 14: Statistics

Exercise 14.1 of class 9 Maths Chapter 14 is based on Primary Data and Secondary Data concepts. Exercise 14.2 emphasize the data in organised way to representation it graphically by using Histogram, Bar Charts or Polygons. Exercise 14.3 is based on questions with miscellaneous type of variation in data. Exercise 14.4 includes the concepts of central tendency of statistical data like Mean, Mode and Median of ungrouped data.

#### Chapter 15: Probability

Class 9 Chapter 15 Probability is based on outcomes of events using favorable outcomes and total number of outcomes in any experiment. Questions based on coins, deck of cards, dice and day to day life experiences are given in the chapter. Some of the questions given in exercise 15.1 are based on statistical data also. Probability is one of the important chapters for further classes also.

##### What are the theorems that may be asked for proving in Exams for class 9 Chapter 6: Lines and Angles?

The following theorems may be asked to prove in exams, rest are for application and motivation:

1. If two lines intersect, vertically opposite angles are equal.

2. The sum of the angles of a triangle is 180.

##### Which theorems may be asked for proving form class 9 Chapter 7: Triangles?

The following theorems may be asked to prove in exams, rest are for application and motivation:

1. 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).

2. The angles opposite to equal sides of a triangle are equal.

##### Is there any theorem in Chapter 10: Quadrilateral to prove?

Yes, there is only one important theorem (The diagonal divides a parallelogram into two congruent triangles), that may be asked to prove.

##### How many theorems are there in Chapter 9: Areas of Parallelograms and Triangles to prove?

There is only one theorem (Parallelograms on the same base and between the same parallels have equal area.) asked for proving. This theorem is quite important one for all examinations.

##### Is there only one theorem in Chapter 10: Circle, which is asked to prove in Examinations?

No, there are two theorems (1. Equal chords of a circle subtend equal angles at the center. 2. The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.) for the examination point of view.

### NCERT Solutions for Class 9 Maths in Hindi

- पाठ 1: संख्या पद्धति
- पाठ 2: बहुपद
- पाठ 3: निर्देशांक ज्यामिति
- पाठ 4: दो चरों वाले रैखिक समीकरण
- पाठ 5: यूक्लिड की ज्यामिति का परिचय
- पाठ 6: रेखाएँ और कोण
- पाठ 7: त्रिभुज
- पाठ 8: चतुर्भुज
- पाठ 9: समांतर चतुर्भुजों और त्रिभुजों के क्षेत्रफल
- पाठ 10: वृत
- पाठ 11: रचनाएँ
- पाठ 12: हीरोन का सूत्र
- पाठ 13: पृष्ठीय क्षेत्रफल और आयतन
- पाठ 14: सांख्यिकी
- पाठ 15: प्रायिकता

#### Class 9 Maths Exemplar Books

Exemplar books are designed to enhance the practice and improve the knowledge and concepts about each chapter. So, students are advised to go through NCERT Exemplar book after doing NCERT books. Vedic Maths is good for improving calculations faster and easier. We are also providing help in solving holiday homework for session 2020-21, if you are facing problem in doing holiday homework, upload in our website and get the solution with a week. The STUDY ONLINE option is given for NCERT solutions, if you don’t want to download. Visit to Discussion Forum to ask your doubts and share your knowledge.

##### Construction for Class 9 in Geometry

Construction of bisectors of line segments and angles of measure 60, 90, 45, 15, 30, 120, 145, etc., and equilateral triangles. Construction of a triangle given its base, sum/difference of the other two sides and one base angle. Construction of a triangle of given perimeter and base angles.

##### Class 9 Maths Solutions Apps

Download NCERT Solutions Online-Offline Apps 2020-21 for class 9 all subjects Free. These Apps works well without internet also. The first draft of solutions of class 9 Maths is prepared in English and then it will be translated into Hindi Medium for the academic session 2020-2021. Considering global thinking, it was decided to prepare the solutions first in English. But as the demand of Hindi Medium is also increasing (according to feedback submitted by students), so we are going to prepare Hindi medium solutions for class 9 all subjects for new session 2020-21. Download NCERT Books for class 9 all subjects.

###### Feedback & Suggestions

A panel of experts from Tiwari Academy held a workshop of two days and thoroughly discussed the content quality and made some suggestions. Amendments were carried out in the English as well as Hindi draft accordingly.

The final draft was thus prepared. The solutions are prepared according to new syllabus for 2019-20 having classical approach of constructions and geometrical chapters. उत्तर प्रदेश में भी NCERT की किताबें लगने के बाद हम लोग UP बोर्ड के लिए भी अध्धयन सामग्री बना रहे हैं। अगर UP Board NCERT Solutions प्राप्त करने में विद्यार्थियों को किसी प्रकार की परेशानी हो रही है तो हमें Contact करें या फोन पर भी बता सकते हैं। UP Board Secondary Education (High School) के लिए अध्ययन सामग्री, Sample Papers, Notes तथा पश्नों के हल यहाँ से प्राप्त करें।

###### Concept of Probability in Class 9 Maths

History about probability, Random Experiments, 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 day to day situations and questions based on statistics as examples used in the chapter.

### Important Questions on Class 9th Maths

= (100 + 3)(100 + 7)

= (100)^2 + (3 + 7)100 + 3 × 7

[∵〖(x + a)(x + b) = x〗^2 + (a + b)x + ab]

= 10000 + 1000 + 21

= 11021

3y = ax + 7.

Putting x = 3 and y = 4,

we have, 3 × 4 = a × 3 + 7

⇒ 12 = 3a + 7

⇒ 12 – 7 = 3a

⇒ a = 5/3

OP = OQ = OR.

S भुजा DA का मध्य-बिंदु हैं [∵ दिया है]

R भुजा DC का मध्य-बिंदु हैं [∵ दिया है]

अतः, SR || AC और SR = 1/2 AC

[∵ मध्य-बिंदु प्रमेय]

Therefore, PO = OR and SO = OQ

In ΔPQS, PO is median. [∵ SO = OQ]

Hence, ar(PSO) = ar(PQO) … (1)

[∵ A median of a triangle divides it into two triangles of equal areas.]

Similarly, in ΔPQR, QO is median. [∵ PO = OR]

Hence, ar(PQO) = ar(QRO) … (2)

And in ΔQRS, RO is median. [∵ SO = OQ]

Hence, ar(QRO) = ar(RSO) … (3)

From the equations (1), (2) and (3), we get

ar(PSO) = ar(PQO) = ar(QRO) = ar(RSO)

Hence, in parallelogram PQRS, diagonals PR and QS divide it into four triangles in equal area.

Because, there are infinite number of equal chords in a circle.

1. To bisect a given angle.

2. To draw the perpendicular bisector of a given line segment.

3. To construct an angle of 60° etc.

4. To construct a triangle given its base, a base angle and the sum of the other two sides.

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

6. To construct a triangle given its perimeter and its two base angles.

a = 28 cm, b = 26 cm और c = 30 cm हैं।

अत:, त्रिभुज का अर्धपरिमाप

s = (a + b + c)/2

= (28 + 26 + 30)/2

= 84/2

= 42 cm

इसलिए, हीरोन के सूत्र से,

त्रिभुज का क्षेत्रफल

= √(s(s – a)(s – b)(s – c) )

= √(42(42 – 28)(42 – 26)(42 – 30) )

= √(42(14)(16)(12) )

= √112896

= 336 cm^2

हम जानते हैं कि समांतर चतुर्भुज का क्षेत्रफल = आधार × संगत ऊँचाई

प्रश्नानुसार, समांतर चतुर्भुज का क्षेत्रफल = त्रिभुज का क्षेत्रफल

⇒ आधार × संगत ऊँचाई = 336

⇒ 28 × संगत ऊँचाई = 336

⇒ संगत ऊँचाई = 336/28 = 12 cm

outer radius R = 28/2 = 14 cm and length h = 35 m

Volume of cylindrical wooden pipe = π(R^2-r^2 )h

= 22/7 × (〖14〗^2-〖12〗^2 )×35

= 22 × (196 – 144) × 5

= 22 × 52 × 5 = 5720 cm^3

Mass of cylindrical wooden pipe

= 5720 × 0.6 g = 3432 g =3.432 kg

[∵1 cm3 of wood has a mass of 0.6 g]

Hence, the volume of cylindrical wooden pipe is 3.432 kg.

height of soup inside the cylindrical bowl h = 4 cm

Volume of cylindrical bowl = πr^2 h

= 22/7 × (3.5)^2 × 4

= 22/7× 3.5 × 3.5 × 4

= 22 × 0.5 × 3.5 × 4

= 154 cm^3

Therefore, the volume of soup per day for 250 patient

= 250 × 154 = 38500 cm^3

Hence, hospital has to prepare 38500 cm^3 soup daily to serve 250 patients.

14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28

बहुलक सबसे अधिक बार आने वाला प्रेक्षण का मान होता है।

क्योंकि प्रेक्षण 14 सबसे अधिक (4 बार) आया है।

अतः, दिए गए प्रेक्षणों का बहुलक 14 है।