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• Number Systems Class 9 Assertion Reason Questions Maths Chapter 1

Last Updated on August 26, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 9 maths. Assertion Reason questions are the new question format that is introduced in CBSE board. The resources for assertion reason questions are very less. So, to help students we have created chapterwise assertion reason questions for class 9 maths. In this article, you will find assertion reason questions for CBSE Class 9 Maths Chapter 1 Number Systems. It is a part of Assertion Reason Questions for CBSE Class 9 Maths Series.

Table of Contents

Assertion Reason Questions on Number Systems

Questions :

Q. 1. Assertion (A): The rationalising factor of $8-\sqrt{7}$ is $8+\sqrt{7}$. Reason (R): If the product of two irrational numbers is rational, then each one is said to be the rationalising factor of the other.

Q. 2. Assertion (A): The sum of two irrational numbers $3-\sqrt{5}$ and $5+\sqrt{5}$ is rational number. Reason (R): The sum of two irrational numbers is always an irrational number.

Q. 3. Assertion (A): The simplified form of $7^4 \times 7^5$ is $7^{20}$. Reason (R): If $a>0$ be a real number and $p$ and $q$ be rational numbers. Then $a^p \times a^q=a^{p+q}$.

1. (a) Assertion (A): It is true that the rationalising factor of $8-\sqrt{7}$ is $8+\sqrt{7}$. Reason (R): It is true to say that each one is rationalising factor in the product of two irrational numbers. Hence, both Assertion (A) and Reason (R) are true and Reason $(R)$ is the correct explanation of Assertion (A).

2. (c) Assertion (A): Here, $3-\sqrt{5}+5+\sqrt{5}=8$, which is a rational number. So, Assertion (A) is true. Reason (R): It is not always true to say that sum of two irrational number is always an irrational number. Hence, Assertion (A) is true but Reason (R) is false.

3. (d) Assertion (A) is false but Reason (R) is true.

Polynomials Class 9 Assertion Reason Questions Maths Chapter 2

Topics from which assertion reason questions may be asked.

• Representation on number line
• Concept of rationalizing the denominator
• Rationalizing the denominator of expressions with square roots
• Applying the laws of exponents to simplify expressions
• Rationalizing surds

The sum or difference of a rational number and an irrational number is irrational. The product or quotient of a non-zero rational number with an irrational number is irrational.

Assertion reason questions from the above given topic may be asked.

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Frequently Asked Questions (FAQs) on Number Systems Assertion Reason Questions Class 9

Q1: what are assertion reason questions.

A1: Assertion-reason questions consist of two statements: an assertion (A) and a reason (R). The task is to determine the correctness of both statements and the relationship between them. The options usually include: (i) Both A and R are true, and R is the correct explanation of A. (ii) Both A and R are true, but R is not the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true. or A is false, and R is also false.

Q2: Why are assertion reason questions important in Maths?

A2: Students need to evaluate the logical relationship between the assertion and the reason. This practice strengthens their logical reasoning skills, which are essential in mathematics and other areas of study.

Q3: How can practicing assertion reason questions help students?

A3: Practicing assertion-reason questions can help students in several ways: Improved Conceptual Understanding:  It helps students to better understand the concepts by linking assertions with their reasons. Enhanced Analytical Skills:  It enhances analytical skills as students need to critically analyze the statements and their relationships. Better Exam Preparation:  These questions are asked in exams and practicing them can improve your performance.

Q4: What strategies should students use to answer assertion reason questions effectively?

A4: Students can use the following strategies: Understand Each Statement Separately:  Determine if each statement is true or false independently. Analyze the Relationship:  If both statements are true, check if the reason correctly explains the assertion.

Q5: What are common mistakes to avoid when answering Assertion Reason questions?

A5: Common mistakes include: Not reading the statements carefully and missing key details. Assuming the Reason explains the Assertion without checking the logical connection. Confusing the order or relationship between the statements. Overthinking and adding information not provided in the question.

Q6: Are all integers also rational numbers?

A6: Yes, all integers are rational numbers because they can be expressed as a fraction where the denominator is 1. For example, 5 can be written as 5/1​, making it a rational number.

Q7: What are the key concepts covered in Chapter 1 of CBSE Class 9 Maths regarding number systems?

A7: Chapter 1 of CBSE Class 9 Maths covers concepts such as understanding rational numbers, irrational numbers and Laws of exponents. (i) Review of representation of natural numbers and Integers on number line (ii) Rational numbers on the number line. (iii) Rational numbers as recurring/ terminating decimals (iv) Operations on real numbers. (v) Definition of nth root of a real number (vi) Law of exponents with integral powers

Q8: Can a number be both rational and irrational?

A8: No, a number cannot be both rational and irrational. A rational number can be expressed as a fraction of two integers, while an irrational number cannot. They are mutually exclusive categories.

Q9: What are the important keywords for CBSE Class 9 Maths Number Systems?

A9: List of important keywords given below – Natural Numbers:  Positive Counting number starting from 1. Whole Number:  All natural numbers together with 0. Integers (Z):  Set of all whole numbers and negative of natural numbers Rational Number:  Numbers which can be expressed in p/q form, where q  ≠  0 and p and q are integers. Fraction:  Numbers which can be expressed in form of p/q but are only positive Equivalent Rational Numbers:  Two rational numbers are said to be equivalent, if numerator and denominators of both rational numbers are in proportion or they are reducible to be equal.

Q10: Are there any online resources or tools available for practicing linear equations in one variable assertion reason questions?

A10: A9: We provide assertion reason questions for CBSE Class 8 Maths on our  website . Students can visit the website and practice sufficient case study questions and prepare for their exams. If you need more case study questions, then you can visit  Physics Gurukul  website. they are having a large collection of case study questions for all classes.

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NCERT Solutions Class 11 Maths Chapter 7 Binomial Theorem

NCERT Solutions for Maths Binomial Theorem Class 11 Chapter 7 - FREE PDF Download

By referring to NCERT Solutions for Binomial Theorem Class 11 Maths Chapter 7, students are able to Understand the topics covered in this chapter in detail according to the CBSE Class 11 Maths Syllabus . The binomial theorem is defined as the process of algebraically expanding the power of sums of two or more binomials. Coefficients of binomial terms in the process of expansion are referred to as binomial coefficients. 

Our subject experts at Vedantu have solved all the sums and explained all the topics covered in this chapter according to the CBSE guidelines in these NCERT Solutions Class 11 Maths . You can download these solutions from Vedantu for free.

Glance on Maths Chapter 7 Class 11 - Binomial Theorem

Chapter 7 Maths Class 11 explains the expansion of binomial expressions using the binomial theorem.

A binomial expression is an algebraic expression containing two terms.

The binomial theorem provides a formula for expanding expressions raised to any positive integer power.

Binomial coefficients are the numerical factors in terms of a binomial expansion.

Pascal's Triangle is a triangular array of numbers that provides the binomial coefficients.

The general term of a binomial expansion represents a specific term in the expanded form.

The middle term is the central term(s) in the binomial expansion when the exponent is even or odd.

This article contains chapter notes, important questions, exemplar solutions, exercises and video links for Chapter 7 -  Binomial Theorem, which you can download as PDFs.

There are two exercises including Miscellaneous Exercise (20 fully solved questions) in Class 11  Maths Chapter 7  Binomial Theorem

Access Exercise wise NCERT Solutions for Chapter 7 Maths Class 11

Exercises Under NCERT Solutions for Class 11 Maths Chapter 7 Binomial Theorem

Exercise 7.1: this exercise consists of 14 questions and introduces students to the concept of binomial theorem. the questions are based on various concepts such as expanding binomial expressions, finding the middle terms of a binomial expansion, and solving problems related to the binomial theorem. students will learn how to expand binomial expressions using the binomial theorem..

Miscellaneous Exercise: This exercise consists of 6 questions and covers a variety of topics related to binomial theorem. The questions are based on various concepts such as solving problems related to the number of terms in a binomial expansion, finding the sum of the terms in a binomial expansion, and solving word problems related to binomial theorem. This exercise will help students to revise and reinforce the concepts learned in the previous exercises.

Overall, the exercises in NCERT Solutions for Class 11 Maths Chapter  7 – Binomial Theorem are designed to help students understand and apply the concepts of binomial theorem in various scenarios. The solutions to these exercises are provided in the textbook, which will help students to check their answers and understand the concepts better.

Access NCERT Solutions for Class 11 Maths Chapter 7 – Binomial Theorem

Exercise 7.1.

1. Expand the expression ${\left( {1 - 2x} \right)^5}$.

Ans. By using Binomial Theorem, the expression ${\left( {1 - 2x} \right)^5}$ can be expanded as

\[\begin{gathered} {\left( {1 - 2x} \right)^5} = {}^5{C_0}{\left( 1 \right)^5} - {}^5{C_1}{\left( 1 \right)^4}\left( {2x} \right) + {}^5{C_2}{\left( 1 \right)^3}{\left( {2x} \right)^2} - {}^5{C_3}{\left( 1 \right)^2}{\left( {2x} \right)^3} + {}^5{C_4}{\left( 1 \right)^1}{\left( {2x} \right)^4} \\ - {}^5{C_5}{\left( {2x} \right)^5} \\ = 1 - 5\left( {2x} \right) + 10\left( {4{x^2}} \right) - 10\left( {8{x^3}} \right) + 5\left( {16{x^4}} \right) - 32{x^5} \\ = 1 - 10x + 40{x^2} - 80{x^3} + 80{x^4} - 32{x^5} \\ \end{gathered}\]

2. Expand the expression ${\left( {\frac{2}{x} - \frac{x}{2}} \right)^5}$.

Ans. By using Binomial Theorem, the expression ${\left( {\frac{2}{x} - \frac{x}{2}} \right)^5}$ can be expanded as

\[\begin{gathered} {\left( {\frac{2}{x} - \frac{x}{2}} \right)^5} = {}^5{C_0}{\left( {\frac{2}{x}} \right)^5} - {}^5{C_1}{\left( {\frac{2}{x}} \right)^4}\left( {\frac{x}{2}} \right) + {}^5{C_2}{\left( {\frac{2}{x}} \right)^3}{\left( {\frac{x}{2}} \right)^2} - {}^5{C_3}{\left( {\frac{2}{x}} \right)^2}{\left( {\frac{x}{2}} \right)^3} + {}^5{C_4}{\left( {\frac{2}{x}} \right)^1}{\left( {\frac{x}{2}} \right)^4} \\ - {}^5{C_5}{\left( {\frac{x}{2}} \right)^5} \\ = \frac{{32}}{{{x^5}}} - 5\left( {\frac{{16}}{{{x^4}}}} \right)\left( {\frac{x}{2}} \right) + 10\left( {\frac{8}{{{x^3}}}} \right)\left( {\frac{{{x^2}}}{4}} \right) - 10\left( {\frac{4}{{{x^2}}}} \right)\left( {\frac{{{x^3}}}{8}} \right) + 5\left( {\frac{2}{x}} \right)\left( {\frac{{{x^4}}}{{16}}} \right) - \frac{{{x^5}}}{{32}} \\ = \frac{{32}}{{{x^5}}} - \frac{{40}}{{{x^3}}} + \frac{{20}}{x} - 5x + \frac{5}{8}{x^3} - \frac{{{x^5}}}{{32}} \\ \end{gathered}\]

3. Expand the expression ${\left( {2x - 3} \right)^6}$.

Ans. By using Binomial Theorem, the expression ${\left( {2x - 3} \right)^6}$ can be expanded as

\[\begin{gathered} {\left( {2x - 3} \right)^6} = {}^6{C_0}{\left( {2x} \right)^6} - {}^6{C_1}{\left( {2x} \right)^5}\left( 3 \right) + {}^6{C_2}{\left( {2x} \right)^4}{\left( 3 \right)^2} - {}^6{C_3}{\left( {2x} \right)^3}{\left( 3 \right)^3} + {}^6{C_4}{\left( {2x} \right)^2}{\left( 3 \right)^4} \\ - {}^6{C_5}\left( {2x} \right){\left( 3 \right)^5} + {}^6{C_6}{\left( 3 \right)^6} \\ = 64{x^6} - 6\left( {32{x^5}} \right)\left( 3 \right) + 15\left( {16{x^4}} \right)\left( 9 \right) - 20\left( {8{x^3}} \right)\left( {27} \right) + 15\left( {4{x^2}} \right)\left( {81} \right) \\ - 6\left( {2x} \right)\left( {243} \right) + 729 \\ = 64{x^6} - 576{x^5} + 2160{x^4} - 4320{x^3} + 4860{x^2} - 2916x + 729 \\ \end{gathered}\]

4. Expand the expression ${\left( {\frac{x}{3} + \frac{1}{x}} \right)^5}$.

Ans. By using Binomial Theorem, the expression ${\left( {\frac{x}{3} + \frac{1}{x}} \right)^5}$ can be expanded as

\[\begin{gathered} {\left( {\frac{x}{3} + \frac{1}{x}} \right)^5} = {}^5{C_0}{\left( {\frac{x}{3}} \right)^5} + {}^5{C_1}{\left( {\frac{x}{3}} \right)^4}\left( {\frac{1}{x}} \right) + {}^5{C_2}{\left( {\frac{x}{3}} \right)^3}{\left( {\frac{1}{x}} \right)^2} + {}^5{C_3}{\left( {\frac{x}{3}} \right)^2}{\left( {\frac{1}{x}} \right)^3} + {}^5{C_4}{\left( {\frac{x}{3}} \right)^1}{\left( {\frac{1}{x}} \right)^4} \\ + {}^5{C_5}{\left( {\frac{1}{x}} \right)^5} \\ = \frac{{{x^5}}}{{243}} + 5\left( {\frac{{{x^4}}}{{81}}} \right)\left( {\frac{1}{x}} \right) + 10\left( {\frac{{{x^3}}}{{27}}} \right)\left( {\frac{1}{{{x^2}}}} \right) + 10\left( {\frac{{{x^2}}}{9}} \right)\left( {\frac{1}{{{x^3}}}} \right) + 5\left( {\frac{x}{3}} \right)\left( {\frac{1}{{{x^4}}}} \right) + \frac{1}{{{x^5}}} \\ = \frac{{{x^5}}}{{243}} + \frac{{5{x^3}}}{{81}} + \frac{{10x}}{{27}} + \frac{{10}}{{9x}} + \frac{5}{{3{x^3}}} + \frac{1}{{{x^5}}} \\ \end{gathered} \]

5. Expand the expression ${\left( {x + \frac{1}{x}} \right)^6}$.

Ans. By using Binomial Theorem, the expression ${\left( {x + \frac{1}{x}} \right)^6}$ can be expanded as

\[\begin{gathered} {\left( {x + \frac{1}{x}} \right)^6} = {}^6{C_0}{\left( x \right)^6} + {}^6{C_1}{\left( x \right)^5}\left( {\frac{1}{x}} \right) + {}^6{C_2}{\left( x \right)^4}{\left( {\frac{1}{x}} \right)^2} + {}^6{C_3}{\left( x \right)^3}{\left( {\frac{1}{x}} \right)^3} + {}^6{C_4}{\left( x \right)^2}{\left( {\frac{1}{x}} \right)^4} \\ + {}^6{C_5}\left( x \right){\left( {\frac{1}{x}} \right)^5} + {}^6{C_6}{\left( {\frac{1}{x}} \right)^6} \\ = {x^6} + 6\left( {{x^5}} \right)\left( {\frac{1}{x}} \right) + 15\left( {{x^4}} \right)\left( {\frac{1}{{{x^2}}}} \right) + 20\left( {{x^3}} \right)\left( {\frac{1}{{{x^3}}}} \right) + 15\left( {{x^2}} \right)\left( {\frac{1}{{{x^4}}}} \right) + 6\left( x \right)\left( {\frac{1}{{{x^5}}}} \right) + \frac{1}{{{x^6}}} \\ = {x^6} + 6{x^4} + 15{x^2} + 20 + \frac{{15}}{{{x^2}}} + \frac{6}{{{x^4}}} + \frac{1}{{{x^6}}} \\ \end{gathered}\]

6. Using Binomial Theorem, evaluate ${\left( {96} \right)^3}$.

Ans. 96 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, the binomial theorem can be applied.

It can be written that, $96 = 100 - 4$ 

\[\begin{gathered} {\left( {96} \right)^3} = {\left( {100 - 4} \right)^3} \\ = {}^3{C_0}{\left( {100} \right)^3} - {}^3{C_1}{\left( {100} \right)^2}\left( 4 \right) + {}^3{C_2}\left( {100} \right){\left( 4 \right)^2} - {}^3{C_3}{\left( 4 \right)^3} \\ = 1000000 - 3\left( {10000} \right)\left( 4 \right) + 3\left( {100} \right)\left( {16} \right) - 64 \\ = 1000000 - 120000 + 4800 - 64 \\ = 884736 \\ \end{gathered}\]

7. Using Binomial Theorem, evaluate ${\left( {102} \right)^5}$.

Ans. 102 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, the binomial theorem can be applied.

It can be written that, $102 = 100 + 2$ 

\[\begin{gathered} {\left( {102} \right)^5} = {\left( {100 + 2} \right)^5} \\ = {}^5{C_0}{\left( {100} \right)^5} + {}^5{C_1}{\left( {100} \right)^4}\left( 2 \right) + {}^5{C_2}{\left( {100} \right)^3}{\left( 2 \right)^2} + {}^5{C_3}{\left( {100} \right)^2}{\left( 2 \right)^3} + {}^5{C_4}\left( {100} \right){\left( 2 \right)^4} \\ + {}^5{C_5}{\left( 2 \right)^5} \\ = 10000000000 + 5\left( {100000000} \right)\left( 2 \right) + 10\left( {1000000} \right)\left( 4 \right) + 10\left( {10000} \right)\left( 8 \right) \\ + 5\left( {100} \right)\left( {16} \right) + 32 \\ = 10000000000 + 1000000000 + 40000000 + 80000 + 8000 + 32 \\ = 11040808032 \\ \end{gathered} \]

8. Using Binomial Theorem, evaluate ${\left( {101} \right)^4}$.

Ans. 101 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, the binomial theorem can be applied.

It can be written that, $101 = 100 + 1$ 

\[\begin{gathered} {\left( {101} \right)^4} = {\left( {100 + 1} \right)^4} \\ = {}^4{C_0}{\left( {100} \right)^4} + {}^4{C_1}{\left( {100} \right)^3}\left( 1 \right) + {}^4{C_2}{\left( {100} \right)^2}{\left( 1 \right)^2} + {}^4{C_3}\left( {100} \right){\left( 1 \right)^3} + {}^4{C_4}{\left( 1 \right)^4} \\ = 100000000 + 4\left( {1000000} \right) + 6\left( {10000} \right) + 4\left( {100} \right) + 1 \\ = 100000000 + 4000000 + 60000 + 400 + 1 \\ = 104060401 \\ \end{gathered} \]

9. Using Binomial Theorem, evaluate ${\left( {99} \right)^5}$.

Ans. 99 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, the binomial theorem can be applied.

It can be written that, $99 = 100 - 1$ 

$\begin{gathered} {\left( {99} \right)^5} = {\left( {100 - 1} \right)^5} \\ = {}^5{C_0}{\left( {100} \right)^5} - {}^5{C_1}{\left( {100} \right)^4}\left( 1 \right) + {}^5{C_2}{\left( {100} \right)^3}{\left( 1 \right)^2} - {}^5{C_3}{\left( {100} \right)^2}{\left( 1 \right)^3} + {}^5{C_4}\left( {100} \right){\left( 1 \right)^4} \\ - {}^5{C_5}{\left( 1 \right)^5} \\ = 10000000000 - 5\left( {100000000} \right) - 10\left( {1000000} \right) - 10\left( {10000} \right) + 5\left( {100} \right) - 1 \\ = 10000000000 - 500000000 - 10000000 - 100000 + 500 - 1 \\ = 9509900499 \\ \end{gathered} $

10. Using Binomial Theorem, indicate which number is larger ${\left( {1.1} \right)^{10000}}$ or $1000$.

Ans. By splitting 1.1 and then applying the Binomial Theorem, the first few terms of ${\left( {1.1} \right)^{10000}}$ be obtained as

${\left( {1.1} \right)^{10000}}$ be obtained as \[\begin{gathered} {\left( {1.1} \right)^{10000}} = {\left( {1 + 0.1} \right)^{10000}} \\ = {}^{10000}{C_0} + {}^{10000}{C_1}\left( {1.1} \right) + {\text{Other positive terms}} \\ = 1 + 10000 \times 1.1 + {\text{Other positive terms}} \\ = 1 + 11000 + {\text{Other positive terms}} \\ > 1000 \\ \end{gathered}$

Hence, \[{\left( {1.1} \right)^{10000}} > 1000\]

11. Find ${\left( {a + b} \right)^4} - {\left( {a - b} \right)^4}$. Hence, evaluate ${\left( {\sqrt 3  + \sqrt 2 } \right)^4} - {\left( {\sqrt 3  - \sqrt 2 } \right)^4}$.

Ans. Using Binomial Theorem, the expressions, ${\left( {a + b} \right)^4}$ and ${\left( {a - b} \right)^4}$ , can be expanded as 

\[\begin{gathered} {\left( {a + b} \right)^4} = {}^4{C_0}{a^4} + {}^4{C_1}{a^3}b + {}^4{C_2}{a^2}{b^2} + {}^4{C_3}a{b^3} + {}^4{C_4}{b^4} \\ {\left( {a - b} \right)^4} = {}^4{C_0}{a^4} - {}^4{C_1}{a^3}b + {}^4{C_2}{a^2}{b^2} - {}^4{C_3}a{b^3} + {}^4{C_4}{b^4} \\ \end{gathered} \]

\[\begin{gathered} {\left( {a + b} \right)^4} - {\left( {a - b} \right)^4} = {}^4{C_0}{a^4} + {}^4{C_1}{a^3}b + {}^4{C_2}{a^2}{b^2} + {}^4{C_3}a{b^3} + {}^4{C_4}{b^4} -  \\ \left[ {{}^4{C_0}{a^4} - {}^4{C_1}{a^3}b + {}^4{C_2}{a^2}{b^2} - {}^4{C_3}a{b^3} + {}^4{C_4}{b^4}} \right] \\ = 2\left( {{}^4{C_1}{a^3}b + {}^4{C_3}a{b^3}} \right) \\ = 2\left( {4{a^3}b + 4a{b^3}} \right) \\ = 8ab\left( {{a^2} + {b^2}} \right) \\ \end{gathered} \]

By putting $a = \sqrt 3 $ and $b = \sqrt 2 $, we obtain

\[\begin{gathered} {\left( {\sqrt 3  + \sqrt 2 } \right)^4} - {\left( {\sqrt 3  - \sqrt 2 } \right)^4} = 8\left( {\sqrt 3 } \right)\left( {\sqrt 2 } \right)\left[ {{{\left( {\sqrt 3 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} \right] \\ = 8\sqrt 6 \left( {3 + 2} \right) \\ = 40\sqrt 6  \\ \end{gathered} \]

12. Find ${\left( {x + 1} \right)^6} + {\left( {x - 1} \right)^6}$. Hence or otherwise evaluate ${\left( {\sqrt 2  + 1} \right)^6} + {\left( {\sqrt 2  - 1} \right)^6}$.

Ans. Using Binomial Theorem, the expressions, ${\left( {x + 1} \right)^6}$ and ${\left( {x - 1} \right)^6}$ , can be expanded as 

\[\begin{gathered} {\left( {x + 1} \right)^6} = {}^6{C_0}{x^6} + {}^6{C_1}{x^5} + {}^6{C_2}{x^4} + {}^6{C_3}{x^3} + {}^6{C_4}{x^2} + {}^6{C_5}x + {}^6{C_6} \hfill \\ {\left( {x - 1} \right)^6} = {}^6{C_0}{x^6} - {}^6{C_1}{x^5} + {}^6{C_2}{x^4} - {}^6{C_3}{x^3} + {}^6{C_4}{x^2} - {}^6{C_5}x + {}^6{C_6} \hfill \\ \end{gathered} \]

\[\begin{gathered} {\left( {x + 1} \right)^6} + {\left( {x - 1} \right)^6} = {}^6{C_0}{x^6} + {}^6{C_1}{x^5} + {}^6{C_2}{x^4} + {}^6{C_3}{x^3} + {}^6{C_4}{x^2} + {}^6{C_5}x + {}^6{C_6} \\ + \left[ {{}^6{C_0}{x^6} - {}^6{C_1}{x^5} + {}^6{C_2}{x^4} - {}^6{C_3}{x^3} + {}^6{C_4}{x^2} - {}^6{C_5}x + {}^6{C_6}} \right] \\ = 2\left( {{}^6{C_0}{x^6} + {}^6{C_2}{x^4} + {}^6{C_4}{x^2} + {}^6{C_6}} \right) \\ = 2\left( {{x^6} + 15{x^4} + 15{x^2} + 1} \right) \\ \end{gathered} \]

By putting $x = \sqrt 2 $, we obtain

\[\begin{gathered} {\left( {\sqrt 2  + 1} \right)^6} + {\left( {\sqrt 2  - 1} \right)^6} = 2\left[ {{{\left( {\sqrt 2 } \right)}^6} + 15{{\left( {\sqrt 2 } \right)}^4} + 15{{\left( {\sqrt 2 } \right)}^2} + 1} \right] \\ = 2\left[ {8 + 15 \cdot 4 + 15 \cdot 2 + 1} \right] \\ = 2\left[ {8 + 60 + 30 + 1} \right] \\ = 2 \times 99 \\ = 198 \\ \end{gathered} \]

13. Show that ${9^{n + 1}} - 8n - 9$ is divisible by 64, whenever n is a positive integer.

Ans. In order to show that ${9^{n + 1}} - 8n - 9$ is divisible by 64, it has to be prove that, ${9^{n + 1}} - 8n - 9 = 64k$, where k is some natural number.

By Binomial Theorem,

${\left( {1 + a} \right)^m} = {}^m{C_0} + {}^m{C_1}a + {}^m{C_2}{a^2} + ... + {}^m{C_m}{a^m}$

For $a = 8$ and $m = n + 1$, we obtain

\[\begin{gathered} {\left( {1 + 8} \right)^{n + 1}} = {}^{n + 1}{C_0} + {}^{n + 1}{C_1}\left( 8 \right) + {}^{n + 1}{C_2}{\left( 8 \right)^2} + ... + {}^{n + 1}{C_{n + 1}}{\left( 8 \right)^{n + 1}} \\ {9^{n + 1}} = 1 + \left( {n + 1} \right)\left( 8 \right) + {8^2}\left[ {{}^{n + 1}{C_2} + {}^{n + 1}{C_3} \times 8 + ... + {}^{n + 1}{C_{n + 1}}{{\left( 8 \right)}^{n - 1}}} \right] \\ {9^{n + 1}} = 9 + 8n + 64\left[ {{}^{n + 1}{C_2} + {}^{n + 1}{C_3} \times 8 + ... + {}^{n + 1}{C_{n + 1}}{{\left( 8 \right)}^{n - 1}}} \right] \\ {9^{n + 1}} - 8n - 9 = 64k,{\text{ where }}k = {}^{n + 1}{C_2} + {}^{n + 1}{C_3} \times 8 + ... + {}^{n + 1}{C_{n + 1}}{\left( 8 \right)^{n - 1}}{\text{ is a natural number}} \\ \end{gathered} \]

Thus, ${9^{n + 1}} - 8n - 9$ is divisible by 64, whenever n is a positive integer .

14. Prove that $\sum\limits_{r = 0}^n {{3^r}{}^n{C_r}}  = {4^n}$.

Ans. By Binomial Theorem,

$\sum\limits_{r = 0}^n {{}^n{C_r}{a^{n - r}}{b^r}}  = {\left( {a + b} \right)^n}$

By putting $b = 3$ and $a = 1$ in the above equation, we obtain

$\begin{gathered} \sum\limits_{r = 0}^n {{}^n{C_r}{{\left( 1 \right)}^{n - r}}{{\left( 3 \right)}^r}}  = {\left( {1 + 3} \right)^n} \\ \sum\limits_{r = 0}^n {{3^r}{}^n{C_r}}  = {4^n} \\ \end{gathered} $

Hence proved.

Miscellaneous Exercise:

1. If a and b are distinct integers, prove that a-b is a factor of ${a^n} - {b^n}$, whenever n is a positive integer

(hint: ${a^n} = {(a - b + b)^n}$)

Ans: To prove to prove that(a-b) is a factor of (${a^n} - {b^n}$), it must be proved that ${a^n} - {b^n}$= \[k(a - b)\], where k is some natural number It can be written that, $a = a - b + b$

${((a - b) + b)^n}{ = ^n}{C_0}{(a - b)^n}{ + ^n}{C_1}{(a - b)^{n - 1}}b{ + ^n}{C_2}{(a - b)^{n - 2}}{b^2} +  \ldots { + ^n}{C_{n - 1}}(a - b){b^{n - 1}}{ + ^n}{C_n}{b^n}$

=${(a - b)^n}{ + ^n}{C_1}{(a - b)^{n - 1}}b{ + ^n}{C_2}{(a - b)^{n - 2}}{b^2} +  \ldots { + ^n}{C_{n - 1}}(a - b){b^{n - 1}} + {b^n}$

=${a^n} - {b^n} = (a - b)$$[{(a - b)^{n - 1}}{ + ^n}{C_1}{(a - b)^{n - 2}}b{ + ^n}{C_2}{(a - b)^{n - 3}}{b^2} +  \ldots { + ^n}{C_{n - 1}}{b^{n - 1}}]$

$ \Rightarrow {a^n} - {b^n} = k(a - b)$

Where k =$[{(a - b)^{n - 1}}{ + ^n}{C_1}{(a - b)^{n - 2}}b{ + ^n}{C_2}{(a - b)^{n - 3}}{b^2} +  \ldots { + ^n}{C_{n - 1}}{b^{n - 1}}]$ is a natural number this shows that \[(a - b)\]is a factor of $({a^n} - {b^n})$,

Where n is a positive integer.

2. Evaluate \[\left( {\sqrt 3 } \right. + {\left. {\sqrt 2 } \right)^6} - \left( {\sqrt 3 } \right. - {\left. {\sqrt 2 } \right)^6}\]

Ans: Firstly, the expression

${\left( {a + b} \right)^6} - {\left( {a - b} \right)^6}$ is simplified by using Binomial Theorem. This can be done as

${\left( {a + b} \right)^6}$=$^6{C_0}{(a)^6}{ + ^6}{C_1}{(a)^5}b{ + ^6}{C_2}{(a)^4}{b^2}{ + ^6}{C_3}{(a)^3}{b^3}{ + ^6}{C_4}{(a)^2}{b^4}{ + ^6}{C_5}(a){b^5}{ + ^6}{C_6}{b^6}$

=${(a)^6} + 6{(a)^5}b + 15{(a)^4}{b^2} + 20{(a)^3}{b^3} + 15{(a)^2}{b^4} + 6a{b^5} + {b^6}$

Putting a=$\sqrt 3 \,and\,$b=$\sqrt 2 $, we obtain

\[\left( {\sqrt 3 } \right. + {\left. {\sqrt 2 } \right)^6} - \left( {\sqrt 3 } \right. - {\left. {\sqrt 2 } \right)^6}\]

\[ = 2\left( {6 \cdot {{(\sqrt 3 )}^5}(\sqrt 2 ) + 20 \cdot {{(\sqrt 3 )}^3}{{(\sqrt 2 )}^3} + 6 \cdot (\sqrt 3 ){{(\sqrt 2 )}^5}} \right)\]

=\[2 \times 198\sqrt 6 \]

\[ = 396\sqrt 6 \]

3. Find the values of${\left( {{a^2} + \sqrt {{a^2} - 1} } \right)^4} + {\left( {{a^2} - \sqrt {{a^2} - 1} } \right)^4}$

Ans: Firstly, the expression is simplified by using the Binomial Theorem.

${\left( {x + y} \right)^4} + {\left( {x - y} \right)^4}$

This can be done as

${\left( {x + y} \right)^4}$=$^4{C_0}{(x)^4}{ + ^4}{C_1}{(x)^3}y{ + ^4}{C_2}{(x)^2}{y^2}{ + ^4}{C_3}x\,{y^3}{ + ^4}{C_4}{y^4}$

=${(x)^4} + 4{(x)^3}y + 6{(x)^2}{y^2} + 4x\,{y^3} + {y^4}$

${\left( {x - y} \right)^4}$=$^4{C_0}{(x)^4}{ - ^4}{C_1}{(x)^3}y{ - ^4}{C_2}{(x)^2}{y^2}{ - ^4}{C_3}x\,{y^3}{ - ^4}{C_4}{y^4}$

=${(x)^4} - 4{(x)^3}y - 6{(x)^2}{y^2} - 4x\,{y^3} - {y^4}$

Putting x=${a^2}$ and $y = \sqrt {{a^2} - 1} ,$ We obtain

${\left( {{a^2} + \sqrt {{a^2} - 1} } \right)^4} + {\left( {{a^2} - \sqrt {{a^2} - 1} } \right)^4}$

\[ = 2\left[ {{{\left( {{a^2}} \right)}^4} + 6{{\left( {{a^2}} \right)}^2}{{\left( {\sqrt {{a^2} - 1} } \right)}^2} + {{\left( {\sqrt {{a^2} - 1} } \right)}^4}} \right]\]

\[ = 2\left[ {\left( {{a^8}} \right) + 6\left( {{a^4}} \right)\left( {{a^2} - 1} \right) + {{\left( {{a^2} - 1} \right)}^2}} \right]\]

\[ = 2\left[ {{a^8} + 6{a^6} - 6{a^4} + {a^4} - 2{a^2} + 1} \right]\]

\[ = 2\left[ {{a^8} + 6{a^6} - 5{a^4} - 2{a^2} + 1} \right]\]

\[ = 2{a^8} + 12{a^6} - 10{a^4} - 4{a^2} + 2\]

4. Find an approximation of ${\left( {0.99} \right)^5}$using the first three terms of its expansion.

Ans: $0.99\, = 1 - 0.01$

${\left( {0.99} \right)^5} = {\left( {1 - 0.01} \right)^5}$

\[^5{C_0}{(1)^5}{ - ^5}{C_1}{(1)^4}\left( {0.01} \right){ - ^5}{C_2}{(1)^3}{\left( {0.01} \right)^2}\]

(Approximately)

$ = 1 - 0.05 + 0.001$

$ = 1.001 - 0.05$

=$ = 0.951$

Thus, the value of ${\left( {0.99} \right)^5}$is approximately 0.951

5. Expand using Binomial Theorem ${\left( {1 + \dfrac{x}{2} - \dfrac{2}{x}} \right)^4},\,x \ne 0$

Ans: ${\left( {1 + \dfrac{x}{2} - \dfrac{2}{x}} \right)^4}$

\[{ = ^n}{C_0}\left( {1 + {{\dfrac{x}{2}}^4}} \right){ - ^n}{C_1}{\left( {1 + {{\dfrac{x}{2}}^4}} \right)^3}\left( {\dfrac{2}{x}} \right) - {\,^n}{C_2}{\left( {1 + {{\dfrac{x}{2}}^4}} \right)^2}{\left( {\dfrac{2}{x}} \right)^2}{ - ^n}{C_3}\left( {1 + {{\dfrac{x}{2}}^4}} \right){\left( {\dfrac{2}{x}} \right)^3}{ - ^n}{C_4}{\left( {\dfrac{2}{x}} \right)^4}\]

\[ = \left( {1 + {{\dfrac{x}{2}}^4}} \right) - 4{\left( {1 + {{\dfrac{x}{2}}^4}} \right)^3}\left( {\dfrac{2}{x}} \right) + \,6\left( {1 + x + {{\dfrac{x}{4}}^2}} \right)\left( {\dfrac{4}{{{x^2}}}} \right) - 4\left( {1 + \dfrac{x}{2}} \right)\left( {\dfrac{8}{{{x^3}}}} \right) + \left( {\dfrac{{16}}{{{x^4}}}} \right)\]

\[ = \left( {1 + {{\dfrac{x}{2}}^4}} \right) - {\left( {1 + {{\dfrac{x}{2}}^4}} \right)^3}\left( {\dfrac{8}{x}} \right) + \,\left( {\dfrac{8}{{{x^2}}}} \right) + \dfrac{{24}}{x} + 6 - \left( {\dfrac{{32}}{{{x^3}}}} \right) + \left( {\dfrac{{16}}{{{x^4}}}} \right)\]…..(1)

Again, by using the Binomial Theorem, we obtain

\[{\left( {1 + \dfrac{x}{2}} \right)^4}{ = ^4}{C_0}{(1)^4}{ + ^4}{C_1}{(1)^3}\left( {\dfrac{x}{2}} \right){ + ^4}{C_2}{(1)^2}{\left( {\dfrac{x}{2}} \right)^2}{ + ^4}{C_3}\,{\left( {\dfrac{x}{2}} \right)^3}{ + ^4}{C_4}{\left( {\dfrac{x}{2}} \right)^4}\]

$ = 1 + 4 \times \dfrac{x}{2} + 6 \times \dfrac{{{x^4}}}{4} + 4 \times \dfrac{{{x^3}}}{8} + \dfrac{{{x^3}}}{{16}}$

$ = 1 + 2x + \dfrac{{3{x^2}}}{2} + \dfrac{{{x^3}}}{2} + \dfrac{{{x^4}}}{{16}}$…..(2)

\[{\left( {1 + \dfrac{x}{2}} \right)^3}{ = ^3}{C_0}{(1)^3}{ + ^3}{C_1}{(1)^2}\left( {\dfrac{x}{2}} \right){ + ^3}{C_2}(1){\left( {\dfrac{x}{2}} \right)^2}{ + ^3}{C_3}\,{\left( {\dfrac{x}{2}} \right)^3}\]

$ = \,1 + \dfrac{{3x}}{2} + \dfrac{{3{x^2}}}{4} + \dfrac{{{x^3}}}{8} + \dfrac{{{x^3}}}{8}$…… (3)

From (1), (2), and (3) we obtain

${\left( {\left( {1 + \dfrac{x}{2}} \right) - \dfrac{2}{x}} \right)^4}$

$ = 1 + 2x + \dfrac{{3{x^2}}}{2} + \dfrac{{{x^3}}}{2} + \dfrac{{{x^4}}}{{16}} - \left( {\dfrac{8}{x}} \right)\left( {1 + \dfrac{{3x}}{2} + \dfrac{{3{x^2}}}{4} + \dfrac{{{x^3}}}{8}} \right) + \dfrac{8}{{{x^2}}} + \dfrac{{24}}{x} + 6 - \dfrac{{32}}{{{x^3}}} + \dfrac{{16}}{{{x^4}}}$

$ = 1 + 2x + \dfrac{{3{x^2}}}{2} + \dfrac{{{x^3}}}{2} + \dfrac{{{x^4}}}{{16}} - \dfrac{8}{x} - 12 - 6x - {x^2} - \dfrac{8}{{{x^2}}} + \dfrac{{24}}{x} + 6 - \dfrac{{32}}{{{x^3}}} + \dfrac{{16}}{{{x^4}}}$

$ = \dfrac{{16}}{x} + \dfrac{8}{{{x^2}}} - \dfrac{{32}}{{{x^3}}} + \dfrac{{16}}{{{x^4}}} - 4x + \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{2} + \dfrac{{{x^4}}}{{16}} - 5$.

6. Find the expansion of ${\left( {3{x^2} - 2ax + 3{a^2}} \right)^3}$using binomial theorem.

Ans: Using the Binomial Theorem, the given expression

${\left( {3{x^2} - 2ax + 3{a^2}} \right)^3}$Can be expanded as

${\left( {3{x^2} - 2ax + 3{a^2}} \right)^3}$

${ = ^3}{C_0}{\left( {3{x^2} - 2ax} \right)^3}{ - ^3}{C_1}{\left( {3{x^2} - 2ax} \right)^2}\left( {3{a^2}} \right){ + ^3}{C_2}\left( {3{x^2} - 2ax} \right){\left( {3{a^2}} \right)^2}{ - ^3}{C_3}{\left( {3{a^2}} \right)^3}$

$ = {\left( {3{x^2} - 2ax} \right)^3} + 3\left( {9{x^4} - 12a{x^3} + 4{a^2}{x^2}} \right)\left( {3{a^2}} \right) + 3\left( {3{x^2} - 2ax} \right)\left( {9{a^4}} \right) + \left( {2{a^6}} \right)$

$ = {\left( {3{x^2} - 2ax} \right)^3} + 81{a^2}{x^4} - 108{a^3}{x^3} + 36{a^4}{x^2} + 81{a^4}{x^2} - 54{a^5}x + 27{a^6}$

$ = {\left( {3{x^2} - 2ax} \right)^3} + 81{a^2}{x^4} - 108{a^3}{x^3} + 117{a^4}{x^2} - 54{a^5}x + 27{a^6}$…. (1)

${\left( {3{x^2} - 2ax} \right)^3}$

${ = ^3}{C_0}{\left( {3{x^2}} \right)^3}{ - ^3}{C_1}{\left( {3{x^2}} \right)^2}\left( {2ax} \right){ + ^3}{C_2}\left( {3{x^2}} \right){\left( {2ax} \right)^2}{ - ^3}{C_3}{\left( {2ax} \right)^3}$

\[ = \left( {27{x^6}} \right) - 3\left( {9{x^4}} \right)\left( {2ax} \right) + 3\left( {3{x^2}} \right)\left( {4{a^2}{x^2}} \right) - 8{a^3}{x^3}\]

\[ = 27{x^6} - 54a{x^5} + 36{a^2}{x^4} - 8{a^3}{x^3}\]……… (2)

From (1) and (2), we obtain

\[ = 27{x^6} - 54a{x^5} + 36{a^2}{x^4} - 8{a^3}{x^3} + 81{a^2}{x^4} - 108{a^3}{x^3} + 117{a^4}{x^2} - 54{a^5}x + 27{a^6}\]

\[ = 27{x^6} - 54a{x^5} + 117{a^2}{x^4} - 116{a^3}{x^3} + 117{a^4}{x^2} - 54{a^5}x + 27{a^6}\].

Overview of Deleted Syllabus for CBSE Class 11 Maths Binomial Theorem

Class 11 Maths Chapter 7: Exercises Breakdown

Binomial Theorem Class 11 NCERT Solutions is essential for mastering the expansion of binomial expressions and understanding binomial coefficients. It is important to focus on the binomial theorem formula, Pascal's Triangle, and the general and middle terms of the expansion. Practicing these concepts through Vedantu's solutions will help reinforce your understanding. In the previous year's CBSE exams, around 3 to 4 questions were asked from this chapter. By thoroughly studying and practicing, you can enhance your problem-solving skills and perform well in your exams.

Other Study Material for CBSE Class 11 Maths Chapter 7

Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions Class 11 Maths Chapter 7 Binomial Theorem

1. What is the Binomial Theorem?

The binomial theorem is defined as the process of algebraically expanding the power of sums of two or more binomials. Coefficients of binomial terms in the process of expansion are referred to as binomial coefficients. The introductory parts of these chapters consist of proper definitions of different aspects of the binomial theorem. 

With the Binomial Theorem Class 11 NCERT Solutions PDF, students can now study with ease and be updated with all information that might appear in their examinations. Learning the concepts of the Binomial Theorem will be easier with the NCERT Solutions available on Vedantu. 

2. How will the NCERT Solutions Class 11 Maths Chapter 7 Binomial Theorem help in understanding the concepts of the Binomial Theorem?

To learn about the expansion procedure, refer to the examples available in binomial theorem class 11 NCERT solutions. These examples have been solved in a step-by-step format that will help students understand the concepts better. 

Students are required to be well-versed in solving these equations if they wish to score well in examinations. There are also exercises in ch 7 maths class 11 that are given in this segment that students can solve on their own. This will allow them to practice what they have learned and clear any doubt that they might have related to the binomial theorem. 

3. What are the properties of positive integers in the Binomial Theorem? 

There are more than 10 properties that are listed under positive integers that students can learn when studying binomial theorem class 11 NCERT solutions. Students are required to study these properties to understand the basic concept of solving such equations. 

Examination papers may target aspects of chapters that seem simple to students but may be tricky to solve, which is why students will need to go through these binomial theorem class 11 NCERT solutions thoroughly if they wish to score well in their upcoming examinations. 

4. Explain the concept of the Binomial Theorem covered in binomial theorem class 11.

The Binomial theorem states, for positive integer n, whenever you add any two numbers, say a and b, the result raised to the power of n can also be written as the sum of (n+10 terms). The coefficients involved are expressed as binomial coefficients. The NCERT solutions of Class 11 Maths Chapter 7 can be accessed on the Vedantu website and the app. Practice all of these judiciously if you want to score well in them and in the other topics related to them.

5. What Chapter is Binomial Theorem Class 11?

The Binomial Theorem is Chapter 7 of the NCERT Mathematics book. It explains in detail the Binomial Theorem and also provides the necessary exercises for a better understanding of the concepts by the students. The solutions of ch 7 maths class 11 can easily be found on the Vedantu site (vedantu.com). The students are advised to go through all topics to efficiently grasp the content and score well in their exams.

6. Which is the best Solution book for NCERT Class 11 Chapter 7 Maths?

The best solution book for Chapter 7-Binomial Theorem is available on Vedantu. NCERT Class 11 Maths Chapter 7 Solutions may be obtained by visiting the Vedantu website. Aside from that, you may access a range of modules that will help you achieve excellent grades in math exams. The link to the solution to the exercise is provided below. Visit the page NCERT Solutions for Class 11 Chapter 7 to download the PDF file free of cost.

7. How can I master Class 11 Maths Chapter 7?

There is no easy or singular motto for performing well or ranking high in any topic. To do well, one must constantly be diligent about the core ideas. It is important to practice as many questions as possible thoroughly so that they have a good understanding of the concepts. Proper practice of Chapter 7 class 11 maths is critical for getting high grades and gaining deeper knowledge. Those interested in accessing NCERT solutions, revision notes and important questions of this chapter, visit the Vedantu website or download the Vedantu app.

8. Do I need to practice all the questions provided in Class 11 Maths Chapter 7 NCERT Solutions?

Yes, it is critical to practice and answer all questions since they cover a variety of subjects and concepts which will give you a good understanding of the kind of questions that might be set from those areas. These questions also help you learn how different questions from the same topic may be set. Each exercise should be thoroughly practised. You can discover modules on the Vedantu site or in the Vedantu app that are relevant to this topic or other topics in this Chapter 7 Class 11 maths.

9. What is Pascal's Triangle in binomial theorem class 11 solutions pdf?

Pascal's Triangle is a triangular array of numbers. Each number is the sum of the two numbers directly above it. This pattern is used to find binomial coefficients. It is helpful in expanding binomial theorem class 11 solutions pdf.

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CBSE Class 9th Maths 2023 : 30 Most Important Case Study Questions with Answers; Download PDF

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CBSE Class 9 Maths exam 2022-23 will have a set of questions based on case studies in the form of MCQs. CBSE Class 9 Maths Question Bank on Case Studies given in this article can be very helpful in understanding the new format of questions.

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CBSE Case Study Questions for Class 9 Maths - Pdf PDF Download

CBSE Case Study Questions for Class  9 Maths

CBSE Case Study Questions for Class 9 Maths are a type of assessment where students are given a real-world scenario or situation and they need to apply mathematical concepts to solve the problem. These types of questions help students to develop their problem-solving skills and apply their knowledge of mathematics to real-life situations.

Chapter Wise Case Based Questions for Class 9 Maths

The CBSE Class 9 Case Based Questions can be accessed from Chapetrwise Links provided below:

Chapter-wise case-based questions for Class 9 Maths are a set of questions based on specific chapters or topics covered in the maths textbook. These questions are designed to help students apply their understanding of mathematical concepts to real-world situations and events.

Chapter 1: Number System

• Case Based Questions: Number System

Chapter 2: Polynomial

• Case Based Questions: Polynomial

Chapter 3: Coordinate Geometry

• Case Based Questions: Coordinate Geometry

Chapter 4: Linear Equations

• Case Based Questions: Linear Equations - 1
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Chapter 5: Introduction to Euclid’s Geometry

• Case Based Questions: Lines and Angles

Chapter 7: Triangles

• Case Based Questions: Triangles

Chapter 8: Quadrilaterals

• Case Based Questions: Quadrilaterals - 1
• Case Based Questions: Quadrilaterals - 2

Chapter 9: Areas of Parallelograms

• Case Based Questions: Circles

Chapter 11: Constructions

• Case Based Questions: Constructions

Chapter 12: Heron’s Formula

• Case Based Questions: Heron’s Formula

Chapter 13: Surface Areas and Volumes

• Case Based Questions: Surface Areas and Volumes

Chapter 14: Statistics

• Case Based Questions: Statistics

Chapter 15: Probability

• Case Based Questions: Probability

Weightage of Case Based Questions in Class 9 Maths

Why are Case Study Questions important in Maths Class  9?

• Enhance critical thinking:  Case study questions require students to analyze a real-life scenario and think critically to identify the problem and come up with possible solutions. This enhances their critical thinking and problem-solving skills.
• Apply theoretical concepts:  Case study questions allow students to apply theoretical concepts that they have learned in the classroom to real-life situations. This helps them to understand the practical application of the concepts and reinforces their learning.
• Develop decision-making skills:  Case study questions challenge students to make decisions based on the information provided in the scenario. This helps them to develop their decision-making skills and learn how to make informed decisions.
• Improve communication skills:  Case study questions often require students to present their findings and recommendations in written or oral form. This helps them to improve their communication skills and learn how to present their ideas effectively.
• Enhance teamwork skills:  Case study questions can also be done in groups, which helps students to develop teamwork skills and learn how to work collaboratively to solve problems.

In summary, case study questions are important in Class 9 because they enhance critical thinking, apply theoretical concepts, develop decision-making skills, improve communication skills, and enhance teamwork skills. They provide a practical and engaging way for students to learn and apply their knowledge and skills to real-life situations.

Class 9 Maths Curriculum at Glance

The Class 9 Maths curriculum in India covers a wide range of topics and concepts. Here is a brief overview of the Maths curriculum at a glance:

• Number Systems:  Students learn about the real number system, irrational numbers, rational numbers, decimal representation of rational numbers, and their properties.
• Algebra:  The Algebra section includes topics such as polynomials, linear equations in two variables, quadratic equations, and their solutions.
• Coordinate Geometry:  Students learn about the coordinate plane, distance formula, section formula, and slope of a line.
• Geometry:  This section includes topics such as Euclid’s geometry, lines and angles, triangles, and circles.
• Trigonometry: Students learn about trigonometric ratios, trigonometric identities, and their applications.
• Mensuration: This section includes topics such as area, volume, surface area, and their applications.
• Statistics and Probability:  Students learn about measures of central tendency, graphical representation of data, and probability.

The Class 9 Maths curriculum is designed to provide a strong foundation in mathematics and prepare students for higher education in the field. The curriculum is structured to develop critical thinking, problem-solving, and analytical skills, and to promote the application of mathematical concepts in real-life situations. The curriculum is also designed to help students prepare for competitive exams and develop a strong mathematical base for future academic and professional pursuits.

Students can also access Case Based Questions of all subjects of CBSE Class 9

• Case Based Questions for Class 9 Science
• Case Based Questions for Class 9 Social Science
• Case Based Questions for Class 9 English
• Case Based Questions for Class 9 Hindi
• Case Based Questions for Class 9 Sanskrit

Frequently Asked Questions (FAQs) on Case Based Questions for Class 9 Maths

What is case-based questions.

Case-Based Questions (CBQs) are open-ended problem solving tasks that require students to draw upon their knowledge of Maths concepts and processes to solve a novel problem. CBQs are often used as formative or summative assessments, as they can provide insights into how students reason through and apply mathematical principles in real-world problems.

What are case-based questions in Maths?

Case-based questions in Maths are problem-solving tasks that require students to apply their mathematical knowledge and skills to real-world situations or scenarios.

What are some common types of case-based questions in class 9 Maths?

Common types of case-based questions in class 9 Maths include word problems, real-world scenarios, and mathematical modeling tasks.

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FAQs on CBSE Case Study Questions for Class 9 Maths - Pdf

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CBSE Class 9 Mathematics Case Study Questions

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If you’re looking for a comprehensive and reliable study resource and case study questions for class 9 CBSE, myCBSEguide is the perfect door to enter. With over 10,000 study notes, solved sample papers and practice questions, it’s got everything you need to ace your exams. Plus, it’s updated regularly to keep you aligned with the latest CBSE syllabus . So why wait? Start your journey to success with myCBSEguide today!

Significance of Mathematics in Class 9

Mathematics is an important subject for students of all ages. It helps students to develop problem-solving and critical-thinking skills, and to think logically and creatively. In addition, mathematics is essential for understanding and using many other subjects, such as science, engineering, and finance.

CBSE Class 9 is an important year for students, as it is the foundation year for the Class 10 board exams. In Class 9, students learn many important concepts in mathematics that will help them to succeed in their board exams and in their future studies. Therefore, it is essential for students to understand and master the concepts taught in Class 9 Mathematics .

Case studies in Class 9 Mathematics

A case study in mathematics is a detailed analysis of a particular mathematical problem or situation. Case studies are often used to examine the relationship between theory and practice, and to explore the connections between different areas of mathematics. Often, a case study will focus on a single problem or situation and will use a variety of methods to examine it. These methods may include algebraic, geometric, and/or statistical analysis.

Example of Case study questions in Class 9 Mathematics

The Central Board of Secondary Education (CBSE) has included case study questions in the Class 9 Mathematics paper. This means that Class 9 Mathematics students will have to solve questions based on real-life scenarios. This is a departure from the usual theoretical questions that are asked in Class 9 Mathematics exams.

The following are some examples of case study questions from Class 9 Mathematics:

Class 9 Mathematics Case study question 1

There is a square park ABCD in the middle of Saket colony in Delhi. Four children Deepak, Ashok, Arjun and Deepa went to play with their balls. The colour of the ball of Ashok, Deepak,  Arjun and Deepa are red, blue, yellow and green respectively. All four children roll their ball from centre point O in the direction of   XOY, X’OY, X’OY’ and XOY’ . Their balls stopped as shown in the above image.

Answer the following questions:

Answer Key:

Class 9 Mathematics Case study question 2

• Now he told Raju to draw another line CD as in the figure
• The teacher told Ajay to mark  ∠ AOD  as 2z
• Suraj was told to mark  ∠ AOC as 4y
• Clive Made and angle  ∠ COE = 60°
• Peter marked  ∠ BOE and  ∠ BOD as y and x respectively

Now answer the following questions:

• 2y + z = 90°
• 2y + z = 180°
• 4y + 2z = 120°
• (a) 2y + z = 90°

Class 9 Mathematics Case study question 3

• (a) 31.6 m²
• (c) 513.3 m³
• (b) 422.4 m²

Class 9 Mathematics Case study question 4

How to Answer Class 9 Mathematics Case study questions

To crack case study questions, Class 9 Mathematics students need to apply their mathematical knowledge to real-life situations. They should first read the question carefully and identify the key information. They should then identify the relevant mathematical concepts that can be applied to solve the question. Once they have done this, they can start solving the Class 9 Mathematics case study question.

Students need to be careful while solving the Class 9 Mathematics case study questions. They should not make any assumptions and should always check their answers. If they are stuck on a question, they should take a break and come back to it later. With some practice, the Class 9 Mathematics students will be able to crack case study questions with ease.

Class 9 Mathematics Curriculum at Glance

At the secondary level, the curriculum focuses on improving students’ ability to use Mathematics to solve real-world problems and to study the subject as a separate discipline. Students are expected to learn how to solve issues using algebraic approaches and how to apply their understanding of simple trigonometry to height and distance problems. Experimenting with numbers and geometric forms, making hypotheses, and validating them with more observations are all part of Math learning at this level.

The suggested curriculum covers number systems, algebra, geometry, trigonometry, mensuration, statistics, graphing, and coordinate geometry, among other topics. Math should be taught through activities that include the use of concrete materials, models, patterns, charts, photographs, posters, and other visual aids.

CBSE Class 9 Mathematics (Code No. 041)

Class 9 Mathematics question paper design

The CBSE Class 9 mathematics question paper design is intended to measure students’ grasp of the subject’s fundamental ideas. The paper will put their problem-solving and analytical skills to the test. Class 9 mathematics students are advised to go through the question paper pattern thoroughly before they start preparing for their examinations. This will help them understand the paper better and enable them to score maximum marks. Refer to the given Class 9 Mathematics question paper design.

QUESTION PAPER DESIGN (CLASS 9 MATHEMATICS)

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Class 9 is an important milestone in a student’s life. It is the last year of high school and the last chance to score well in the CBSE board exams. myCBSEguide is the perfect platform for students to get started on their preparations for Class 9 Mathematics. myCBSEguide provides comprehensive study material for all subjects, including practice questions, sample papers, case study questions and mock tests. It also offers tips and tricks on how to score well in exams. myCBSEguide is the perfect door to enter for class 9 CBSE preparations.

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16 thoughts on “CBSE Class 9 Mathematics Case Study Questions”

This method is not easy for me

aarti and rashika are two classmates. due to exams approaching in some days both decided to study together. during revision hour both find difficulties and they solved each other’s problems. aarti explains simplification of 2+ ?2 by rationalising the denominator and rashika explains 4+ ?2 simplification of (v10-?5)(v10+ ?5) by using the identity (a – b)(a+b). based on above information, answer the following questions: 1) what is the rationalising factor of the denominator of 2+ ?2 a) 2-?2 b) 2?2 c) 2+ ?2 by rationalising the denominator of aarti got the answer d) a) 4+3?2 b) 3+?2 c) 3-?2 4+ ?2 2+ ?2 d) 2-?3 the identity applied to solve (?10-?5) (v10+ ?5) is a) (a+b)(a – b) = (a – b)² c) (a – b)(a+b) = a² – b² d) (a-b)(a+b)=2(a² + b²) ii) b) (a+b)(a – b) = (a + b

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All questions was easy but search ? hard questions. These questions was not comparable with cbse. It was totally wastage of time.

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I am Riddhi Shrivastava… These questions was very good.. That’s it.. ..

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Case Study Questions for Class 9 Maths Chapter 12 Herons Formula

Case study questions for class 9 maths chapter 9 areas of parallelograms and triangles, case study questions for class 9 maths chapter 6 lines and angles, case study questions for class 9 maths chapter 7 triangles, case study questions for class 9 maths chapter 5 introduction to euclid’s geometry, case study and passage based questions for class 9 maths chapter 14 statistics, case study questions for class 9 maths chapter 1 real numbers, case study questions for class 9 maths chapter 4 linear equations in two variables, case study questions for class 9 maths chapter 3 coordinate geometry, case study questions for class 9 maths chapter 15 probability, case study questions for class 9 maths chapter 13 surface area and volume, case study questions for class 9 maths chapter 10 circles, case study questions for class 9 maths chapter 9 quadrilaterals, case study questions for class 9 maths chapter 2 polynomials.

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles

NCERT Solutions for Class 9 Maths Chapter 7 Triangles are provided here. Our NCERT Maths solutions contain all the questions of the NCERT textbook that are solved and explained beautifully. Here you will get complete NCERT Solutions for Class 9 Maths Chapter 7 all exercises Exercise in one place. These solutions are prepared by the subject experts and as per the latest NCERT syllabus and guidelines. CBSE Class 9 Students who wish to score good marks in the maths exam must practice these questions regularly.

Class 9 Maths Chapter 7 Triangles NCERT Solutions

Below we have provided the solutions of each exercise of the chapter. Go through the links to access the solutions of exercises you want. You should also check out our NCERT Class 9 Solutions for other subjects to score good marks in the exams.

NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.1

NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.2

NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.3

NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.4

NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.5

NCERT Solutions for Class 9 Maths Chapter 7 – Topic Discussion

Below we have listed the topics that have been discussed in this chapter.

• Congruence of triangles.
• Criteria for congruence of triangles (SAS congruence rule, ASA congruence rule, SSS congruence rule, RHS congruence rule).
• Properties of triangles.
• Inequalities of triangle.

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CBSE Case Study Questions Class 9 Maths Chapter 12 Heron’s Formula PDF Download

CBSE Case Study Questions Class 9 Maths Chapter 12  are very important to solve for your exam. Class 9 Maths Chapter 12 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving  case study-based   questions for Class 9 Maths Chapter 12  Heron’s Formula

Case Study Questions Class 9 Maths Chapter 12

Case Study 1: A group of students is learning about Heron’s Formula for finding the area of a triangle. They encountered the following scenario:

Rohan and Kavya came across a triangular field in their village. They made the following observations:

• The lengths of the three sides of the triangular field are 8 meters, 12 meters, and 15 meters.
• The perimeter of the triangular field is 35 meters.

Based on this information, the students were asked to apply Heron’s Formula to find the area of the triangular field. Let’s see if you can answer the questions correctly:

MCQ Questions:

Q1. The semiperimeter of the triangular field is: (a) 8 meters (b) 12 meters (c) 15 meters (d) 17.5 meters

Answer: (d) 17.5 meters

Q2. Using Heron’s Formula, the area of the triangular field is: (a) 24 square meters (b) 30 square meters (c) 36 square meters (d) 40 square meters

Answer: (b) 30 square meters

Q3. The type of triangle formed by the sides of the field is: (a) Equilateral (b) Isosceles (c) Scalene (d) Right-angled

Answer: (c) Scalene

Q4. The length of the altitude corresponding to the side of 15 meters is: (a) 2 meters (b) 4 meters (c) 6 meters (d) 8 meters

Answer: (c) 6 meters

Q5. The lengths of the altitudes corresponding to the sides of 8 meters and 12 meters are: (a) 4 meters and 6 meters (b) 6 meters and 8 meters (c) 8 meters and 10 meters (d) 10 meters and 12 meters

Answer: (a) 4 meters and 6 meters

Case Study 2: A group of students is studying Heron’s Formula for finding the area of a triangle. They encountered the following scenario:

Neha and Mohan went on a field trip to a riverbank. They noticed a triangular piece of land that they wanted to measure and calculate its area. They made the following observations:

• Neha measured the lengths of the three sides of the triangular piece of land as 7 meters, 9 meters, and 11 meters.
• Mohan measured the lengths of the three sides of the same triangular piece of land as 10 meters, 12 meters, and 15 meters.

Based on this information, the students were asked to apply Heron’s Formula to find the area of the triangular piece of land. Let’s see if you can answer the questions correctly:

Q1. Using Neha’s measurements, the semiperimeter of the triangular piece of land is: (a) 13 meters (b) 16 meters (c) 19 meters (d) 23 meters

Answer: (c) 19 meters

Q2. Using Neha’s measurements, the area of the triangular piece of land is: (a) 24 square meters (b) 26 square meters (c) 28 square meters (d) 30 square meters

Answer: (a) 24 square meters

Q3. Using Mohan’s measurements, the semiperimeter of the triangular piece of land is: (a) 16 meters (b) 18 meters (c) 21 meters (d) 25 meters

Answer: (c) 21 meters

Q4. Using Mohan’s measurements, the area of the triangular piece of land is: (a) 40 square meters (b) 42 square meters (c) 45 square meters (d) 48 square meters

Answer: (b) 42 square meters

Q5. The measurements taken by Neha represent a triangle that is: (a) Equilateral (b) Isosceles (c) Scalene (d) Right-angled

Hope the information shed above regarding Case Study and Passage Based Questions for Case Study Questions Class 9 Maths Chapter 12 Heron’s Formula with Answers Pdf free download has been useful to an extent. If you have any other queries about Case Study Questions Class 9 Maths Chapter 12 Heron’s Formula and Passage-Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible.

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles

Ncert solutions for class 9 maths chapter 7 triangles| pdf download.

Page No: 118

• Exercise 7.1 Chapter 7 Class 9 Maths NCERT Solutions
• Exercise 7.2 Chapter 7 Class 9 Maths NCERT Solutions
• Exercise 7.3 Chapter 7 Class 9 Maths NCERT Solutions
• Exercise 7.4 Chapter 7 Class 9 Maths NCERT Solutions
• Exercise 7.5 Chapter 7 Class 9 Maths NCERT Solutions

NCERT Solutions for Class 9 Maths Chapters:

How many exercises in Chapter 7 Triangles?

Each of the equal angles of an isosceles triangle is 38°, what is the measure of the third angle, find the measure of each of acute angle in a right angle isosceles triangle., if two angles are (30 ∠ a)º and (125 + 2a)º and they are supplement of each other. find the value of ‘a’., contact form.

CBSE Class 9 Maths 30 Most Important Case Study Questions with Answers

Cbse class 9 maths 30 most important case study questions with answers download here free in pdf format..

CBSE Class 9 Maths exam 2023 will have a set of questions based on case studies in the form of MCQs. CBSE Class 9 Maths Question Bank on Case Studies given in this webpage can be very helpful in understanding the new format of questions.

Each question has five sub-questions, each followed by four options and one correct answer. Candidates can easily download these questions in PDF format and refer to them for exam preparation 2023.

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Class 9 Maths Case Study Questions of Chapter 8 Quadrilaterals PDF Download

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Class 9 Maths Case Study Questions Chapter 8  are very important to solve for your exam. Class 9 Maths Chapter 8 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 9 Maths Chapter 8 Quadrilaterals

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

These case study questions challenge students to apply their knowledge of quadrilaterals in practical scenarios, enhancing their problem-solving abilities. This article provides the Class 9 Maths Case Study Questions of Chapter 8: Quadrilaterals, enabling students to practice and excel in their examinations.

Quadrilaterals Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 9 Maths Chapter 8 Quadrilaterals

Case Study/Passage-Based Questions

Case Study 1: Laveena’s class teacher gave students some colorful papers in the shape of quadrilaterals. She asked students to make a parallelogram from it using paper folding. Laveena made the following parallelogram.

How can a parallelogram be formed by using paper folding? (a) Joining the sides of a quadrilateral (b) Joining the mid-points of sides of a quadrilateral (c) Joining the various quadrilaterals (d) None of these

Answer: (b) Joining the mid-points of sides of quadrilateral

Which of the following is true? (a) PQ = BD (b) PQ = 1/2 BD (c) 3PQ = BD (d) PQ = 2BD

Answer: (b) PQ = 1/2 BD

Which of the following is correct combination? (a) 2RS = BD (b) RS = 1/3 BD (c) RS = BD (d) RS = 2BD

Answer: (a) 2RS = BD

Which of the following is correct? (a) SR = 2PQ (b) PQ = SR (c) SR = 3PQ (d) SR = 4PQ

Answer: (b) PQ = SR

Case Study/Passage Based Questions

Case Study 2: Anjali and Meena were trying to prove mid-point theorem. They draw a triangle ABC, where D and E are found to be the midpoints of AB and AC respectively. DE was joined and extended to F such that DE = EF and FC is also joined.

▲ADE and ▲CFE are congruent by which criterion? (a) SSS (b) SAS (c) RHS (d) ASA

Answer: (b) SAS

∠EFC is equal to which angle? (a) ∠DAE (b) ∠EDA (c) ∠AED (d) ∠DBC

Answer: (b)∠EDA

∠ECF is equal to which angle? (a) ∠EAD (b) ∠ADE (c) ∠AED (d) ∠B

Answer: (a) ∠EAD

CF is equal to (a) EC (b) BE (c) BC (d) AD

Answer: (d) AD

CF is parallel to (a) AE (b) CE (c) BD (d) AC

Answer: (c) BD

Case Study 3. A group of students is exploring different types of quadrilaterals. They encountered the following scenario:

Four friends, Aryan, Bhavana, Chetan, and Divya, participated in a geometry project. They constructed a figure with four sides and made the following observations:

• The opposite sides of the figure are parallel.
• The opposite angles of the figure are congruent.
• The figure has two pairs of congruent adjacent sides.
• The sum of the measures of the interior angles of the figure is 360 degrees.

Based on this information, the students were asked to analyze the properties of the quadrilateral they constructed. Let’s see if you can answer the questions correctly:

MCQ Questions:

Q1. The type of quadrilateral formed by their figure is: (a) Parallelogram (b) Rhombus (c) Rectangle (d) Square

Answer: (a) Parallelogram

Q2. The measure of each angle in the figure is: (a) 90 degrees (b) 120 degrees (c) 135 degrees (d) 180 degrees

Answer: (d) 180 degrees

Q3. The figure is an example of a quadrilateral that satisfies the: (a) Opposite sides are equal condition (b) Opposite angles are congruent condition (c) Diagonals bisect each other condition (d) None of the above

Answer: (b) Opposite angles are congruent condition

Q4. The sum of the measures of the exterior angles of the figure is: (a) 90 degrees (b) 180 degrees (c) 270 degrees (d) 360 degrees

Answer: (d) 360 degrees

Q5. The figure has rotational symmetry of: (a) Order 1 (b) Order 2 (c) Order 3 (d) Order 4

Answer: (a) Order 1

Hope the information shed above regarding Case Study and Passage Based Questions for Class 9 Mathematics Chapter 8 Quadrilaterals with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 9 Maths Quadrilaterals Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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• NCERT Solutions
• NCERT Class 9
• NCERT 9 Maths
• Chapter 7: Geometry Of Triangles

NCERT Solutions for Class 9 Maths Chapter 7 - Triangles

Ncert solutions for class 9 maths chapter 7 – cbse free pdf download.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles provides the answers and questions related to the chapter as included in the CBSE Syllabus for 2023-24. The word triangle itself describes its meaning. “Tri” means “three”,; so a closed figure formed by three intersecting lines is known as a Triangle. Students must have already studied the angle sum property of a triangle in Chapter 6 of NCERT Class 9 Maths. Now, in continuation to it, Chapter 7 of NCERT Solutions for Class 9 Maths will brief students about the congruence of triangles and the rules of congruence. Also, they will learn a few more properties of triangles and inequalities in a triangle.

Download Exclusively Curated Chapter Notes for Class 9 Maths Chapter – 7 Goemetry of Triangles

Download most important questions for class 9 maths chapter – 7 goemetry of triangles.

Here we have provided the complete Class 9 Maths NCERT Solutions Chapter 7  Triangles in PDF format solved by experienced teachers. Students can download the free PDF of these NCERT Solutions for Class 9 by clicking on the link below to keep it handy for future reference.

• Chapter 1 Number System
• Chapter 2 Polynomials
• Chapter 3 Coordinate Geometry
• Chapter 4 Linear Equations in Two Variables
• Chapter 5 Introduction to Euclids Geometry
• Chapter 6 Lines and Angles
• Chapter 7 Triangles
• Chapter 8 Quadrilaterals
• Chapter 9 Areas of Parallelograms and Triangles
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Heron’s Formula
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
• Chapter 15 Inroduction to Probability

NCERT Solutions for Class 9 Maths Chapter 7 – Triangles

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List of Exercises in NCERT Class 9 Maths Chapter 7

Exercise 7.1 Solution 8 Questions (6 Short Answer Questions, 2 Long Answer Question) Exercise 7.2 Solution  8 Questions (6 Short Answer Questions, 2 Long Answer Question) Exercise 7.3 Solution  5 Questions (3 Short Answer Questions, 2 Long Answer Question) Exercise 7.4 Solution 6 Questions (5 Short Answer Questions, 1 Long Answer Question) Exercise 7.5 (Optional) Solution 4 Questions

Access Answers of Maths NCERT Class 9 Chapter 7 – Triangles

Exercise: 7.1 (Page No: 118)

1. In quadrilateral ACBD, AC = AD and AB bisect ∠A (see Fig. 7.16). Show that ΔABC≅ ΔABD. What can you say about BC and BD?

It is given that AC and AD are equal i.e. AC = AD and the line segment AB bisects ∠A.

We will have to now prove that the two triangles ABC and ABD are similar i.e. ΔABC ≅ ΔABD

Consider the triangles ΔABC and ΔABD,

(i) AC = AD (It is given in the question)

(ii) AB = AB (Common)

(iii) ∠CAB = ∠DAB (Since AB is the bisector of angle A)

So, by SAS congruency criterion , ΔABC ≅  ΔABD.

For the 2 nd part of the question, BC and BD are of equal lengths by the rule of C.P.C.T.

2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see Fig. 7.17). Prove that

(i) ΔABD ≅  ΔBAC

(ii) BD = AC

(iii) ∠ ABD = ∠ BAC.

The given parameters from the questions are ∠ DAB = ∠ CBA and AD = BC.

(i) ΔABD and ΔBAC are similar by SAS congruency as

AB = BA (It is the common arm)

∠ DAB = ∠ CBA and AD = BC (These are given in the question)

So, triangles ABD and BAC are similar i.e. ΔABD ≅  ΔBAC. (Hence proved).

(ii) It is now known that ΔABD ≅  ΔBAC so,

BD = AC (by the rule of CPCT).

(iii) Since ΔABD ≅  ΔBAC so,

Angles ∠ ABD = ∠ BAC (by the rule of CPCT).

3. AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.

It is given that AD and BC are two equal perpendiculars to AB.

We will have to prove that CD is the bisector of AB

Triangles ΔAOD and ΔBOC are similar by AAS congruency since:

(i) ∠A = ∠B (They are perpendiculars)

(ii) AD = BC (As given in the question)

(iii) ∠AOD = ∠BOC (They are vertically opposite angles)

∴ ΔAOD ≅ ΔBOC.

So, AO = OB (by the rule of CPCT).

Thus, CD bisects AB (Hence proved).

4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ΔABC ≅ ΔCDA.

It is given that p || q and l || m

Triangles ABC and CDA are similar i.e. ΔABC ≅ ΔCDA

Consider the ΔABC and ΔCDA,

(i) ∠BCA = ∠DAC and ∠BAC = ∠DCA Since they are alternate interior angles

(ii) AC = CA as it is the common arm

So, by ASA congruency criterion, ΔABC ≅ ΔCDA.

5. Line l is the bisector of an angle ∠A and B is any point on l . BP and BQ are perpendiculars from B to the arms of ∠A (see Fig. 7.20). Show that:

(i) ΔAPB ≅ ΔAQB

(ii) BP = BQ or B is equidistant from the arms of ∠A.

It is given that the line “ l ” is the bisector of angle ∠A and the line segments BP and BQ are perpendiculars drawn from l .

(i) ΔAPB and ΔAQB are similar by AAS congruency because:

∠P = ∠Q (They are the two right angles)

AB = AB (It is the common arm)

∠BAP = ∠BAQ (As line l is the bisector of angle A)

So, ΔAPB ≅ ΔAQB.

(ii) By the rule of CPCT, BP = BQ. So, it can be said the point B is equidistant from the arms of ∠A.

6. In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

It is given in the question that AB = AD, AC = AE, and ∠BAD = ∠EAC

The line segment BC and DE are similar i.e. BC = DE

We know that ∠BAD = ∠EAC

Now, by adding ∠DAC on both sides we get,

∠BAD + ∠DAC = ∠EAC +∠DAC

This implies, ∠BAC = ∠EAD

Now, ΔABC and ΔADE are similar by SAS congruency since:

(i) AC = AE (As given in the question)

(ii) ∠BAC = ∠EAD

(iii) AB = AD (It is also given in the question)

∴ Triangles ABC and ADE are similar i.e. ΔABC ≅ ΔADE.

So, by the rule of CPCT, it can be said that BC = DE.

7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see Fig. 7.22). Show that

(i) ΔDAP ≅ ΔEBP

(ii) AD = BE

In the question, it is given that P is the mid-point of line segment AB. Also, ∠BAD = ∠ABE and ∠EPA = ∠DPB

(i) It is given that ∠EPA = ∠DPB

Now, add ∠DPE on both sides,

∠EPA +∠DPE = ∠DPB+∠DPE

This implies that angles DPA and EPB are equal i.e. ∠DPA = ∠EPB

Now, consider the triangles DAP and EBP.

∠DPA = ∠EPB

AP = BP (Since P is the mid-point of the line segment AB)

∠BAD = ∠ABE (As given in the question)

So, by ASA congruency , ΔDAP ≅ ΔEBP.

(ii) By the rule of CPCT, AD = BE.

8. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see Fig. 7.23). Show that:

(i) ΔAMC ≅ ΔBMD

(ii) ∠DBC is a right angle.

(iii) ΔDBC ≅ ΔACB

(iv) CM = ½ AB

It is given that M is the mid-point of the line segment AB, ∠C = 90°, and DM = CM

(i) Consider the triangles ΔAMC and ΔBMD:

AM = BM (Since M is the mid-point)

CM = DM (Given in the question)

∠CMA = ∠DMB (They are vertically opposite angles)

So, by SAS congruency criterion , ΔAMC ≅ ΔBMD.

(ii) ∠ACM = ∠BDM (by CPCT)

∴ AC || BD as alternate interior angles are equal.

Now, ∠ACB +∠DBC = 180° (Since they are co-interiors angles)

⇒ 90° +∠B = 180°

∴ ∠DBC = 90°

(iii) In ΔDBC and ΔACB,

BC = CB (Common side)

∠ACB = ∠DBC (They are right angles)

DB = AC (by CPCT)

So, ΔDBC ≅ ΔACB by SAS congruency .

(iv) DC = AB (Since ΔDBC ≅ ΔACB)

⇒ DM = CM = AM = BM (Since M the is mid-point)

So, DM + CM = BM+AM

Hence, CM + CM = AB

⇒ CM = (½) AB

Exercise: 7.2 (Page No: 123)

1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:

(i) OB = OC (ii) AO bisects ∠A

AB = AC and

the bisectors of ∠B and ∠C intersect each other at O

(i) Since ABC is an isosceles with AB = AC,

½ ∠B = ½ ∠C

⇒ ∠OBC = ∠OCB (Angle bisectors)

∴ OB = OC (Side opposite to the equal angles are equal.)

(ii) In ΔAOB and ΔAOC,

AB = AC (Given in the question)

AO = AO (Common arm)

OB = OC (As Proved Already)

So, ΔAOB ≅ ΔAOC by SSS congruence condition.

BAO = CAO (by CPCT)

Thus, AO bisects ∠A.

2. In ΔABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that ΔABC is an isosceles triangle in which AB = AC.

It is given that AD is the perpendicular bisector of BC

In ΔADB and ΔADC,

AD = AD (It is the Common arm)

∠ADB = ∠ADC

BD = CD (Since AD is the perpendicular bisector)

So, ΔADB ≅ ΔADC by SAS congruency criterion .

AB = AC (by CPCT)

3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.

(i) BE and CF are altitudes.

(ii) AC = AB

Triangles ΔAEB and ΔAFC are similar by AAS congruency since

∠A = ∠A (It is the common arm)

∠AEB = ∠AFC (They are right angles)

∴ ΔAEB ≅ ΔAFC and so, BE = CF (by CPCT).

4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that

(i) ΔABE ≅ ΔACF

(ii) AB = AC, i.e., ABC is an isosceles triangle.

It is given that BE = CF

(i) In ΔABE and ΔACF,

∠A = ∠A (It is the common angle)

BE = CF (Given in the question)

∴ ΔABE ≅ ΔACF by AAS congruency condition .

(ii) AB = AC by CPCT and so, ABC is an isosceles triangle.

5. ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ∠ABD = ∠ACD.

In the question, it is given that ABC and DBC are two isosceles triangles.

We will have to show that ∠ABD = ∠ACD

Triangles ΔABD and ΔACD are similar by SSS congruency since

AD = AD (It is the common arm)

AB = AC (Since ABC is an isosceles triangle)

BD = CD (Since BCD is an isosceles triangle)

So, ΔABD ≅ ΔACD.

∴ ∠ABD = ∠ACD by CPCT.

6. ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠BCD is a right angle.

It is given that AB = AC and AD = AB

We will have to now prove ∠BCD is a right angle.

Consider ΔABC,

AB = AC (It is given in the question)

Also, ∠ACB = ∠ABC (They are angles opposite to the equal sides and so, they are equal)

Now, consider ΔACD,

Also, ∠ADC = ∠ACD (They are angles opposite to the equal sides and so, they are equal)

∠CAB + ∠ACB + ∠ABC = 180°

So, ∠CAB + 2∠ACB = 180°

⇒ ∠CAB = 180° – 2∠ACB — (i)

Similarly, in ΔADC,

∠CAD = 180° – 2∠ACD — (ii)

∠CAB + ∠CAD = 180° (BD is a straight line.)

Adding (i) and (ii) we get,

∠CAB + ∠CAD = 180° – 2∠ACB+180° – 2∠ACD

⇒ 180° = 360° – 2∠ACB-2∠ACD

⇒ 2(∠ACB+∠ACD) = 180°

⇒ ∠BCD = 90°

7. ABC is a right-angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

In the question, it is given that

∠A = 90° and AB = AC

⇒ ∠B = ∠C (They are angles opposite to the equal sides and so, they are equal)

∠A+∠B+∠C = 180° (Since the sum of the interior angles of the triangle)

∴ 90° + 2∠B = 180°

⇒ 2∠B = 90°

So, ∠B = ∠C = 45°

8. Show that the angles of an equilateral triangle are 60° each.

Let ABC be an equilateral triangle as shown below:

Here, BC = AC = AB (Since the length of all sides is same)

⇒ ∠A = ∠B =∠C (Sides opposite to the equal angles are equal.)

Also, we know that

∠A+∠B+∠C = 180°

⇒ 3∠A = 180°

∴ ∠A = ∠B = ∠C = 60°

So, the angles of an equilateral triangle are always 60° each.

Exercise: 7.3 (Page No: 128)

1. ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that

(i) ΔABD ≅ ΔACD

(ii) ΔABP ≅ ΔACP

(iii) AP bisects ∠A as well as ∠D.

(iv) AP is the perpendicular bisector of BC.

In the above question, it is given that ΔABC and ΔDBC are two isosceles triangles.

(i) ΔABD and ΔACD are similar by SSS congruency because:

AB = AC (Since ΔABC is isosceles)

BD = CD (Since ΔDBC is isosceles)

∴ ΔABD ≅ ΔACD.

(ii) ΔABP and ΔACP are similar as:

AP = AP (It is the common side)

∠PAB = ∠PAC (by CPCT since ΔABD ≅ ΔACD)

So, ΔABP ≅ ΔACP by SAS congruency condition.

(iii) ∠PAB = ∠PAC by CPCT as ΔABD ≅ ΔACD.

AP bisects ∠A. — (i)

Also, ΔBPD and ΔCPD are similar by SSS congruency as

PD = PD (It is the common side)

BD = CD (Since ΔDBC is isosceles.)

BP = CP (by CPCT as ΔABP ≅ ΔACP)

So, ΔBPD ≅ ΔCPD.

Thus, ∠BDP = ∠CDP by CPCT. — (ii)

Now by comparing (i) and (ii) it can be said that AP bisects ∠A as well as ∠D.

(iv) ∠BPD = ∠CPD (by CPCT as ΔBPD ΔCPD)

and BP = CP — (i)

∠BPD +∠CPD = 180° (Since BC is a straight line.)

⇒ 2∠BPD = 180°

⇒ ∠BPD = 90° —(ii)

Now, from equations (i) and (ii), it can be said that

AP is the perpendicular bisector of BC.

2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

(i) AD bisects BC (ii) AD bisects ∠A.

It is given that AD is an altitude and AB = AC. The diagram is as follows:

(i) In ΔABD and ΔACD,

∠ADB = ∠ADC = 90°

AD = AD (Common arm)

∴ ΔABD ≅ ΔACD by RHS congruence condition.

Now, by the rule of CPCT,

So, AD bisects BC

(ii) Again, by the rule of CPCT, ∠BAD = ∠CAD

Hence, AD bisects ∠A.

3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see Fig. 7.40). Show that:

(i) ΔABM ≅ ΔPQN

(ii) ΔABC ≅ ΔPQR

Given parameters are:

BC = QR and

(i) ½ BC = BM and ½ QR = QN (Since AM and PN are medians)

Also, BC = QR

So, ½ BC = ½ QR

In ΔABM and ΔPQN,

AM = PN and AB = PQ (As given in the question)

BM = QN (Already proved)

∴ ΔABM ≅ ΔPQN by SSS congruency.

(ii) In ΔABC and ΔPQR,

AB = PQ and BC = QR (As given in the question)

∠ABC = ∠PQR (by CPCT)

So, ΔABC ≅ ΔPQR by SAS congruency.

4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

It is known that BE and CF are two equal altitudes.

Now, in ΔBEC and ΔCFB,

∠BEC = ∠CFB = 90° (Same Altitudes)

BE = CF (Common side)

So, ΔBEC ≅ ΔCFB by RHS congruence criterion.

Also, ∠C = ∠B (by CPCT)

Therefore, AB = AC as sides opposite to the equal angles is always equal.

5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

In the question, it is given that AB = AC

Now, ΔABP and ΔACP are similar by RHS congruency as

∠APB = ∠APC = 90° (AP is altitude)

AP = AP (Common side)

So, ΔABP ≅ ΔACP.

∴ ∠B = ∠C (by CPCT)

Exercise: 7.4 (Page No: 132)

1. Show that in a right-angled triangle, the hypotenuse is the longest side.

It is known that ABC is a triangle right angled at B.

We know that,

∠A +∠B+∠C = 180°

Now, if ∠B+∠C = 90° then ∠A has to be 90°.

Since A is the largest angle of the triangle, the side opposite to it must be the largest.

So, AB is the hypotenuse which will be the largest side of the above right-angled triangle i.e. ΔABC.

2. In Fig. 7.48, sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

It is given that ∠PBC < ∠QCB

We know that ∠ABC + ∠PBC = 180°

So, ∠ABC = 180°-∠PBC

∠ACB +∠QCB = 180°

Therefore ∠ACB = 180° -∠QCB

Now, since ∠PBC < ∠QCB,

∴ ∠ABC > ∠ACB

Hence, AC > AB as sides opposite to the larger angle is always larger.

3. In Fig. 7.49, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

In the question, it is mentioned that angles B and angle C is smaller than angles A and D respectively i.e. ∠B < ∠A and ∠C < ∠D.

Since the side opposite to the smaller angle is always smaller

AO < BO — (i)

And OD < OC —(ii)

By adding equation (i) and equation (ii) we get

AO+OD < BO + OC

So, AD < BC

4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50).

Show that ∠A > ∠C and ∠B > ∠D.

In ΔABD, we see that

AB < AD < BD

So, ∠ADB < ∠ABD — (i) (Since angle opposite to longer side is always larger)

Now, in ΔBCD,

BC < DC < BD

Hence, it can be concluded that

∠BDC < ∠CBD — (ii)

Now, by adding equation (i) and equation (ii) we get,

∠ADB + ∠BDC < ∠ABD + ∠CBD

∠ADC < ∠ABC

Similarly, In triangle ABC,

∠ACB < ∠BAC — (iii) (Since the angle opposite to the longer side is always larger)

Now, In ΔADC,

∠DCA < ∠DAC — (iv)

By adding equation (iii) and equation (iv) we get,

∠ACB + ∠DCA < ∠BAC+∠DAC

⇒ ∠BCD < ∠BAD

∴ ∠A > ∠C

5. In Fig 7.51, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR > ∠PSQ.

It is given that PR > PQ and PS bisects ∠QPR

Now we will have to prove that angle PSR is smaller than PSQ i.e. ∠PSR > ∠PSQ

∠QPS = ∠RPS — (ii) (As PS bisects ∠QPR)

∠PQR > ∠PRQ — (i) (Since PR > PQ as angle opposite to the larger side is always larger)

∠PSR = ∠PQR + ∠QPS — (iii) (Since the exterior angle of a triangle equals to the sum of opposite interior angles)

∠PSQ = ∠PRQ + ∠RPS — (iv) (As the exterior angle of a triangle equals to the sum of opposite interior angles)

By adding (i) and (ii)

∠PQR +∠QPS > ∠PRQ +∠RPS

Thus, from (i), (ii), (iii) and (iv), we get

∠PSR > ∠PSQ

6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

First, let “ l ” be a line segment and “B” be a point lying on it. A line AB perpendicular to l is now drawn. Also, let C be any other point on l . The diagram will be as follows:

In ΔABC, ∠B = 90°

Now, we know that

∴ ∠A +∠C = 90°

Hence, ∠C must be an acute angle which implies ∠C < ∠B

So, AB < AC (As the side opposite to the larger angle is always larger)

Summary of NCERT Solutions for Class 9 Maths Chapter 7 Triangles

Triangle is a part of Geometry. However, the complete Geometry of Class 9 constitutes a weightage of 22 marks out of 80 marks. Have a look at the types of questions that are expected from Geometry in the annual exam of class 9 Maths paper.

Students should practise all the questions from the exercise to score high marks in the Class 9 Maths paper. The step-by-step solution to all the exercises of NCERT Class 9 Maths Chapter 7 is provided below:

In previous classes, students must have used the properties of triangles for solving the questions, but now in the  NCERT Textbook of Class 9 Maths Chapter 7, students will learn how to prove these properties. For solving the questions related to this topic, it is very necessary that students should know the congruence rule. First, go through the theory and then look at the solved examples that are already there in the book. After that, start solving the exercise problems.

Students can also have a look at the NCERT Solutions of Class 9 for Science subject to know the answers of all the chapters with a detailed explanation.

Important Concepts Learned in NCERT Class 9 Maths Chapter 7 – Triangles

The aim of including this Triangle Chapter in the Class 9 Maths NCERT textbook is to make students know the following concepts:

• Congruence of triangles.
• Criteria for congruence of triangles (SAS congruence rule, ASA congruence rule, SSS congruence rule, RHS congruence rule).
• Properties of triangles.
• Inequalities of triangles.

We hope this information on “ NCERT Solution Class 9 Maths Chapter 7 ” is useful for students. Click on NCERT Solutions to get the solved answers from the NCERT book for all the classes. Stay tuned for further updates on CBSE and other competitive exams. To access interactive Maths and Science Videos, download BYJU’S App and subscribe to our YouTube Channel.

Triangles is a very important chapter as the fundamental concepts introduced here are carried forward to higher levels of education. For this purpose, after completing the questions from the NCERT Textbook, students are recommended to solve the other textbooks which are prescribed by the CBSE Board.

• RD Sharma Solutions for Class 9 Maths Chapter 9 Triangle and its Angles

Disclaimer:

Dropped Topics –  7.6 Inequalities in triangles.

Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 7

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