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Case Study Questions for Class 9 Maths Chapter 7 Triangles

• Last modified on: 10 months ago
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Here we are providing case study questions for Class 9 Maths Chapter 7 Triangles. Students are suggested to solve the questions by themselves first and then check the answers. This will help students to check their grasp on this particular chapter Triangles.

Case Study Questions:

Questions 1:

In a park, there are two triangular flower beds. Flower bed ABC has sides AB = 8 cm, BC = 6 cm, and CA = 10 cm. Flower bed PQR has sides PQ = 8 cm, QR = 10 cm, and RP = 6 cm.

(i) Using the given information, can we conclude that flower bed ABC is congruent to flower bed PQR? Why or why not? a) Yes, because all three sides of ABC are equal to the corresponding sides of PQR. b) Yes, because all three angles of ABC are equal to the corresponding angles of PQR. c) No, because the sides and angles of ABC and PQR are not equal. d) Yes, because ABC and PQR are both triangular flower beds.

(ii) If angle A = 40° and angle B = 60° in flower bed ABC, what is the measure of angle C? a) 40° b) 60° c) 80° d) 100°

(iii) In flower bed ABC, a new side AD is added such that AD = 8 cm. Now, can we conclude that triangle ABD is congruent to triangle PQR? Why or why not? a) Yes, because all three sides of ABD are equal to the corresponding sides of PQR. b) Yes, because all three angles of ABD are equal to the corresponding angles of PQR. c) No, because the sides and angles of ABD and PQR are not equal. d) Yes, because both triangles have a side of 8 cm.

(iv) If angle P = 50° and angle Q = 70° in flower bed PQR, what is the measure of angle R? a) 50° b) 70° c) 90° d) 110°

(v) Suppose flower bed ABC is shifted to a new location within the park without changing its shape or size. In this new location, is flower bed ABC congruent to its original position? Why or why not? a) Yes, because the shape and size of ABC remain the same. b) No, because the location of ABC has changed. c) Yes, because the park remains the same. d) No, because the other flower bed PQR is not shifted.

(i) c) No, because the sides and angles of ABC and PQR are not equal.

(ii) c) 80°

(iii) b) Yes, because all three angles of ABD are equal to the corresponding angles of PQR.

(iv) d) 110°

(v) a) Yes, because the shape and size of ABC remain the same.

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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.

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

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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
• Case Based Questions: Linear Equations -2

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

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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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14 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

MATHS PAAGAL HAI

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

Where is search ? bar

maths is love

Can I have more questions without downloading the app.

I love math

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Case Study Questions for Class 9 Maths

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Are you preparing for your Class 9 Maths board exams and looking for an effective study resource? Well, you’re in luck! In this article, we will provide you with a collection of Case Study Questions for Class 9 Maths specifically designed to help you excel in your exams. These questions are carefully curated to cover various mathematical concepts and problem-solving techniques. So, let’s dive in and explore these valuable resources that will enhance your preparation and boost your confidence.

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

CBSE Class 9 Maths Board Exam will have a set of questions based on case studies in the form of MCQs. The CBSE Class 9 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

If you want to want to prepare all the tough, tricky & difficult questions for your upcoming exams, this is where you should hang out.  CBSE Case Study Questions for Class 9  will provide you with detailed, latest, comprehensive & confidence-inspiring solutions to the maximum number of Case Study Questions covering all the topics from your  NCERT Text Books !

Table of Contents

CBSE Class 9th – MATHS: Chapterwise Case Study Question & Solution

Case study questions are a form of examination where students are presented with real-life scenarios that require the application of mathematical concepts to arrive at a solution. These questions are designed to assess students’ problem-solving abilities, critical thinking skills, and understanding of mathematical concepts in practical contexts.

Chapterwise Case Study Questions for Class 9 Maths

Case study questions play a crucial role in the field of mathematics education. They provide students with an opportunity to apply theoretical knowledge to real-world situations, thereby enhancing their comprehension of mathematical concepts. By engaging with case study questions, students develop the ability to analyze complex problems, make connections between different mathematical concepts, and formulate effective problem-solving strategies.

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

The above  Case studies for Class 9 Mathematics will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 9 Maths Case Studies have been developed by experienced teachers of schools.studyrate.in for benefit of Class 10 students.

• Class 9 Science Case Study Questions
• Class 9 Social Science Case Study Questions

How to Approach Case Study Questions

When tackling case study questions, it is essential to adopt a systematic approach. Here are some steps to help you approach and solve these types of questions effectively:

• Read the case study carefully: Understand the given scenario and identify the key information.
• Identify the mathematical concepts involved: Determine the relevant mathematical concepts and formulas applicable to the problem.
• Formulate a plan: Devise a plan or strategy to solve the problem based on the given information and mathematical concepts.
• Solve the problem step by step: Apply the chosen approach and perform calculations or manipulations to arrive at the solution.
• Verify and interpret the results: Ensure the solution aligns with the initial problem and interpret the findings in the context of the case study.

Tips for Solving Case Study Questions

Here are some valuable tips to help you effectively solve case study questions:

• Read the question thoroughly and underline or highlight important information.
• Break down the problem into smaller, manageable parts.
• Visualize the problem using diagrams or charts if applicable.
• Use appropriate mathematical formulas and concepts to solve the problem.
• Show all the steps of your calculations to ensure clarity.
• Check your final answer and review the solution for accuracy and relevance to the case study.

Benefits of Practicing Case Study Questions

Practicing case study questions offers several benefits that can significantly contribute to your mathematical proficiency:

• Enhances critical thinking skills
• Improves problem-solving abilities
• Deepens understanding of mathematical concepts
• Develops analytical reasoning
• Prepares you for real-life applications of mathematics
• Boosts confidence in approaching complex mathematical problems

Case study questions offer a unique opportunity to apply mathematical knowledge in practical scenarios. By practicing these questions, you can enhance your problem-solving abilities, develop a deeper understanding of mathematical concepts, and boost your confidence for the Class 9 Maths board exams. Remember to approach each question systematically, apply the relevant concepts, and review your solutions for accuracy. Access the PDF resource provided to access a wealth of case study questions and further elevate your preparation.

Q1: Can case study questions help me score better in my Class 9 Maths exams?

Yes, practicing case study questions can significantly improve your problem-solving skills and boost your performance in exams. These questions offer a practical approach to understanding mathematical concepts and their real-life applications.

Q2: Are the case study questions in the PDF resource relevant to the Class 9 Maths syllabus?

Absolutely! The PDF resource contains case study questions that align with the Class 9 Maths syllabus. They cover various topics and concepts included in the curriculum, ensuring comprehensive preparation.

Q3: Are the solutions provided for the case study questions in the PDF resource?

Yes, the PDF resource includes solutions for each case study question. You can refer to these solutions to validate your answers and gain a better understanding of the problem-solving process.

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Unit 10: Triangles

Triangles part 1.

• Intro to medians of a triangle (Opens a modal)
• Angles in a triangle sum to 180° proof (Opens a modal)
• Triangle angle challenge problem 2 (Opens a modal)
• Identify medians and altitudes Get 3 of 4 questions to level up!
• Triangle exterior angle property problems Get 3 of 4 questions to level up!
• Finding angle measures using triangles Get 5 of 7 questions to level up!

Triangles part 2

• Finding angles in isosceles triangles (example 2) (Opens a modal)
• Triangle inequality theorem (Opens a modal)
• Find angles in isosceles triangles Get 3 of 4 questions to level up!
• Triangle side length rules Get 3 of 4 questions to level up!

Pythagorean theorem

• Intro to the Pythagorean theorem (Opens a modal)
• Pythagorean theorem example (Opens a modal)
• Use Pythagorean theorem to find right triangle side lengths Get 5 of 7 questions to level up!
• Right triangle side lengths Get 3 of 4 questions to level up!
• Use area of squares to visualize Pythagorean theorem Get 3 of 4 questions to level up!

Congruence part 1

• Corresponding parts of congruent triangles are congruent (Opens a modal)
• Triangle congruence postulates/criteria (Opens a modal)
• Corresponding parts of congruent triangles Get 3 of 4 questions to level up!

Congruence part 2

• Determining congruent triangles (Opens a modal)
• Why SSA isn't a congruence postulate/criterion (Opens a modal)
• Determine congruent triangles Get 5 of 7 questions to level up!
• Find angles in congruent triangles Get 3 of 4 questions to level up!

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Chapter 7 Class 9 Triangles

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NCERT Solutions of Chapter 7 Class 9 Triangles is available free at teachoo. Solutions to all exercise questions, examples and theorems is provided with video of each and every question.

Let's see what we will learn in this chapter. The topics in the chapter are - 

• What is congruency of figures
• Naming of triangles when two triangles are congruent
• What is CPCT - Corresponding Parts of Congruent Triangles
• SAS (Side Angle Side) Congruence Rule (this is without proof)
• ASA (Angle Side Angle) congruence rule with proof (Theorem 7.1)
• Is ASA and AAS congruency the same ?
• Angles opposite to equal sides is equal (Isosceles Triangle Property)
• SSS (Side Side Side) congruence rule with proof (Theorem 7.4)
• RHS (Right angle Hypotenuse Side) congruence rule with proof (Theorem 7.5)
• Angle opposite to longer side is larger, and Side opposite to larger angle is longer
• Triangle Inequality - Sum of two sides of a triangle is always greater than the third side

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CBSE Case Study Questions Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles PDF Download

CBSE Case Study Questions Class 9 Maths Chapter 9  are very important to solve for your exam. Class 9 Maths Chapter 9 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 9  Areas of Parallelograms and Triangles

Case Study Questions Class 9 Maths Chapter 9

Case Study 1. A group of students is learning about the areas of parallelograms and triangles. They came across the following scenario:

Rahul and Sana participated in a gardening project where they needed to create rectangular plots in their backyard. They made the following observations:

• Rahul’s rectangular plot has a length of 10 meters and a height of 5 meters.
• Sana’s rectangular plot has a length of 12 meters and a height of 6 meters.
• Both plots have the same base and height.

Based on this information, the students were asked to analyze the areas of the parallelograms formed by the rectangular plots. Let’s see if you can answer the questions correctly:

Q1. The area of Rahul’s parallelogram is: (a) 25 square meters (b) 50 square meters (c) 60 square meters (d) 120 square meters

Answer: (b) 50 square meters

Q2. The area of Sana’s parallelogram is: (a) 25 square meters (b) 50 square meters (c) 60 square meters (d) 120 square meters

Answer: (c) 60 square meters

Q3. The type of parallelogram formed by Rahul’s plot is: (a) Rectangle (b) Square (c) Rhombus (d) None of the above

Answer: (a) Rectangle

Q4. The type of parallelogram formed by Sana’s plot is: (a) Rectangle (b) Square (c) Rhombus (d) None of the above

Answer: (c) Rhombus

Q5. The area of the triangle formed by the diagonal of Rahul’s plot is: (a) 25 square meters (b) 30 square meters (c) 50 square meters (d) It cannot be determined

Answer: (a) 25 square meters

Case Study 2. A group of students is exploring the concept of the area of parallelograms and triangles. They encountered the following scenario:

Riya and Ravi went on a school trip to a park. They noticed a triangular garden in the shape of an equilateral triangle. They made the following observations:

• Each side of the triangular garden measures 8 meters.
• The height of the triangular garden is 6 meters.
• The garden is divided into two equal parts by a diagonal.

Based on this information, the students were asked to analyze the areas of the parallelograms and triangles formed by the garden. Let’s see if you can answer the questions correctly:

MCQ Questions:

Q1. The area of the triangular garden is: (a) 12 square meters (b) 24 square meters (c) 32 square meters (d) 48 square meters

Answer: (b) 24 square meters

Q2. The area of each triangular part formed by the diagonal is: (a) 12 square meters (b) 16 square meters (c) 24 square meters (d) 48 square meters

Answer: (c) 24 square meters

Q3. The type of parallelogram formed by the triangular garden is: (a) Rectangle (b) Square (c) Rhombus (d) None of the above

Q4. The length of the diagonal of the parallelogram formed by the triangular garden is: (a) 8 meters (b) 12 meters (c) 16 meters (d) 24 meters

Answer: (b) 12 meters

Q5. The area of the parallelogram formed by the triangular garden is: (a) 24 square meters (b) 32 square meters (c) 48 square meters (d) 64 square meters

Answer: (c) 48 square meters

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• CBSE Maths Important Questions
• Class 9 Maths
• Chapter 7: Triangles

Important Questions CBSE Class 9 Maths Chapter 7- Triangles

Important Questions for CBSE Class 9 Chapter 7 -Triangles are provided here by our experts, along with their solutions. These questions are extracted from the NCERT book as per CBSE syllabus . Students who are preparing for standard 9 Maths final exam (2022 – 2023)  should practise these questions to score excellent marks.

Solve extra questions along with important questions for class 9 Maths to have a better practice and brief revision before the examination. The questions in Chapter 7, triangles, are majorly based on congruency or similarity of triangles. Solving these questions will give students an idea of the types of questions asked in the exam.

Also Check:

• Important 2 Marks Questions for CBSE 9th Maths
• Important 3 Marks Questions for CBSE 9th Maths
• Important 4 Marks Questions for CBSE 9th Maths

Important Questions & Solutions For CBSE Class 9 Chapter 7 (Triangles)

Q.1: ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that

(i) ΔABD ≅ ΔBAC

(ii) BD = AC

(iii) ∠ABD = ∠BAC.

As per given in the question,

∠DAB = ∠CBA and AD = BC.

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

AB = BA (common arm)

∠DAB = ∠CBA and AD = BC (given)

So, triangles ABD and BAC are similar

i.e. ΔABD ≅ ΔBAC. (Hence proved).

(ii) As it is already proved,

ΔABD ≅ ΔBAC

BD = AC (by CPCT)

(iii) Since ΔABD ≅ ΔBAC

So, the angles,

∠ABD = ∠BAC (by CPCT).

Q.2: AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.

Given, AD and BC are two equal perpendiculars to AB.

To prove: CD is the bisector of AB

Triangles ΔAOD and ΔBOC are similar by AAS congruency

(i) ∠A = ∠B (perpendicular angles)

(ii) AD = BC (given)

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

∴ ΔAOD ≅ ΔBOC.

So, AO = OB ( by CPCT).

Thus, CD bisects AB (Hence proved).

Q.3: 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. 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 (both are right angles)

AB = AB (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, we can say point B is equidistant from the arms of ∠A.

Q.4: 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. Show that

(i) ΔDAP ≅ ΔEBP

(ii) AD = BE

Given, P is the mid-point of line segment AB.

Also, ∠BAD = ∠ABE and ∠EPA = ∠DPB

(i) Given, ∠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 (given)

So, by ASA congruency criterion,

ΔDAP ≅ ΔEBP.

(ii) By the rule of CPCT,

Q.5: 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 the figure). Show that:

(i) ΔAMC ≅ ΔBMD

(ii) ∠DBC is a right angle.

(iii) ΔDBC ≅ ΔACB

(iv) CM = 1/2 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)

∠CMA = ∠DMB (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 (Both 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

Q.6: ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. 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 (common arm)

∠AEB = ∠AFC (both are right angles)

AB = AC (Given)

∴ ΔAEB ≅ ΔAFC

and BE = CF (by CPCT).

Q.7: ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that ∠BCD is a right angle.

Given, AB = AC and AD = AB

To prove: ∠BCD is a right angle.

Consider ΔABC,

Also, ∠ACB = ∠ABC (Angles opposite to equal sides)

Now, consider ΔACD,

Also, ∠ADC = ∠ACD (Angles opposite to equal sides)

∠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°

Q.8: 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 the figure). 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:

AD = AD (It is the common arm)

AB = AC (Since ΔABC is isosceles)

BD = CD (Since ΔDBC is isosceles)

∴ ΔABD ≅ ΔACD.

(ii) ΔABP and ΔACP are similar as:

AP = AP (common side)

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

So, ΔABP ≅ ΔACP by SAS congruency.

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

AP bisects ∠A. ………… (1)

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. ……………. (2)

Now by comparing equation (1) and (2) it can be said that AP bisects ∠A as well as ∠D.

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

and BP = CP — (1)

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

⇒ 2∠BPD = 180°

⇒ ∠BPD = 90° —(2)

Now, from equations (1) and (2), it can be said that

AP is the perpendicular bisector of BC.

Q.9: 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 the figure). Show that:

(i) ΔABM ≅ ΔPQN

(ii) ΔABC ≅ ΔPQR

BC = QR and

(i) 1/2 BC = BM and 1/2QR = QN (Since AM and PN are medians)

Also, BC = QR

So, 1/2 BC = 1/2QR

In ΔABM and ΔPQN,

AM = PN and AB = PQ (Given)

BM = QN (Already proved)

∴ ΔABM ≅ ΔPQN by SSS congruency.

(ii) In ΔABC and ΔPQR,

AB = PQ and BC = QR (Given)

∠ABC = ∠PQR (by CPCT)

So, ΔABC ≅ ΔPQR by SAS congruency.

Q.10: In the Figure, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR > ∠PSQ.

Given, PR > PQ and PS bisects ∠QPR

To prove: ∠PSR > ∠PSQ

∠QPS = ∠RPS — (1) (PS bisects ∠QPR)

∠PQR > ∠PRQ — (2) (Since PR > PQ as angle opposite to the larger side is always larger)

∠PSR = ∠PQR + ∠QPS — (3) (Since the exterior angle of a triangle equals the sum of opposite interior angles)

∠PSQ = ∠PRQ + ∠RPS — (4) (As the exterior angle of a triangle equals to the sum of opposite interior angles)

By adding (1) and (2)

∠PQR + ∠QPS > ∠PRQ + ∠RPS

Now, from (1), (2), (3) and (4), we get

∠PSR > ∠PSQ.

Video Lesson on Congruent Triangles

Extra Questions for Class 9 Maths Triangles Chapter 7

• Two adjacent angles on a straight line are in a ratio 5:4. Find the measure of each one of these angles. (Answer: 100 ° and 80 °degree).
• Prove that bisectors of two adjacent supplementary angle include a right angle.
• If two straight lines are perpendicular to the same line, prove that they are parallel to each other.
• The angles of the triangle are in the ratio 2:3:7. Find the measure of each angle of the triangle. (Answer: 30°, 45° and 105°)
• In triangle ABC , ∠A – ∠B = 33 degree and ∠B – ∠C = 180 degree.Find the measure of each angle of the triangle. ( Answer: ∠A = 88° and ∠B = 55° and ∠C = 37°).
• Of the three angles of the triangle, one is twice the smallest and another one is thrice the smallest. Find the angles.  (Answer: 60°, 90°, 30°)
• If each of a triangle is less than the sum of the other two, show that the triangle is acute-angled.
• The sum of two angles of a triangle is 116° and their difference is 24. Find the measure of each angle of the triangle. (Answer: 46°, 70 °, 64°) .
• Two straight lines AB and CD cut each other at O. if angle BOD = 63° then find angle BOC. (Answer: 117°) .
• Two complementary angles are such that twice the measure of the one is equal to three times the measure of the other. The larger of the two measures is. (Answer: 54°) .
• Show that of all line segments drawn from a given point, not on it, the perpendicular line segment is the shortest.
•  If the diagonals of the rectangle ABCD intersect at O and  ∠COD = 78°, then ∠OAB is:

        d) 110 o

     13.  The measure of an angle is twice the measure of supplementary angle. What is the measure of angles?

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles

• NCERT Solutions for Class 10 Maths Chapter 6 Triangles
• NCERT Solutions for Class 9 Maths Chapter 14 Statistics
• Class 9 NCERT Solutions- Chapter 7 Triangles - Exercise 7.5
• Class 9 NCERT Solutions- Chapter 7 Triangles - Exercise 7.4
• Class 9 NCERT Solutions- Chapter 7 Triangles - Exercise 7.1
• Class 9 NCERT Solutions- Chapter 7 Triangles - Exercise 7.3
• Class 9 NCERT Solutions- Chapter 7 Triangles - Exercise 7.2
• NCERT Solutions for Class 9 Maths: Chapter Wise PDF 2024
• NCERT Solutions for Class 10 Maths: Chapter Wise PDF
• NCERT Solutions for Class 11 Maths - Chapter Wise PDF
• NCERT Solutions for Class 9 Maths Chapter 1 Number Systems
• NCERT Solutions for Class 10 Maths Chapter 14 Statistics
• NCERT Solutions for Class 9 Maths Chapter 10 Circles
• Class 10 NCERT Solutions- Chapter 6 Triangles - Exercise 6.3
• Class 10 NCERT Solutions- Chapter 6 Triangles - Exercise 6.1
• Class 10 NCERT Solutions- Chapter 6 Triangles - Exercise 6.4
• Class 10 NCERT Solutions- Chapter 6 Triangles - Exercise 6.2
• Class 10 NCERT Solutions- Chapter 6 Triangles - Exercise 6.6
• NCERT Solutions for Class 8 Maths: Chapter Wise Solution PDF

NCERT Solutions for Class 9 Maths Chapter 7 Triangles – This article contains detailed NCERT Solutions for Class 9 Maths Chapter 7 Triangles curated by the team of subject matter experts at GFG, to help students understand how to solve the NCERT problems easily.

Chapter 7 Triangles of NCERT for Class 9 Maths helps students understand the basic concepts of congruence of triangles and the rules of congruence. It is also helpful in understanding a few more properties of triangles and the inequalities in a triangle.

NCERT Chapter 7 Triangles in Class 9 covers topics such as :

• Acute Angled Triangle
• Right Angled Triangle
• Obtuse Angled Triangle
• Equilateral Triangle
• Isosceles Triangle
• Scalene Triangle
• Congruence of Triangles
• Triangle Inequalities

This article provides solutions to all the problems asked in Class 9 Maths Chapter 7 Triangles of your NCERT textbook in a step-by-step manner as per the latest CBSE Syllabus 2023-24 and guidelines. Solutions to all the exercises in the NCERT Class 9 Maths Chapter 7 Triangles are regularly revised to check errors and updated according to the latest CBSE Syllabus 2023-24 and guidelines.

NCERT Solutions for Class 9 Maths Chapter 7 Triangles: Exercise 7.1

Question 1. in quadrilateral acbd ac = ad and ab bisects ∠ a (see fig. 7.16). show that ∆ abc ≅ ∆ abd what can you say about bc and bd.

Given that AC and AD are equal  i.e. AC = AD and the line AB bisects ∠A. Considering the two triangles ΔABC and ΔABD, Where,  AC = AD { As given}………………………………………… (i)  ∠CAB = ∠DAB ( As AB  bisects of ∠A)……………. (ii)  AB { Common side of both the triangle} …….. …(iii) From above three equation both the triangle satisfies “SAS” congruency criterion So, ΔABC ≅ ΔABD. Also,  BC and BD will be of equal lengths as they are corresponding parts of congruent triangles(CPCT).  So BC = BD. 

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

(i) ∆ ABD ≅ ∆ BAC

(ii) BD = AC

(iii) ∠ ABD = ∠ BAC.

(i) Given that AD = BC,  And ∠ DAB = ∠ CBA.  Considering two triangles ΔABD and ΔBAC.  Where,  AD = BC { As given }………………………………………….. (i)  ∠ DAB = ∠ CBA { As given also}……………………….. (ii)  AB  {Common side of both the triangle)…………. (iii)  From above three equation two triangles ABD and BAC satisfies “SAS” congruency criterion So, ΔABD ≅ ΔBAC (ii) Also,  BD and AC  will be equal as they are corresponding parts of congruent triangles(CPCT).  So BD = AC (iii) Similarly, ∠ABD and ∠BAC will be equal as they are corresponding parts of congruent triangles(CPCT).  So,  ∠ABD = ∠BAC. 

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

Given that AD and BC are two equal perpendiculars to a line segment AB Considering two triangles ΔAOD and ΔBOC Where,  ∠ AOD = ∠ BOC {Vertically opposite angles}………………. (i)  ∠ OAD = ∠ OBC {Given that they are perpendiculars}…. (ii)  AD = BC {As given}………………………………………………… (iii)  From above three equation both the triangle satisfies “AAS” congruency criterion So, ΔAOD ≅ ΔBOC AO and BO will be equal as they are corresponding parts of congruent triangles(CPCT).  So, AO = BO Hence, CD bisects AB at O. 

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

Given that l and m are two parallel lines p and q are another pair of parallel lines Considering two triangles ΔABC and ΔCDA Where,  ∠ BCA = ∠DAC  {Alternate interior angles}…. (i)  ∠ BAC = ∠ DCA {Alternate interior angles}…. (ii)   AC  {Common side of two triangles}………….(iii) From above three equation both the triangle satisfies “ASA” congruency criterion So, ΔABC ≅ ΔCDA

Question 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) ∆ APB ≅ ∆ AQB

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

Given that, Line l is the bisector of an angle ∠ A and B BP and BQ are perpendiculars from angle A.  Considering two triangles ΔAPB and ΔAQB Where,  ∠ APB = ∠ AQB { Two right angles as given }…… (i)  ∠BAP = ∠BAQ (As line l  bisects  angle A }……… (ii)  AB  { Common sides of both the triangle }……… (iii)  From above three equation both the triangle satisfies “AAS” congruency criterion So, ΔAPB≅ ΔAQB. (ii) Also we can say BP and BQ are equal as they are corresponding parts of congruent triangles(CPCT). So, BP = BQ

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

Given that AC = AE, AB = AD  And ∠BAD = ∠EAC As given that ∠BAD = ∠EAC Adding ∠DAC on both the sides  We get, ∠BAD + ∠DAC = ∠EAC + ∠ DAC  ∠BAC = ∠EAD Considering two triangles ΔABC and ΔADE  Where,   AC = AE  { As given }…………………… (i)  ∠BAC = ∠EAD { Hence proven }…….. (ii)  AB = AD {As also given }……………….. (iii)  From above three equation both the triangle satisfies “SAS” congruency criterion So, ΔABC ≅ ΔADE (ii) Also we can say BC and  DE are equal as they are corresponding parts of congruent triangles(CPCT). So that BC = DE.

Question 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) ∆ DAP ≅ ∆ EBP

(ii) AD = BE

Given that P is the mid-point of line  AB, So AP = BP Also, ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB Now adding ∠DPE on both the sides of two equal angle ∠ EPA = ∠ DPB ∠ EPA + ∠ DPE = ∠ DPB + ∠DPE Which implies that two angles ∠ DPA = ∠ EPB Considering two triangles ∆ DAP and ∆ EBP ∠ DPA = ∠ EPB { Hence proven }…… (i)  AP = BP { Hence Given }……………… (ii)  ∠ BAD = ∠ ABE { As given }…………..(iii) From above three equation both the triangle satisfies “ASA” congruency criterion So, ΔDAP ≅ ΔEBP (ii) Also we can say AD and BE are equal as they are corresponding parts of congruent triangles(CPCT). So that, AD = BE

Question 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) ∆ AMC ≅ ∆ BMD

(ii) ∠ DBC is a right angle.

(iii) ∆ DBC ≅ ∆ ACB

(iv) CM =  1/2 AB

Given that M is the mid-point of  AB  So AM = BM ∠ ACB = 90° and DM = CM (i) Considering two triangles ΔAMC and ΔBMD: AM = BM { As given }………………………………………. (i)  ∠ CMA = ∠ DMB { Vertically opposite angles }…. (ii)  CM = DM { As given also}……………………………….. (iii)  From above three equation both the triangle satisfies “SAS” congruency criterion So, ΔAMC ≅ ΔBMD (ii) From above congruency we can say  ∠ ACD = ∠ BDC  Also alternate interior angles of two parallel lines AC and DB.  Since sum of two co-interiors angles results to 180°. So, ∠ ACB + ∠ DBC = 180° ∠ DBC = 180° – ∠ ACB ∠ DBC = 90° { As ∠ACB =90° } (iii) In ΔDBC and ΔACB, BC  { Common side of both the triangle }……. (i)  ∠ ACB = ∠ DBC { As both are right angles }….(ii)  DB = AC (by CPCT)………………………………….. (iii)  From above three equation both the triangle satisfies “SAS” congruency criterion So, ΔDBC ≅ ΔACB (iv) As M is the mid point so we can say  DM = CM = AM = BM  Also we can say that AB = CD ( By CPCT )  As M is the mid point of CD we can write CM + DM = AB Hence, CM + CM = AB  (As DM = CM )  Hence, CM = (½) AB

NCERT Solutions for Class 9 Maths Chapter 7 Triangles : Exercise 7.2

Question 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.

Given: (i) An isosceles ∆ABC in which AB=AC             (ii) bisects of ∠B and ∠C each other at O. Show: (i) OB=OC           (ii) AO bisects ∠A (∠1=∠2) (i) In ∆ABC, AB = AC ∠B =∠C                          [angles opposite to equal sides are equal] 1/2 ∠B = 1/2∠C  ∠OBC=∠OCB ∴OB = OC                      [sides opposite equal ∠ are equal] (ii) In ∆AOB and ADC  AB = AC                         [given side]  1/2 ∠B = 1/2∠C  ∠ABO = ∠ACO               [Angle] BO = OC                        [proved above side] ∴∆AOB ≅ AOC  Thus ∠1 = ∠2 Therefore, AO bisects ∠A

Question 2. In Δ ABC, AD is the perpendicular bisector of BC (see fig.). Show that Δ ABC is an isosceles triangle in which AB = AC.

Given: AD is ⊥ bisector of BC Show: AB=BC In ∆ABD and ∆ACD BD=DC                    [AD is ⊥ bisector side]  ∠ADB=∠ADC           [Each 90° angle]  AD=AD                    [common side]  ∴∆ABD≅∆ACD        [S.A.S] AB=AC                     [C.P.C.T]

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

Given: AB=AC,BE and CF are altitudes  Show: BE=CF In ∆AEB and ∆AFC, ∠E=∠F                     [Each 90° angle]  ∠A=∠A                    [common angle]  AB=AC                    [given side]  ∴∆AEB≅∆AFC        [A.A.S] BE=CF                    [C.P.C.T]

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

(i) δ abe ≅ δ acf, (ii) ab = ac, i.e., abc is an isosceles triangle..

Given: Altitudes BE and CF to sides AC and AB are equal Show: (i) ΔABE ≅ ΔACF              (ii) AB = AC (i) In ∆ABF and ∆ACF, ∠E=∠F                    [Each 90° angle]  ∠A=∠A                   [common angle]  AB=AC                   [given] S ∴∆AEB≅∆AFC       [A.A.S] (ii) AB=AC             [C.P.C.T]

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

Given: AB=AC,BD=DC Show: ∠ABD = ∠ACD In ∆ABD, AD=AC ∴∠1=∠2                  [angle opposite to equal sides are equal] [1] In ∆BDC, BD=DC ∴∠3=∠4                  [angle opposite to equal sides are equal] [2] Adding 1 and 2 ∠1+∠2= ∠2+∠4 ∠ABD=∠ACD

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

In ∆ABC, AB=AC ∠ACB=∠ABC           [1] In ∆ACD, AC=AD ∠ACD=∠ADC          [2] Adding 1 and 2 ∠ACB+∠ACD=∠ABC+∠ADC  ∠BCD=∠ABC+∠BDC Adding ∠BCD on both side ∠BCD+∠BCD=∠ABC+∠BDC+∠BCD 2∠BCD=180° ∠BCD=(180°)/2=90°

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

Find: ∠B=? and ∠C? In ∆ABC, AB=AC ∴∠B=∠C                           [angle opposite to equal side are equal] ∠A+∠B+∠C=180°            [angle sum property of triangle] 90°+∠B+∠B=180° 2∠B=180°-90° ∠B=(90°)/2=45° Therefore, ∠B=45° and ∠C =45°

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

Given: Let ∆ABC is an equilateral ∆  Show: ∠A=∠B=∠C=60° In ∆ABC, AB=AC ∠B=∠C                    [1] Also AC=BC ∠B=∠A                    [2] From 1,2 and 2 ∠A=∠B=∠C In ∆ABC, ∠A+∠B+∠C=180°      [angle sum property of triangle] ∠A+∠A+∠A=180° 3∠A=180° ∠A=(180°)/3=60° ∠A=60°  ∴∠B=60° and ∠C=60°

NCERT Solutions for Class 9 Maths Chapter 7 Triangles: Exercise 7.3

Question 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 figure). 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.

Given: ∆ABC and ∆DCB are isosceles ∆on the same base BC. To show:   ΔABD ≅ ΔACD ΔABP ≅ ΔACP AP bisects ∠A as well as ∠D. AP is the perpendicular bisector of BC. i) in ∆ABD and ∆ACB AB=AC BD=CD AD=AD ∆ABD≅∆ACD ————-(S.S.S) ii) in ∆ABP and ∆ACP AB=AC ∠ BAP≅∠CAP    [∆ABD≅∆ACD   BY C.P.CT] AP=AP   ———[common] ∴[∆ABD≅∆ACD         ———–[S.A.S] iii) [∆ABD≅∆ACD         ———–[S.A.S] ∠BAD=∠CAD AD, bisects ∠A AP, bisects ∠A   —————–1 In ∆ BDP and ∆DPB BD=CD   —————(GIVEN) DP=PC     ———-[∆AB≅ ∆ACP   C.P.C.T] DP=DP         ———–[common] ∴∆BDP≅∆CDP    (S.S.S)   ∠BDP=∠CDP  (C.P.C.T) DP bisects ∠D AP bisects ∠D  ——————-2 From 1 and 2, AP bisects ∠ A as well as ∠ D iv) ∠ AP +∠APC =180°        ————[linear pair] ∠APB=∠APC    ————-[∆ABP≅∆ACP       C.P.CT] ∠APB + ∠APC=180° 2 ∠ APB=180° ∠APB=180/2=90° BP=PC            (FROM ii) ∴AP is ⊥ bisects of BC.

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

(i) ad bisects bc                      (ii) ad bisects ∠a..

Given: AB=AC,  AD altitude To Show:  (i) AD bisects BC                      (ii) AD bisects ∠A. In ∆ADB and ∆ADC ∠ADB=∠ADC   ——– ———–[each 90°]    R AB=AC               ——————–[given] S AD=AD             ——–[common] S ∴ ∆ADB ≅∆ADC BD=DC             ————-[c.p.c.t] ∴AD bisects BC ∠1=∠2               ————-[c.p.c.t] ∴AD bisects ∠A

Question 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 Δ PQR (see Fig. 7.40). Show that:

(i) ΔABM ≅ ΔPQN

(ii) ΔABC ≅ PQR

Given:       AB=PQ BC=QR AM=PN AM and PN are medians To show :(i) ΔABM ≅ ΔPQN           (ii) ΔABC ≅ PQR Solution: In  ΔABM and ΔPQN   AB=PQ AM=PN because AM and PN are medians BC=QR therefore   =1/2BC=1/2QR   ∴BM=QN ∴) ΔABM ≅ ΔPQN                    ———[S.S.S]         ∠B=∠Q             ——–[c.p.c.t] ii)now in       ΔABC and  ΔPQR     AB=PQ                ———-[given] ∠B=∠Q             from (i) BC=QR               —————-[given] ∴ ΔABC ≅ PQR                 [S.A.S]

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

Given: altitude BE and CF are equal   To prove: ΔABC is an isosceles  Δ In ΔBEC and ΔCEB ∠E=∠F        —————-[each 90°] R BC=BC       —————–[common]  H BF=CF       —————-[given]  S # ΔBEC ≅ ΔCEB     [R.H.S] ∠C=∠B        ————-[C.P.C.T]  In ΔABC, ∠C=∠B        

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

Given:   In ∆ABC, AB=BC AP ⊥ BC to show that: ∠B = ∠C. Solution:   In ∆APB and ∆APC ∠APB = ∠APC         —————[ each 90°]   R AB=AC               ——————-[given]       H AP=AP         ——————–[common]      S ∴∆APB ≅ ∆APC            ———-[R.H.S] ∠B = ∠C  —————[C.P.C.T]          

NCERT Solutions for Class 9 Maths Chapter 7 Triangles: Exercise 7.4

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

Given: Right angle triangle intersect ∠B=90° To show: AC>AB and AC>BC Solution:∠A+∠B+∠C=180°           —————-[angle sum property] ∠A+90°+∠C=180°         ∠A+∠C=180°=90°   ∠A+∠C=90°   ∴∠A<90°         and  ∠C<90°   BC<AC                         AB<AC                   ———-[sides opposite to longer angle is greater] ∴Hypotenuse  is the longest side.

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

Given : ∠PBC < ∠QCB       LET this be  [∠1 < ∠2] To Show : AD < BC Solution: ∠1 < ∠2                       ——–[given] -∠1 > -∠2 180-∠1>180-∠2                         ∠3>∠4                                  ———[linear pair] In ∆ABC, ∠3>∠4   AC>AB

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

Given: ∠B < ∠A and ∠C < ∠D To show: AD<BC Solution: In  ∆BOA ∠B < ∠A ∴AO<BO————-1 In ∆COD ∠C < ∠D ∴OD<OC————-2 Adding 1 and 2   AO+OD+<BO+OC AD<BC

Question 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.

Given: AB is smaller side CD is longest side To show: ∠A > ∠C and ∠B > ∠D. Solution : In ∆ABC BC>AB   ∠1>∠2              ———-[angle opposite to greater side is greatest]-1 In ∆ABC CD>AD ∠3>∠4             ———[ angle opposite to greater side is greatest]-2 Adding 1 and 2 ∠1+∠2+∠3+∠4 ∠A>∠C ii) In ∆ABD AD>AB ∠5>∠6             ——————-[ angle opposite to greater side is greatest]-3 In ∆BCD CD>BC ∠7>∠8             ——————-[ angle opposite to greater side is greatest]-4 Adding  3 and 4 ∠5+∠6+∠7+∠8 ∠B > ∠D

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

Given: PR>PQ PS is angle bisects ∠QPR To show: ∠PSR > ∠PSQ Solution: PR>PQ ∴∠3+∠4          ————[angle opposite to greater side is larger] ∠3+∠1+x=180° ————-[angle sum property of ∆]   ∠3=180°-∠1-x ————1 In ∆PSR 4+∠2+x=180° ————-[angle sum property of ∆]   ∠4=180°-∠2-x ————2 Because ∠3>∠4 180°-∠1-x >180°-∠2-x -∠1>-∠2   ∠1<∠2   ∠PSQ<∠PSR  OR ∠PSR>∠PSQ

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

Given: Let P be any point not lying on a line  L. PM ⊥ L Now, ∠ is any point another than M lying on line=L In ∆PMN ∠M90°                  ———-[ PM ⊥ L] ∠L<90°                 ——-[∴∠M90° ∠L<90°  ∠L<90°]   ∠L<∠M   AM<PL                ———[side opposite to greater is greater ]

NCERT Solutions for Class 9 Maths Chapter 7 Triangles : Exercise 7.5 (Optional)

Question 1. abc is a triangle. locate a point in the interior of δabc which is equidistant from all the vertices of δabc..

To obtain a point which is equidistant from all vertices of a triangle we construct perpendicular bisectors of all sides (AB, BC, CA) of the triangle (ΔABC). The point of intersection of these bisectors is known as Circumcenter(O) which is equidistant from all vertices.

Question 2. In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

To obtain a point which is equidistant from all sides of a triangle we construct angle bisectors of all angles present in ΔABC i.e ∠BAC, ∠ABC, ∠ACB. The point of intersection of these bisectors is called Incentre(I) which is equidistant from all sides.

Question 3. In a huge park, people are concentrated at three points :

A: where there are different slides and swings for children,, b: near which a man-made lake is situated,, c: which is near to a large parking and exit., where should an ice cream parlour be set up so that the maximum number of persons can approach it, (hint : the parlour should be equidistant from a, b and c).

The ice-cream parlour must be set somewhere so that it’s easily available for the public. So for such point it should be at a equal distance from point A, B, C & such point is termed as circumcenter.

Question 4. Complete the hexagonal and star-shaped Rangolies by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?

We need to find the number of triangles that can get fit the above figures i.e the hexagon and the star. So, Area of hexagon = (Area of small triangle inside hexagon) * 6 Area of small equilateral triangle = √3/4 * a 2  = √3/4 * 5 2 = √3/4 * 25 = 25√3/4 So, Area of hexagon = 25√3/4 * 6     = 150√3/4 cm 2    Area of Star = Area of 6 triangles and 1 hexagon = 6 * 25√3/4 + 150√3/4   = 300√3/4 cm 2  Area of triangles of 1cm side that are to be fitted = √3/4 * 1 2 = √3/4 cm 2 Number of triangles that can be accommodated inside hexagon and stars : a. For Hexagon : Area of hexagon/ Area of 1cm side triangle = 150√3/4 cm 2 / √3/4 cm2  = 150 triangles               b. For Star : Area of star/ Area of 1cm triangle = 300√3/4 cm 2 / √3/4 cm 2 = 300 triangles         Hence, the star can accommodate 150 more triangles than the hexagon.

Key Features of NCERT Solutions for Class 9 Maths Chapter 7 Triangles:

• These NCERT solutions are developed by the GfG team, with a focus on students’ benefit .
• NCERT Solutions are developed for each chapter of class 9 including Triangles.
• NCERT Solutions provides accurate and complete solutions for problems given in NCERT textbooks.

FAQs on NCERT Solutions for Class 9 Maths Chapter 7-Triangles

1. why is it important to learn about triangles .

One of the fundamental geometric forms is the triangle. Understanding polygonal characteristics and connections is based on them. Students build a solid geometric foundation by studying triangles, and they also gain understanding into the concepts of congruence, resemblance, and symmetry.

2. What topics are covered in NCERT Solutions for Class 9 Maths Chapter 7 Triangles?

NCERT Solutions for Class 9 Maths Chapter 7-Triangles covers topics such as congruence of triangles, their rules, attributes, and triangle inequalities.

3. How can NCERT Solutions for Class 9 Maths Chapter 7 Triangleshelp me?

NCERT Solutions for Class 9 Maths Chapter 7-Triangles can help you solve the NCERT exercise without any limitations. If you are stuck on a problem, you can find its solution in these solutions and free yourself from the frustration of being stuck on some question.

4. How many exercises are there in NCERT Solutions for Class 9 Maths Chapter 7 Triangles?

There are 3 exercises in the Class 9 Maths Chapter 7-Triangles which covers all the important topics and sub-topics.

5. Where can I find NCERT Solutions for Class 9 Maths Chapter 7 Triangles?

You can find these NCERT Solutions in this article created by our team of experts at GeeksforGeeks.

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CBSE MCQ for Class 9 Maths Chapter 7 Triangles Free PDF

Guys, we are working very hard to provide you with TOPIC-WISE MCQs (as listed below). Till then, attached below is the Master PDF having all the topics. Hope you understand. Enjoy your preparation! All the Best!

CBSE MCQ for Class 9 Maths Chapter 7 Triangles PDF

The CBSE MCQ for Class 9 Maths Chapter 7 Triangles are provided above, in detailed and free to download PDF format. The solutions are latest , comprehensive , confidence inspiring , with easy to understand explanation . To download NCERT Class 9 Solutions PDF for Free, just click ‘ Download pdf ’.

Other MCQ Questions for Maths Class 9th CBSE

• CBSE MCQ for Class 9 Maths Chapter 5 Euclid’s Geometry
• CBSE MCQ for Class 9 Maths Chapter 6 Lines and Angles
• CBSE MCQ for Class 9 Maths Chapter 8 Quadrilaterals
• CBSE MCQ for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles

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Solve every question of NCERT by hand, without looking at the solution.

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Practice Assertion Reason & Case Study Based Questions

Sit through FULLY INVIGILATED TESTS involving MCQs. Assertion reason & Case Study Based Questions

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