case study based questions class 9 maths herons formula

• Heron’s Formula Class 9 Case Study Questions Maths Chapter 10

Last Updated on September 8, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 9 maths. Case study questions are the new question format that is introduced in CBSE board. The resources for case study questions are very less. So, to help students we have created chapterwise case study questions for class 9 maths. In this article, you will find case study questions for CBSE Class 9 Maths Chapter 10 Heron’s Formula. It is a part of Case Study Questions for CBSE Class 9 Maths Series.

	Heron’s Formula
	Case Study Questions
	Competency Based Questions
	CBSE
	9
	Maths
	Class 9 Studying Students
	Yes
	Mentioned

Table of Contents

Case Study Questions on Heron’s Formula

Mayank bought a triangle shape field and wants to grow potato and wheat on his field. He divided his field by joining opposite sides. On the largest park he grew wheat and on the rest part he grew potato. The dimensions of a park are shown in the park.

On the basis of the above information, solve the following questions:

Q 1. Find the length of AC in ΔABC.

Q 2. Find the area of ΔABC.

Q 3. If the cost of ploughing park is ₹5 per cm 2 , then find the total cost of ploughing the park.

1. In right angled $\triangle A B C$, use Pythagoras theorem,

$$ \begin{aligned} A C & =\sqrt{(A B)^2+(B C)^2}=\sqrt{(12)^2+(5)^2} \\ \vdots & =\sqrt{144+25}=\sqrt{169}=13 \mathrm{~m} \end{aligned} $$

Hence, length of $A C$ is 13 m .

2. Area of $\triangle \mathrm{ABC}$

$$ \begin{aligned} & =\frac{1}{2} \times A B \times B C \\ & =\frac{1}{2} \times 12 \times 5=30 \mathrm{~m}^2 \end{aligned} $$

3. Since, the total area of the park $=30 \mathrm{~m}^2$ $\because$ The cost of ploughing the park in $1 \mathrm{~m}^2=5$ $\therefore$ The cost of ploughing the park in $30 \mathrm{~m}^2$

$$ \begin{aligned} & = 5\times 30\\ & =150 \end{aligned} $$

Understanding Heron’s Formula

Area of Triangle: The total space occupied inside the boundary of the triangle is said to be an area of triangle.

Perimeter of Triangle: Sum of lengths of all three sides of a triangle.

$\begin{aligned} 2 s & =a+b+c \\ s & =\frac{a+b+c}{2}\end{aligned}$

Right-angled Triangle: It is a triangle with one right angle.

1. Area $=\frac{1}{2} \times a \times b$ 2. Altitude $=a$ 3. Perimeter$=a+b+\sqrt{a^2+b^2}$

where ‘a’ and ‘b’ are the sides that includes to the right angle.

Isosceles Triangle: Triangle that has two equal sides and corresponding two equal angles.

1. Area $=\frac{b}{4} \sqrt{4 a^2-b^2}$ 2. Perimeter $=2 a+b$ 3. Altitude $=\frac{1}{2} \sqrt{4 a^2-b^2}$

where ‘a’ is length of two equal sides and ‘b’ is base.

Equilateral Triangle: Triangle with all sides and all angles equal (each being 60°)

1. Area $=\frac{\sqrt{3}}{4} a^2$ 2. Perimeter $=3 a$ 3. Altitude $=\frac{\sqrt{3}}{2} a$

where ‘a’ is side.

Heron‘s Formula: The formula given by Heron about the area of a triangle.

Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

where a, b and c are the sides of triangle and s is its semi-perimeter.

Boost your knowledge

(i) The length of longest altitude is the perpendicular distance from the opposite vertex to the smallest side of a triangle. (ii) The length of smallest altitude is the perpendicular distance from the opposite vertex to the largest side of a triangle. (iii) Heron‘s formula is helpful when it is not possible to find the height of the triangle easily. (iv) Heron‘s formula is applicable to all types of triangles whether it is a right triangle or an isosceles or an equilateral triangle

• Circles Class 9 Case Study Questions Maths Chapter 9
• Quadrilaterals Class 9 Case Study Questions Maths Chapter 8
• Triangles Class 9 Case Study Questions Maths Chapter 7
• Lines and Angles Class 9 Case Study Questions Maths Chapter 6
• Introduction to Euclid’s Geometry Class 9 Case Study Questions Maths Chapter 5
• Linear Equations in Two Variables Class 9 Case Study Questions Maths Chapter 4
• Coordinate Geometry Class 9 Case Study Questions Maths Chapter 3

Polynomials Class 9 Case Study Questions Maths Chapter 2

Number systems class 9 case study questions maths chapter 1, topics from which case study questions may be asked.

• Quadrilaterals
• Parallelograms
• Properties of a parallelogram
• Mid-point theorem
• Converse of Mid-point theorem

A trapezium is not a parallelogram (as only one pair of opposite sides is parallel in a trapezium and we require both pairs to be parallel in a parallelogram)

Case study questions from the above given topic may be asked.

Download Customised White Label Study Materials in MS Word Format

We are providing teaching resources to teachers and coaching institute looking for customised study materials in MS word format. Our High-quality editable study material which is prepared by the expert faculties are Highly useful for Teachers, Mentors, Tutors, Faculties, Coaching Institutes, Coaching Experts, Tuition Centers.

Frequently Asked Questions (FAQs) on Heron’s Formula Case Study

Q1: what is heron’s formula.

A1: Heron’s Formula is used to calculate the area of a triangle when the lengths of all three sides are known. The formula is: Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

Q2: How is Heron’s Formula derived?

A2: Heron’s formula is derived from the general area formula of a triangle. It simplifies the calculation of the area without needing the height. This is particularly useful for triangles where the height is difficult to determine directly. The formula is based on the triangle’s semi-perimeter and each of its sides.

Q3: When should Heron’s Formula be used?

A3: Heron’s Formula should be used when the lengths of all three sides of a triangle are given, and the height is unknown. It allows you to find the area of any triangle (scalene, isosceles, or equilateral) as long as you know the sides.

Q4: Can Heron’s Formula be used for right-angled triangles?

A4: Yes, Heron’s Formula can be used for right-angled triangles, although for right-angled triangles, a simpler method involving the base and height (half the product of the base and height) can also be used to find the area.

Q5: What is the significance of the semi-perimeter in Heron’s Formula?

A5: The semi-perimeter (s) is half the perimeter of the triangle. It acts as a key variable in the formula, simplifying the calculation of the area by incorporating all three sides of the triangle in a single expression.

Q6: Can Heron’s Formula be applied to a quadrilateral?

A6: No, Heron’s Formula specifically applies to triangles. However, if a quadrilateral can be divided into two triangles, the area of each triangle can be found using Heron’s Formula, and the sum of the areas will give the total area of the quadrilateral.

Q7: How is Heron’s Formula applied in real-life problems?

A7: Heron’s Formula is used in fields like civil engineering, architecture, and land surveying where determining the area of irregular land plots or structures is necessary, especially when only the lengths of the boundaries are known.

Q8: Are there any online resources or tools available for practicing Circles case study questions?

A9: We provide case study questions for CBSE Class 9 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.

Related Posts

Gurukul of Excellence

Classes for Physics, Chemistry and Mathematics by IITians

Join our Telegram Channel for Free PDF Download

Case Study Questions for Class 9 Maths Chapter 12 Herons Formula

• Last modified on: 8 months ago
• Reading Time: 1 Minute

Here we are providing case study questions for Class 9 Maths Chapter 12 Herons Formula. 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:

Related Posts

Category lists (all posts).

All categories of this website are listed below with number of posts in each category for better navigation. Visitors can click on a particular category to see all posts related to that category.

• Full Form (1)
• Biography of Scientists (1)
• Assertion Reason Questions in Biology (37)
• Case Study Questions for Class 12 Biology (14)
• DPP Biology for NEET (12)
• Blog Posts (35)
• Career Guidance (1)
• Assertion Reason Questions for Class 10 Maths (14)
• Case Study Questions for Class 10 Maths (15)
• Extra Questions for Class 10 Maths (12)
• Maths Formulas for Class 10 (1)
• MCQ Questions for Class 10 Maths (15)
• NCERT Solutions for Class 10 Maths (4)
• Quick Revision Notes for Class 10 Maths (14)
• Assertion Reason Questions for Class 10 Science (16)
• Case Study Questions for Class 10 Science (14)
• Evergreen Science Book Solutions for Class 10 (17)
• Extra Questions for Class 10 Science (23)
• HOTS for Class 10 Science (17)
• Important Questions for Class 10 Science (10)
• Lakhmir Singh Class 10 Biology Solutions (4)
• Lakhmir Singh Class 10 Chemistry Solutions (5)
• Lakhmir Singh Class 10 Physics Solutions (5)
• MCQ Questions for Class 10 Science (20)
• NCERT Exemplar Solutions for Class 10 Science (16)
• NCERT Solutions for Class 10 Science (15)
• Quick Revision Notes for Class 10 Science (4)
• Study Notes for Class 10 Science (17)
• Assertion Reason Questions for Class 10 Social Science (14)
• Case Study Questions for Class 10 Social Science (24)
• MCQ Questions for Class 10 Social Science (3)
• Topicwise Notes for Class 10 Social Science (4)
• CBSE CLASS 11 (1)
• Assertion Reason Questions for Class 11 Chemistry (14)
• Case Study Questions for Class 11 Chemistry (11)
• Free Assignments for Class 11 Chemistry (1)
• MCQ Questions for Class 11 Chemistry (8)
• Very Short Answer Questions for Class 11 Chemistry (7)
• Assertion Reason Questions for Class 11 Entrepreneurship (8)
• Important Questions for CBSE Class 11 Entrepreneurship (1)
• Assertion Reason Questions for Class 11 Geography (24)
• Case Study Questions for Class 11 Geography (24)
• Assertion Reason Questions for Class 11 History (12)
• Case Study Questions for Class 11 History (12)
• Assertion and Reason Questions for Class 11 Maths (16)
• Case Study Questions for Class 11 Maths (16)
• Formulas for Class 11 Maths (6)
• MCQ Questions for Class 11 Maths (17)
• NCERT Solutions for Class 11 Maths (8)
• Case Study Questions for Class 11 Physical Education (11)
• Assertion Reason Questions for Class 11 Physics (15)
• Case Study Questions for Class 11 Physics (12)
• Class 11 Physics Study Notes (5)
• Concept Based Notes for Class 11 Physics (2)
• Conceptual Questions for Class 11 Physics (10)
• Derivations for Class 11 Physics (3)
• Extra Questions for Class 11 Physics (13)
• MCQ Questions for Class 11 Physics (16)
• NCERT Solutions for Class 11 Physics (16)
• Numerical Problems for Class 11 Physics (4)
• Physics Formulas for Class 11 (7)
• Revision Notes for Class 11 Physics (11)
• Very Short Answer Questions for Class 11 Physics (11)
• Assertion Reason Questions for Class 11 Political Science (20)
• Case Study Questions for Class 11 Political Science (20)
• CBSE CLASS 12 (8)
• Extra Questions for Class 12 Biology (14)
• MCQ Questions for Class 12 Biology (13)
• Case Studies for CBSE Class 12 Business Studies (13)
• MCQ Questions for Class 12 Business Studies (1)
• Revision Notes for Class 12 Business Studies (10)
• Assertion Reason Questions for Class 12 Chemistry (15)
• Case Study Based Questions for Class 12 Chemistry (14)
• Extra Questions for Class 12 Chemistry (5)
• Important Questions for Class 12 Chemistry (15)
• MCQ Questions for Class 12 Chemistry (8)
• NCERT Solutions for Class 12 Chemistry (16)
• Revision Notes for Class 12 Chemistry (7)
• Assertion Reason Questions for Class 12 Economics (9)
• Case Study Questions for Class 12 Economics (9)
• MCQ Questions for Class 12 Economics (1)
• MCQ Questions for Class 12 English (2)
• Assertion Reason Questions for Class 12 Entrepreneurship (7)
• Case Study Questions for Class 12 Entrepreneurship (7)
• Case Study Questions for Class 12 Geography (18)
• Assertion Reason Questions for Class 12 History (8)
• Case Study Questions for Class 12 History (13)
• Assertion Reason Questions for Class 12 Informatics Practices (13)
• Case Study Questions for Class 12 Informatics Practices (11)
• MCQ Questions for Class 12 Informatics Practices (5)
• Assertion and Reason Questions for Class 12 Maths (14)
• Case Study Questions for Class 12 Maths (13)
• Maths Formulas for Class 12 (5)
• MCQ Questions for Class 12 Maths (14)
• Problems Based on Class 12 Maths (1)
• RD Sharma Solutions for Class 12 Maths (1)
• Assertion Reason Questions for Class 12 Physical Education (11)
• Case Study Questions for Class 12 Physical Education (11)
• MCQ Questions for Class 12 Physical Education (10)
• Assertion Reason Questions for Class 12 Physics (16)
• Case Study Based Questions for Class 12 Physics (14)
• Class 12 Physics Conceptual Questions (16)
• Class 12 Physics Discussion Questions (1)
• Class 12 Physics Latest Updates (2)
• Derivations for Class 12 Physics (8)
• Extra Questions for Class 12 Physics (4)
• Important Questions for Class 12 Physics (8)
• MCQ Questions for Class 12 Physics (14)
• NCERT Solutions for Class 12 Physics (18)
• Numerical Problems Based on Class 12 Physics (16)
• Physics Class 12 Viva Questions (1)
• Revision Notes for Class 12 Physics (7)
• Assertion Reason Questions for Class 12 Political Science (16)
• Case Study Questions for Class 12 Political Science (16)
• Notes for Class 12 Political Science (1)
• Assertion Reason Questions for Class 6 Maths (13)
• Case Study Questions for Class 6 Maths (13)
• Extra Questions for Class 6 Maths (1)
• Worksheets for Class 6 Maths (1)
• Assertion Reason Questions for Class 6 Science (16)
• Case Study Questions for Class 6 Science (16)
• Extra Questions for Class 6 Science (1)
• MCQ Questions for Class 6 Science (9)
• Assertion Reason Questions for Class 6 Social Science (1)
• Case Study Questions for Class 6 Social Science (26)
• NCERT Exemplar for Class 7 Maths (13)
• NCERT Exemplar for Class 7 Science (19)
• NCERT Exemplar Solutions for Class 7 Maths (12)
• NCERT Exemplar Solutions for Class 7 Science (18)
• NCERT Notes for Class 7 Science (18)
• Assertion Reason Questions for Class 7 Maths (14)
• Case Study Questions for Class 7 Maths (14)
• Extra Questions for Class 7 Maths (5)
• Assertion Reason Questions for Class 7 Science (18)
• Case Study Questions for Class 7 Science (17)
• Extra Questions for Class 7 Science (19)
• Assertion Reason Questions for Class 7 Social Science (1)
• Case Study Questions for Class 7 Social Science (30)
• Assertion Reason Questions for Class 8 Maths (7)
• Case Study Questions for Class 8 Maths (17)
• Extra Questions for Class 8 Maths (1)
• MCQ Questions for Class 8 Maths (6)
• Assertion Reason Questions for Class 8 Science (16)
• Case Study Questions for Class 8 Science (11)
• Extra Questions for Class 8 Science (2)
• MCQ Questions for Class 8 Science (4)
• Numerical Problems for Class 8 Science (1)
• Revision Notes for Class 8 Science (11)
• Assertion Reason Questions for Class 8 Social Science (27)
• Case Study Questions for Class 8 Social Science (23)
• CBSE Class 9 English Beehive Notes and Summary (2)
• Assertion Reason Questions for Class 9 Maths (14)
• Case Study Questions for Class 9 Maths (14)
• MCQ Questions for Class 9 Maths (11)
• NCERT Notes for Class 9 Maths (6)
• NCERT Solutions for Class 9 Maths (12)
• Revision Notes for Class 9 Maths (3)
• Study Notes for Class 9 Maths (10)
• Assertion Reason Questions for Class 9 Science (16)
• Case Study Questions for Class 9 Science (15)
• Evergreen Science Book Solutions for Class 9 (15)
• Extra Questions for Class 9 Science (22)
• MCQ Questions for Class 9 Science (11)
• NCERT Solutions for Class 9 Science (15)
• Revision Notes for Class 9 Science (1)
• Study Notes for Class 9 Science (15)
• Topic wise MCQ Questions for Class 9 Science (2)
• Topicwise Questions and Answers for Class 9 Science (15)
• Assertion Reason Questions for Class 9 Social Science (15)
• Case Study Questions for Class 9 Social Science (19)
• CHEMISTRY (8)
• Chemistry Articles (2)
• Daily Practice Problems (DPP) (3)
• Books for CBSE Class 9 (1)
• Books for ICSE Class 10 (3)
• Editable Study Materials (8)
• Exam Special for CBSE Class 10 (3)
• H. C. Verma (Concepts of Physics) (13)
• Study Materials for ICSE Class 10 Biology (14)
• Extra Questions for ICSE Class 10 Chemistry (1)
• Study Materials for ICSE Class 10 Chemistry (5)
• Study Materials for ICSE Class 10 Maths (16)
• Important Questions for ICSE Class 10 Physics (13)
• MCQ Questions for ICSE Class 10 Physics (4)
• Study Materials for ICSE Class 10 Physics (8)
• Study Materials for ICSE Class 9 Maths (7)
• Study Materials for ICSE Class 9 Physics (10)
• Topicwise Problems for IIT Foundation Mathematics (4)
• Challenging Physics Problems for JEE Advanced (2)
• Topicwise Problems for JEE Physics (1)
• DPP for JEE Main (1)
• Integer Type Questions for JEE Main (1)
• Integer Type Questions for JEE Chemistry (6)
• Chapterwise Questions for JEE Main Physics (1)
• Integer Type Questions for JEE Main Physics (8)
• Physics Revision Notes for JEE Main (4)
• JEE Mock Test Physics (1)
• JEE Study Material (1)
• JEE/NEET Physics (6)
• CBSE Syllabus (1)
• Maths Articles (2)
• NCERT Books for Class 12 Physics (1)
• NEET Chemistry (13)
• Important Questions for NEET Physics (17)
• Topicwise DPP for NEET Physics (5)
• Topicwise MCQs for NEET Physics (32)
• NTSE MAT Questions (1)
• Physics (1)
• Alternating Current (1)
• Electrostatics (6)
• Fluid Mechanics (2)
• PowerPoint Presentations (13)
• Previous Years Question Paper (3)
• Products for CBSE Class 10 (15)
• Products for CBSE Class 11 (10)
• Products for CBSE Class 12 (6)
• Products for CBSE Class 6 (2)
• Products for CBSE Class 7 (5)
• Products for CBSE Class 8 (1)
• Products for CBSE Class 9 (3)
• Products for Commerce (3)
• Products for Foundation Courses (2)
• Products for JEE Main & Advanced (10)
• Products for NEET (6)
• Products for ICSE Class 6 (1)
• Electrostatic Potential and Capacitance (1)
• Topic Wise Study Notes (Physics) (2)
• Topicwise MCQs for Physics (2)
• Uncategorized (138)

Test series for students preparing for Engineering & Medical Entrance Exams are available. We also provide test series for School Level Exams. Tests for students studying in CBSE, ICSE or any state board are available here. Just click on the link and start test.

Download Books – Exam Special

Sample Papers for CBSE 2025 Exams

• Sample Question Papers for CBSE Class 8 All Subjects (for 2025 Exams)
• Sample Question Papers for CBSE Class 9 All Subjects (for 2025 Exams)
• Sample Question Papers for CBSE Class 10 All Subjects (for 2025 Exams)
• Sample Question Papers for CBSE Class 12 All Subjects (for 2025 Exams)

CBSE Class 10 Most Downloaded Books

• CBSE Important Numerical Problems Class 10 Physics Board Exams
• CBSE Practical Based Questions for Class 10 Science Board Exams
• CBSE Important Diagram Based Questions Class 10 Physics Board Exams
• CBSE Most Repeated Questions for Class 10 Science Board Exams

CBSE Class 12 Most Downloaded Books

• CBSE Important Diagrams & Graphs Asked in Board Exams Class 12 Physics
• CBSE Important Numericals Class 12 Physics Board Exams
• CBSE Important Laws & Principles Class 12 Physics Board Exams
• CBSE Important Definitions Class 12 Physics Board Exams
• Master Organic Conversions CBSE Class 12 Chemistry Board Exams
• CBSE Class 12 Physics Chapterwise Important Questions

CBSE Class 8 Most Downloaded Books

• Worksheets for CBSE Class 8 Maths – Chapterwise

ICSE Class 10

• ICSE Important Numericals Class 10 Physics BOARD Exams (215 Numericals)
• ICSE Important Figure Based Questions Class 10 Physics BOARD Exams (230 Questions)
• ICSE Mole Concept and Stoichiometry Numericals Class 10 Chemistry (65 Numericals)
• ICSE Reasoning Based Questions Class 10 Chemistry BOARD Exams (150 Qs)
• ICSE Important Functions and Locations Based Questions Class 10 Biology
• ICSE Reasoning Based Questions Class 10 Biology BOARD Exams (100 Qs)
• ICSE Reasoning Based Questions Class 10 Geography BOARD Exams
• ICSE Revision Notes for Class 10 Chemistry BOARD Exams
• ICSE Revision Notes for Class 10 Physics BOARD Exams

ICSE Class 9

• ICSE Important Figure Based Questions Class 9 Physics Exams
• ICSE Important Numerical Problems for Class 9 Physics Exams
• ICSE Reasoning Based Questions Class 9 Geography BOARD Exams (150 Qs)

CBSE Chapter-Wise Test Papers

• CBSE Class 9 Science Chapterwise Test Papers
• CBSE Class 10 Science Chapterwise Test Papers
• CBSE Class 10 Maths Chapterwise Test Papers
• CBSE Class 10 Social Science Chapterwise Test Papers
• CBSE Class 12 Physics Chapterwise Test Papers
• CBSE Class 12 Chemistry Chapterwise Test papers

✨ Join our Online NEET Test Series for 499/- Only for 1 Year

Leave a Reply Cancel reply

Editable Study Materials for Your Institute - CBSE, ICSE, State Boards (Maharashtra & Karnataka), JEE, NEET, FOUNDATION, OLYMPIADS, PPTs

Discover more from Gurukul of Excellence

Subscribe now to keep reading and get access to the full archive.

Type your email…

Continue reading

CBSE Expert

CBSE Class 9 Maths Case Study Questions PDF Download

Download Class 9 Maths Case Study Questions to prepare for the upcoming CBSE Class 9 Exams 2023-24. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 9 so that they can score 100% in Exams.

Case study questions play a pivotal role in enhancing students’ problem-solving skills. By presenting real-life scenarios, these questions encourage students to think beyond textbook formulas and apply mathematical concepts to practical situations. This approach not only strengthens their understanding of mathematical concepts but also develops their analytical thinking abilities.

Table of Contents

CBSE Class 9th MATHS: Chapterwise Case Study Questions

Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked. For Class 9 Maths Case Study Questions, there would be 5 case-based sub-part questions, wherein a student has to attempt 4 sub-part questions.

Chapterwise Case Study Questions of Class 9 Maths

• 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 Quadrilaterals
• 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

Checkout: Class 9 Science Case Study Questions

And for mathematical calculations, tap Math Calculators which are freely proposed to make use of by calculator-online.net

The above  Class 9 Maths Case Study Question s will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 9 Maths Case Study Questions have been developed by experienced teachers of cbseexpert.com for the benefit of Class 10 students.

Class 9 Maths Syllabus 2023-24

UNIT I: NUMBER SYSTEMS

1. REAL NUMBERS (18 Periods)

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

2. Examples of non-recurring/non-terminating decimals. Existence of non-rational numbers (irrational numbers) such as √2, √3 and their representation on the number line. Explaining that every real number is represented by a unique point on the number line and conversely, viz. every point on the number line represents a unique real number.

3. Definition of nth root of a real number.

4. Rationalization (with precise meaning) of real numbers of the type

(and their combinations) where x and y are natural number and a and b are integers.

5. Recall of laws of exponents with integral powers. Rational exponents with positive real bases (to be done by particular cases, allowing learner to arrive at the general laws.)

UNIT II: ALGEBRA

1. POLYNOMIALS (26 Periods)

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

RELATED STORIES

and their use in factorization of polynomials.

2. LINEAR EQUATIONS IN TWO VARIABLES (16 Periods)

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

UNIT III: COORDINATE GEOMETRY COORDINATE GEOMETRY (7 Periods)

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

UNIT IV: GEOMETRY

1. INTRODUCTION TO EUCLID’S GEOMETRY (7 Periods)

History – Geometry in India and Euclid’s geometry. Euclid’s method of formalizing observed phenomenon into rigorous Mathematics with definitions, common/obvious notions, axioms/postulates and theorems. The five postulates of Euclid. Showing the relationship between axiom and theorem, for example: (Axiom)

1. Given two distinct points, there exists one and only one line through them. (Theorem)

2. (Prove) Two distinct lines cannot have more than one point in common.

2. LINES AND ANGLES (15 Periods)

1. (Motivate) If a ray stands on a line, then the sum of the two adjacent angles so formed is 180O and the converse.

2. (Prove) If two lines intersect, vertically opposite angles are equal.

3. (Motivate) Lines which are parallel to a given line are parallel.

3. TRIANGLES (22 Periods)

1. (Motivate) Two triangles are congruent if any two sides and the included angle of one triangle is equal to any two sides and the included angle of the other triangle (SAS Congruence).

2. (Prove) Two triangles are congruent if any two angles and the included side of one triangle is equal to any two angles and the included side of the other triangle (ASA Congruence).

3. (Motivate) Two triangles are congruent if the three sides of one triangle are equal to three sides of the other triangle (SSS Congruence).

4. (Motivate) Two right triangles are congruent if the hypotenuse and a side of one triangle are equal (respectively) to the hypotenuse and a side of the other triangle. (RHS Congruence)

5. (Prove) The angles opposite to equal sides of a triangle are equal.

6. (Motivate) The sides opposite to equal angles of a triangle are equal.

4. QUADRILATERALS (13 Periods)

1. (Prove) The diagonal divides a parallelogram into two congruent triangles.

2. (Motivate) In a parallelogram opposite sides are equal, and conversely.

3. (Motivate) In a parallelogram opposite angles are equal, and conversely.

4. (Motivate) A quadrilateral is a parallelogram if a pair of its opposite sides is parallel and equal.

5. (Motivate) In a parallelogram, the diagonals bisect each other and conversely.

6. (Motivate) In a triangle, the line segment joining the mid points of any two sides is parallel to the third side and in half of it and (motivate) its converse.

5. CIRCLES (17 Periods)

1. (Prove) Equal chords of a circle subtend equal angles at the center and (motivate) its converse.

2. (Motivate) The perpendicular from the center of a circle to a chord bisects the chord and conversely, the line drawn through the center of a circle to bisect a chord is perpendicular to the chord.

3. (Motivate) Equal chords of a circle (or of congruent circles) are equidistant from the center (or their respective centers) and conversely.

4. (Prove) The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

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

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

7. (Motivate) The sum of either of the pair of the opposite angles of a cyclic quadrilateral is 180° and its converse.

UNIT V: MENSURATION 1.

1. AREAS (5 Periods)

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

2. SURFACE AREAS AND VOLUMES (17 Periods)

Surface areas and volumes of spheres (including hemispheres) and right circular cones.

UNIT VI: STATISTICS & PROBABILITY

STATISTICS (15 Periods)

 Bar graphs, histograms (with varying base lengths), and frequency polygons.

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.

Benefits of Practicing CBSE Class 9 Maths Case Study Questions

Regular practice of CBSE Class 9 Maths case study questions offers several benefits to students. Some of the key advantages include:

• Deeper Understanding : Case study questions foster a deeper understanding of mathematical concepts by connecting them to real-world scenarios. This improves retention and comprehension.
• Practical Application : Students learn to apply mathematical concepts to practical situations, preparing them for real-life problem-solving beyond the classroom.
• Critical Thinking : Case study questions require students to think critically, analyze data, and devise appropriate solutions. This nurtures their critical thinking abilities, which are valuable in various academic and professional domains.
• Exam Readiness : By practicing case study questions, students become familiar with the question format and gain confidence in their problem-solving abilities. This enhances their readiness for CBSE Class 9 Maths exams.
• Holistic Development: Solving case study questions cultivates not only mathematical skills but also essential life skills like analytical thinking, decision-making, and effective communication.

Tips to Solve CBSE Class 9 Maths Case Study Questions Effectively

Solving case study questions can be challenging, but with the right approach, you can excel. Here are some tips to enhance your problem-solving skills:

• Read the case study thoroughly and understand the problem statement before attempting to solve it.
• Identify the relevant data and extract the necessary information for your solution.
• Break down complex problems into smaller, manageable parts to simplify the solution process.
• Apply the appropriate mathematical concepts and formulas, ensuring a solid understanding of their principles.
• Clearly communicate your solution approach, including the steps followed, calculations made, and reasoning behind your choices.
• Practice regularly to familiarize yourself with different types of case study questions and enhance your problem-solving speed.Class 9 Maths Case Study Questions

Remember, solving case study questions is not just about finding the correct answer but also about demonstrating a logical and systematic approach. Now, let’s explore some resources that can aid your preparation for CBSE Class 9 Maths case study questions.

Q1. Are case study questions included in the Class 9 Maths Case Study Questions syllabus?

Yes, case study questions are an integral part of the CBSE Class 9 Maths syllabus. They are designed to enhance problem-solving skills and encourage the application of mathematical concepts to real-life scenarios.

Q2. How can solving case study questions benefit students ?

Solving case study questions enhances students’ problem-solving skills, analytical thinking, and decision-making abilities. It also bridges the gap between theoretical knowledge and practical application, making mathematics more relevant and engaging.

Q3. How do case study questions help in exam preparation?

Case study questions help in exam preparation by familiarizing students with the question format, improving analytical thinking skills, and developing a systematic approach to problem-solving. Regular practice of case study questions enhances exam readiness and boosts confidence in solving such questions.

Leave a Comment Cancel reply

Save my name, email, and website in this browser for the next time I comment.

Download India's best Exam Preparation App Now.

Key Features

• Revision Notes
• Important Questions
• Previous Years Questions
• Case-Based Questions
• Assertion and Reason Questions

No thanks, I’m not interested!

Test: Heron`s Formula- Case Based Type Questions - Class 9 MCQ

10 questions mcq test - test: heron`s formula- case based type questions, direction: isosceles triangles were used to construct a bridge in which the base (unequal side) of an isosceles triangle is 4 cm and its perimeter is 20 cm. q. what is the length of equal sides.

So, x + x + 4 = 20

2x + 4 = 20

2x = 20 – 4

x = 16/2 = 8 cm.

Direction: Isosceles triangles were used to construct a bridge in which the base (unequal side) of an isosceles triangle is 4 cm and its perimeter is 20 cm. Q. If the sides of a triangle are in the ratio 3 : 5 : 7 and its perimeter is 300 m. Find its area.

100√2 m 2

500√2 m 2

1500√3 m 2

200√3 m 2

Let the sides of a triangle are a = 3x, b = 5x, c = 7x

Then a + b + c = 300

3x + 5x + 7x = 300

So, a = 60, b = 100, c = 140

= 300/2 = 150

1 Crore+ students have signed up on EduRev. Have you?

Direction: Isosceles triangles were used to construct a bridge in which the base (unequal side) of an isosceles triangle is 4 cm and its perimeter is 20 cm. Q. What is the Heron's formula for the area of?

{Area} = area

s = semi-perimeter

a = length of side a

b = length of side b

c = length of side c

Isosceles triangles were used to construct a bridge in which the base (unequal side) of an isosceles triangle is 4 cm and its perimeter is 20 cm. What is the semi perimeter of the Isosceles triangle?

Required semi perimeter = Perimeter/2 = 20/2 = 10 m.

Direction: Isosceles triangles were used to construct a bridge in which the base (unequal side) of an isosceles triangle is 4 cm and its perimeter is 20 cm.

Q. What is the area of highlighted triangle ?

Thus, area of the triangle

Direction: Shakshi prepared a Rangoli in triangular shape on Diwali. She makes a small triangle under a big triangle as shown in figure.

Sides of big triangle are 25 cm, 26 cm and 28 cm. Also, ΔPQR is formed by joining mid points of sides of ΔABC.

Use the above data to help her in resolving below doubts.

Q. What is the semi-perimeter of ΔABC?

= (25 + 26 + 28) cm = 79 cm

Q. Area of ΔPQR =

where s is the semi-perimeter of ΔPQR.

Q. ½ of AB =

Q. What is the length of RQ?

Q. If colourful rope is to be placed along the sides of small ΔPQR. What is the length of the rope?

= (12.5 + 13 + 14) cm

--> and get INR 200 additional OFF

Top Courses for Class 9

Important Questions for Heron`s Formula- Case Based Type Questions

Heron`s formula- case based type questions mcqs with answers, online tests for heron`s formula- case based type questions.

cation olution

Join the 10M+ students on EduRev

Welcome Back

Create your account for free.

Forgot Password

• Skip to main content
• Skip to secondary menu
• Skip to primary sidebar
• Skip to footer

Learn Insta

RD Sharma Solutions , RS Aggarwal Solutions and NCERT Solutions

Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Solutions Answers

May 28, 2022 by Prasanna

Here we are providing Heron’s Formula Class 9 Extra Questions Maths Chapter 12 with Answers Solutions, Extra Questions for Class 9 Maths  was designed by subject expert teachers.

Extra Questions for Class 9 Maths Heron’s Formula with Answers Solutions

Extra Questions for Class 9 Maths Chapter 12 Heron’s Formula with Solutions Answers

Heron’s Formula Class 9 Extra Questions Very Short Answer Type

Question 1. Find the area of an equilateral triangle having side 6 cm. Solutioin: Area of an equilateral triangle = \(\frac{\sqrt{3}}{4}\) × (side) 2 = \(\frac{\sqrt{3}}{4}\) × 6 × 6 = 9√3 cm 2

Heron’s Formula Class 9 Extra Questions Short Answer Type 2

CBSE Class 10 Maths Formulas for Chapter 4 Quadratic Equations are mentioned in this article.

Heron’s Formula Class 9 Extra Questions Long Answer Type

Heron’s Formula Class 9 Extra Questions HOTS

Now, area of orange shaded paper in kite = Area of ∆AOD + Area of ∆CEF = 450 cm 2 + 198.4 cm 2 = 648.4 cm 2 Area of blue shaded paper in kite = Area of ∆AOB + Area of ∆COD = 450 cm 2 + 450 cm 2 = 900 cm 2 Area of black shaded paper in kite = Area of ∆BOC = 450 cm 2 .

Heron’s Formula Class 9 Extra Questions Value Based (VBQs)

• Class 6 Maths
• Class 6 Science
• Class 6 Social Science
• Class 6 English
• Class 7 Maths
• Class 7 Science
• Class 7 Social Science
• Class 7 English
• Class 8 Maths
• Class 8 Science
• Class 8 Social Science
• Class 8 English
• Class 9 Maths
• Class 9 Science
• Class 9 Social Science
• Class 9 English
• Class 10 Maths
• Class 10 Science
• Class 10 Social Science
• Class 10 English
• Class 11 Maths
• Class 11 Computer Science (Python)
• Class 11 English
• Class 12 Maths
• Class 12 English
• Class 12 Economics
• Class 12 Accountancy
• Class 12 Physics
• Class 12 Chemistry
• Class 12 Biology
• Class 12 Computer Science (Python)
• Class 12 Physical Education
• GST and Accounting Course
• Excel Course
• Tally Course
• Finance and CMA Data Course
• Payroll Course

Interesting

• Learn English
• Learn Excel
• Learn Tally
• Learn GST (Goods and Services Tax)
• Learn Accounting and Finance
• GST Tax Invoice Format
• Accounts Tax Practical
• Tally Ledger List
• GSTR 2A - JSON to Excel

Are you in school ? Do you love Teachoo?

We would love to talk to you! Please fill this form so that we can contact you

You are learning...

Chapter 10 Class 9 Herons Formula

Click on any of the links below to start learning from Teachoo ...

Updated for new NCERT - 2023-24 Curriculmn

Get NCERT Solutions of all exercise questions and examples of Chapter 10 Class 9 Herons Formula. Answers to all question have been solved in a step-by-step manner, with videos of all questions available.

We have studied that

Area of triangle  = 1/2 × Base × Height

In questions where Height and Base is given, we can find the area of triangle easily.

But, in cases where all 3 sides are given, how will we find the area?

If all 3 sides are given, we find Area of Triangle using Herons (or Hero's) Formula

By Hero's Formula

Area of triangle = Square root (s (s-a) (s-b) (s-c))

where a,b, c are sides of the triangle

and s = Semi-Perimeter of Triangle

i.e. s = (a+b+c)/2

In this chapter, we will find Area of Triangle using Herons formula

We will also find Area of Quadrilateral by dividing it into two triangles, and then finding Area of triangle using Hero's Formula

Click on an exercise link or a topic link below to start doing the chapter.

Serial order wise

Concept wise.

What's in it?

Hi, it looks like you're using AdBlock :(

Please login to view more pages. it's free :), solve all your doubts with teachoo black.

• CBSE- Herons Formula
• Sample Questions

Herons Formula-Sample Questions

• STUDY MATERIAL FOR CBSE CLASS 9 MATH
• Chapter 1 - Area of Parallelograms and Triangles
• Chapter 2 - Circles
• Chapter 3 - Constructions
• Chapter 4 - Coordinate Geometry
• Chapter 5 - Herons Formula
• Chapter 6 - Introduction to Euclids Geometry
• Chapter 7 - Linear Equations in two variables
• Chapter 8 - Lines and Angles
• Chapter 9 - Number Systems
• Chapter 10 - Polynomials
• Chapter 11 - Probability
• Chapter 12 - Quadrilaterals
• Chapter 13 - Statistics
• Chapter 14 - Surface Areas and Volumes
• Chapter 15 - Triangles

NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

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

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1 are part of NCERT Solutions for Class 9 Maths . Here we have given NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1.

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.2

For surface IV and V: Surface V is a right-angled triangle with base 6cm arid height 1.5 cm. Also, area of surface IV = area of surface V = \(\frac { 1 }{ 2 }\) x base x height = (\(\frac { 1 }{ 2 }\) x 6 x 15) cm 2 = 4.5 cm 2 Thus, the total area of the paper used = (area of surface I) + (area of surface II) + (area of surface III) + (area of surface IV) + (area of surface V) = [2.475 + 6.5 + 1.3 + 4.5 + 4.5] cm 2 = 19.275 cm 2 = 19.3 cm 2 (approx.)

For triangle II: Since, diagonal of a square divides it into two congruent triangles. So, area of triangle II = area of triangle I ∴ Area of triangle II = 256 cm 2

(ii) For parallelogram ABED: Let the height of the ∆BCE corresponding to the side EC be h m. Area of a triangle = \(\frac { 1 }{ 2 }\) x base x height ∴ \(\frac { 1 }{ 2 }\) x 15 x h = 84 ⇒ (10 + \(\frac { 82\times 2 }{ 15 }\) = \(\frac { 56 }{ 5 }\) Now, area of a parallelogram = base x height = (10 x \(\frac { 56 }{ 5 }\)) = (2 x 56) m 2 = 112 m 2 So, area of the field = area of ∆BCE + area of parallelogram ABED = 84 m 2 + 112 m 2 = 196 m 2

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula (हीरोन सूत्र) (Hindi Medium) Ex 12.1

NCERT Solutions for Class 9 Maths

• Chapter 1 Number systems
• Chapter 2 Polynomials
• Chapter 3 Coordinate Geometry
• Chapter 4 Linear Equations in Two Variables
• Chapter 5 Introduction to Euclid 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 Probability
• Class 9 Maths (Download PDF)

We hope the NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula Ex 12.1, drop a comment below and we will get back to you at the earliest.

Free Resources

NCERT Solutions

Quick Resources

NCERT Solutions Class 9 Maths Chapter 12 Heron's Formula

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula is a fundamental math concept applied in many fields. Therefore, it is necessary to learn this topic along with understanding its applications. One of the reliable resources to gain this knowledge is by referring to the NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula. The solutions are designed in an efficient way to cover these concepts in detail. By practicing questions and sample problems composed in these solutions, students will quickly gain the key skills required for advanced math studies.

There are some ways and formulas to calculate the area of triangles . Heron’s formula is a useful technique to calculate the area of a triangle when the length of all three sides is given. These Class 9 maths NCERT solutions Chapter 12 Heron’s Formula will help students to understand this concept in detail. For more such facts and formulas , read the detailed solution given below and also find some of these in the exercises given below.

• NCERT Solutions Class 9 Maths Chapter 12 Ex 12.1
• NCERT Solutions Class 9 Maths Chapter 12 Ex 12.2

NCERT Solutions for Class 9 Maths Chapter 12 PDF

Triangles are a foundational shape used in various fields of mathematics and other subjects like physics and geography. Learning to find the area of a triangle using Heron’s formula will enable students to gain a better understanding of the related topics and their applications. More details of this topic can be found in the NCERT solutions Class 9 maths chapter 12 Heron's Formula given below:

☛ Download Class 9 Maths NCERT Solutions Chapter 12 Heron’s Formula

NCERT Class 9 Maths Chapter 12   Download PDF

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

The area of a triangle refers to the space closed within the boundary of a triangle. These solutions will enable students to derive Heron’s formula step-by-step. The concepts explained in these solutions are noteworthy and hold great importance in various spheres of life. The exercise-wise detailed analysis of the NCERT Solutions Class 9 Maths Chapter 12 Heron's Formula is shown below:

• Class 9 Maths Chapter 12 Ex 12.1 - 6 Questions
• Class 9 Maths Chapter 12 Ex 12.2 - 19 Questions

☛ Download Class 9 Maths Chapter 12 NCERT Book

Topics Covered: The topics covered in the Class 9 maths NCERT solutions chapter 12 are as follows: Introduction to Heron’s Formula , Area of a triangle based on its height and base, Area of a triangle using Heron’s Formula, and applications of Heron’s Formula in finding the area of quadrilaterals.

Total Questions: The Class 9 maths chapter 12 Heron's Formula Chapter 12 consists of a total of 15 questions. The students will find some to be easy (3 sums), while others will fall in the moderate (7 sums) and tougher categories (2 sums). 

List of Formulas in NCERT Solutions Class 9 Maths Chapter 12

The NCERT solutions Class 9 maths Chapter 12 involves studying the area of triangles and quadrilaterals , which requires applying different formulas explained in this chapter. Some of the important concepts and formulas listed in this chapter are given below:

• Area of a triangle using Heron’s Formula = A = √{s(s-a)(s-b)(s-c)} , where a, b and c are the length of the three sides of a triangle and s is the semi-perimeter of the triangle given by (a + b + c)/2.
• The area of a quadrilateral whose sides and one diagonal are given can be calculated by

dividing the quadrilateral into two triangles and using Heron’s formula.

Important Questions for Class 9 Maths NCERT Solutions Chapter 12

CBSE Important Questions for Class 9 Maths Chapter 12 Exercise 12.1

CBSE Important Questions for Class 9 Maths Chapter 12 Exercise 12.2

Video Solutions for Class 9 Maths NCERT Chapter 12

NCERT Video Solutions for Class 9 Maths Chapter 12
Video Solutions for Class 9 Maths Exercise 12.1
Video Solutions for Class 9 Maths Exercise 12.2

FAQs on NCERT Solutions Class 9 Maths Chapter 12

Why are ncert solutions class 9 maths chapter 12 important.

NCERT Solutions for Class 9 Maths Chapter 12 explains all the concepts and formulas in detail to quickly gain deep knowledge of this chapter. These solutions provide examples, sample problems, and illustrations that are highly beneficial for students to understand the concepts clearly. The solutions are well-organized guides prepared by a team of experts to deliver precise and accurate knowledge of this lesson.

Do I Need to Practice all Questions Provided in NCERT Solutions Class 9 Maths Heron's Formula?

Learning maths requires practice and perseverance. With the practice of a wide range of questions included in the NCERT solutions class 9 maths chapter 12, students will gain knowledge of the core concepts and learn some creative ways to memorize formulas and data. It will enable them to employ their knowledge of particular formulas in various situations, which is highly useful for facing competitive exams.

What are the Important Topics Covered in Class 9 Maths NCERT Solutions Chapter 12?

The important topics covered in the Class 9 maths NCERT solutions chapter 12 are an introduction to triangles, finding the area of a triangle based on its height and base, and calculating the area of a triangle using Heron’s Formula. Additionally, it also covers the topic of applications of Heron’s Formula in finding the area of quadrilaterals.

How Many Questions are there in NCERT Solutions Class 9 Maths Chapter 12 Heron's Formula?

The Class 9 maths chapter 12 Heron's Formula Chapter 12 consists of a total of 15 questions. All of them are based on this formula and its applications. Therefore, the students should practice and memorize it carefully.

How CBSE Students can utilize NCERT Solutions Class 9 Maths Chapter 12 effectively?

NCERT Solutions Class 9 Maths Chapter 12 comprises interactive illustrations and exercises that will assist students in gaining a better understanding of Heron's Formula in practical situations. The students should read the entire chapter carefully as each concept included in these solutions is vital for understanding facts and formulas applied in this lesson.

Why Should I Practice Class 9 Maths NCERT Solutions Heron's Formula chapter 12?

The NCERT Solutions Class 9 Maths Heron's Formula chapter 12 has been prepared by experts in their respective fields to deliver comprehensive knowledge of each and every concept. The facts and data incorporated in these solutions are compiled to promote accurate knowledge in an easy-to-understand manner. Thus, it is very necessary to practice all questions in the NCERT textbook.

Talk to our experts

1800-120-456-456

NCERT Solutions for Class 9 Maths Chapter 10 Heron's Formula

NCERT Solutions for Maths Chapter 10 Class 9 Heron’s Formula - FREE PDF Download

NCERT Solutions for heron's formula class 9 maths ch 10 Heron’s Formula curated by our subject experts to facilitate a practical and smooth understanding of the concepts related to Heron's Formula. These NCERT Solutions can be accessed anytime and anywhere, at your convenience, to understand the concepts in a better way.  These solutions to each exercise question in the PDF are explained using a clear step-by-step method. It acts as an essential tool for you to prepare the chapter quickly and efficiently during exams. You can download and practice these NCERT Solutions for herons formula Class 9 Maths Chapter 10 to thoroughly understand the concepts covered in the chapter.

Glance on Maths Chapter 10 Class 9 - Heron's Formula

This article deals with Heron's Formula, which is a method to calculate the area of a triangle when the lengths of all three sides are known.

The questions cover topics such as finding the area of a triangle when the sides are given, finding the missing side when the area and two sides are given, and finding the height of a triangle when the area and the base are given.

Chapter 10 Maths Class 9 recalls how to calculate the perimeter of various figures and shapes.

This article contains chapter notes, exercises, explanation videos, links, and important questions for Chapter 10 - Heron's Formula where you can download a FREE PDF.

There is one exercise (6 fully solved questions) in class 9th maths chapter 10 Heron's Formula.

Access Exercise Wise NCERT Solutions for Chapter 10 Maths Class 9

S.No.	Current Syllabus Exercises of Class 9 Maths Chapter 10
1

Exercises Under NCERT Solutions for Class 9 Maths Chapter 10 Heron's Formula

Exercise 10.1: This exercise consists of six questions that are based on the concept of Heron's Formula. The questions cover topics such as finding the area of a triangle when the sides are given, finding the missing side when the area and two sides are given, and finding the height of a triangle when the area and the base are given. The exercise also includes word problems that require the application of Heron's Formula to find the area of triangles.

Access NCERT Solutions Maths Chapter 10 - Heron’s formula

Exercise 10.1 .

1. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side $'a'$. Find the area of the signal board, using Heron’s formula. If its perimeter is $180\ \text{cm}$, what will be the area of the signal board?

Length of the side of traffic signal board $=a$ 

Perimeter of traffic signal board which is an equilateral triangle $=3\times a$

We know that,

$2s=$ Perimeter of the triangle, 

So, $2s=3a$

$\Rightarrow s=\dfrac{3}{2}a$

 Area of triangle can be evaluated by Heron’s formula:

$A=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

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

\[s=\dfrac{a+b+c}{2}\]

Substituting $s=\dfrac{3}{2}a$ in Heron’s formula, we get:

Area of given triangle:

 $A=\sqrt{\dfrac{3}{2}a\left( \dfrac{3}{2}a-a \right)\left( \dfrac{3}{2}a-a \right)\left( \dfrac{3}{2}a-a \right)}$ 

$A=\dfrac{\sqrt{3}}{2}{{a}^{2}}\ \ ......\text{(1)}$

Perimeter of traffic signal board:

$P=180\ \text{cm}$

Hence, side of traffic signal board

$a=\left( \dfrac{180}{3} \right)$

$a=60\ \ ......\text{(2)}$

Substituting Equation (2) in Equation (1), we get:

Area of traffic signal board is $A=\dfrac{\sqrt{3}}{2}{{\left( 60\,cm \right)}^{2}}$

$\Rightarrow A=\left( \dfrac{3600}{4}\sqrt{3} \right)\,c{{m}^{2}}$

$\Rightarrow A=900\sqrt{3}\,c{{m}^{2}}$

Hence, the area of the signal board is $900\sqrt{3}\text{ c}{{\text{m}}^{2}}$.

2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $122m,\,22m,$ and $120m$. The advertisements yield on earning of Rs. $5000\,per\,{{m}^{2}}$ per year. A company hired one of its walls for $3$ months. How much rent did it pay?

Ans: The length of the sides of the triangle are (say $a$, $b$ and $c$)

$a=122\ \text{m}$

$b=22\ \text{m}$

$c=120\ \text{m}$

Perimeter of triangle = sum of the length of all sides

Perimeter of triangle is:

$P=122+22+120$

$P=264\ \text{m}$

$2s=264\ \text{m}$

$s=132\ \text{m}$

Area of triangle can be evaluated by Heron’s formula:

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

So, in this question,

$s=\dfrac{122+22+140}{2}$

Substituting values of $s$ $a$, $b$, $c$ in Heron’s formula, we get:

Area of given triangle $=\left[ \sqrt{132\left( 132-122 \right)\left( 132-22 \right)\left( 132-120 \right)} \right]{{m}^{2}}$

$=\left[ \sqrt{132\left( 10 \right)\left( 110 \right)\left( 12 \right)} \right]\,{{m}^{2}}=1320\,{{m}^{2}}$ 

It is given that:

Rent of $1{{m}^{2}}$ area per year is:

$R=Rs.\text{ 5000/}{{\text{m}}^{2}}$ 

Rent of $1{{m}^{2}}$ area per month will be:

$R=Rs.\ \dfrac{5000}{12}/{{m}^{2}}$

Rent of $1320{{m}^{2}}$ area for $3$ months:

$R=\left( \dfrac{5000}{12}\times 3\times 1320 \right)/{{m}^{2}}$

$\Rightarrow R=Rs.\ 1650000$

Therefore, the total cost rent that company must pay is Rs. $1650000$.

3. There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN”. If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Ans: It is given that the sides of the wall are 15 m, 11 m and 6 m.

So, the semi perimeter of triangular wall (s) = (15+11+6)/2 m = 16 m

Using Heron’s formula,

$A=\sqrt{16(16 - 15)(16 - 11)(16 - 6)}$

$A=\sqrt800 m^2$

$A=20\sqrt2 m^2$

4. Find the area of a triangle two sides of which are $18\mathbf{cm}$ and $10\mathbf{cm}$ and the perimeter is $42cm$.

Ans: Let the length of the third side of the triangle be $x$. 

Perimeter of the given triangle:

Let the sides of the triangle be $a$, $b$ and $c$.

Perimeter of the triangle = sum of all sides

$18+10+x=42$

$\Rightarrow 28+x=42$

$\Rightarrow x=14$

$\Rightarrow s=\dfrac{18+10+14}{2}$

$\Rightarrow s=21cm$

Substituting values of  $s$ $a$, $b$, $c$ in Heron’s formula, we get:

$A=\left[ \sqrt{21\left( 21-18 \right)\left( 21-10 \right)\left( 21-14 \right)} \right]$

$\Rightarrow A=\left[ \sqrt{21\left( 3 \right)\left( 11 \right)\left( 7 \right)} \right]$

$\Rightarrow A=21\sqrt{11}\text{ }c{{m}^{2}}$

Hence, the area of the given triangle is $21\sqrt{11}\text{ }c{{m}^{2}}$.

5. Sides of a triangle are in the ratio of $12:17:25$ and its perimeter is $540cm$. Find its area.

Ans: Let the common ratio between the sides of the given triangle be $x$. 

Therefore, the side of the triangle will be $12x$, $17x$, and $25x$.

It is given that,

Perimeter of this triangle $=540cm$

Perimeter = sum of the length of all sides

$12x+17x+25x=540$

$\Rightarrow 54x=540$

$\Rightarrow x=10$

Sides of the triangle will be:

\[12\times 10=120cm\]

$17\times 10=170cm$

$25\times 10=250cm$

$\Rightarrow s=\dfrac{120+170+250}{2}$

$\Rightarrow s=270cm$

$A=\left[ \sqrt{270\left( 270-120 \right)\left( 270-170 \right)\left( 270-250 \right)} \right]$

$\Rightarrow A=\left[ \sqrt{270\left( 150 \right)\left( 100 \right)\left( 20 \right)} \right]$

$\Rightarrow A=9000c{{m}^{2}}$

Therefore, the area of this triangle is $9000\,c{{m}^{2}}.$

6. An isosceles triangle has perimeter $30cm$ and each of the equal sides is $12cm$. Find the area of the triangle.

Ans: Let the third side of this triangle be $x$.

Measure of equal sides is $12cm$ as the given triangle is an isosceles triangle.

Perimeter of triangle, $P=30cm$

Perimeter of triangle = Sum of the sides 

$12+12+x=30$

$\Rightarrow x=6cm$

$\Rightarrow s=\dfrac{12+12+6}{2}$

$\Rightarrow s=15cm$

Substituting values of $s$ $a$, $b$, $c$ in Heron’s formula we get:$A=\left[ \sqrt{15\left( 15-12 \right)\left( 15-12 \right)\left( 15-6 \right)} \right]$

$\Rightarrow A=\left[ \sqrt{15\left( 3 \right)\left( 3 \right)\left( 9 \right)} \right]$

$\Rightarrow A=9\sqrt{15}c{{m}^{2}}$

Hence, the area of the given isosceles triangle is $9\sqrt{15}c{{m}^{2}}$.

Overview of Deleted Syllabus for CBSE Class 9 Maths Heron's Formula

Chapter	Dropped Topics
Heron’s Formula	10.1 Introduction
	10.3 Application of Heron’s formula in finding areas of quadrilaterals.

Class 9 Maths Chapter 10: Exercises Breakdown

Exercise	Number of Questions
Exercise 10.1	6 Questions & Solutions (2 Short Answers, 2 Long Answers, 2 Very Long Answers)

NCERT Solutions for Class 9 Maths Chapter 10 - Heron's Formula provides a necessary resource for students diving into the world of geometry and trigonometry. In this chapter, students are equipped with Heron's Formula, a powerful method for easily calculating triangle areas. Mastering this chapter not only enhances their understanding of geometry but also improves their problem-solving abilities. Heron's Formula is a valuable addition to their mathematical toolkit, setting a solid foundation for future studies in mathematics and various real-life scenarios where the calculation of triangle areas is essential.

Other Study Material for CBSE Class 9 Maths Chapter 10

S. No.	Important Links for Chapter 10 Circles
1
2
3
4

Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

S.No.	NCERT Solutions Class 9 Chapter-wise Maths PDF
1
2
3
4
5
6
7
8
9
10
11

FAQs on NCERT Solutions for Class 9 Maths Chapter 10 Heron's Formula

1. How to find altitude in Heron's Formula class 9?

In class 9 maths herons formula , Heron's formula itself doesn't directly calculate the altitude of a triangle. However, it can be used along with the concept of area to find the altitude in a scalene triangle (one with all sides unequal). Here's the process:

Heron's formula gives the area (A) of a triangle with sides a, b, and c as:

A = $\sqrt{s(s-a)(s-b)(s-c)}$

where s is the semi-perimeter (s = (a + b + c) / 2)

The area of a triangle can also be calculated as 1/2 * base * height (where base is the side along which the altitude is drawn, and the height is the altitude itself).

By equating these two expressions for area and solving for the height (h) with base (b) known, you can get the formula for altitude using Heron's formula:

h = 2 * $\sqrt{s(s-a)(s-b)(s-c)}$ / b

2. Is Heron's formula applicable for all triangles in chapter 10 class 9 maths solutions pdf?

Yes, In class 9 ch 10 , Heron's formula is applicable to all triangles, irrespective of their type (scalene, isosceles, or equilateral). As long as you know the lengths of all three sides, you can use the formula to find the area of the triangle.

3. What is the real-life application of Heron's formula?

Suppose you have to calculate the area of a triangular land. What is the probability that the area is of a regular shape? It is impossible that you will come across lands with regular spaces and sizes, and this is where Heron’s formula comes into use. To calculate the area of real-life objects, the best way to find the exact area of the land is to use Heron’s formula.

4. What is Heron's formula?

Triangle is a three-dimensional closed shape. Heron’s formula calculates the area of a triangle when the length of all three sides is given. Using Heron's formula, we can calculate the area of any triangle, be it a scalene, isosceles or equilateral triangle. For example, the sides of a triangle are given as a, b, and c. Using Heron’s formula, the area of the triangle can be calculated by Area = √S (S-a)(S-b)(S-c) where s is the semi-perimeter of the triangle.

5. How do you solve Heron's formula questions?

To solve questions based on Heron’s formula, you need to remember Heron’s formula, Area= Area= √S (S-a)(S-b)(S-c). Here, ‘s’ is the semi-perimeter of the triangle, and a, b, and c are the lengths of the sides of the triangle. The semi perimeter is denoted by S. It can be calculated by using the formula: S = a+b+c/2. By substituting the values given in these formulas, you can calculate the area of a triangle.

6. What is the meaning of s in Heron's formula?

In Heron’s formula, Area = √S (S-a)(S-b)(S-c), where ‘s’ stands for the semi-perimeter of the triangle whose area we need to calculate. Semi-perimeter can be calculated by the given formula: S = a+b+c/2. To learn more about the semi-perimeter and its usage in calculating the area of a triangle, you can download the Vedantu app or check out the official website of  Vedantu .

7. What is a Semi-Perimeter?

In Geometry, the Semi-perimeter of any polygon is half of its perimeter. In Class 9, Chapter 12, Heron’s Formula explains the semi-perimeter of a triangle. Semi-perimeter is denoted by ‘s’ in Heron’s formula, which is Area= √s  (s-a)(s-b)(s-c) ‘s’ stands for semi-perimeter, which can be calculated by the given formula: s = a+b+c/2, where a, b, and c are the sides of the triangle of which the area has to be calculated.

8. What is a Semi-Perimeter?

In Geometry, the Semi-perimeter of any polygon is half of its perimeter. In Class 9, Chapter 12, Heron’s Formula explains the semi-perimeter of a triangle. Semi-perimeter is denoted by ‘s’ in Heron’s formula, which is Area= √s  (s-a)(s-b)(s-c)

‘s’ stands for semi-perimeter, which can be calculated by the given formula:

s = a+b+c/2, where a, b and c are the sides of the triangle of which the area has to be calculated.

NCERT Solutions for Class 9 Maths

Ncert solutions for class 9.

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

Ncert solutions for class 9 maths chapter 12 heron's formula| pdf download.

• Exercise 12.1 Chapter 12 Class 9 Maths NCERT Solutions
• Exercise 12.2 Chapter 12 Class 9 Maths NCERT Solutions

NCERT Solutions for Class 9 Maths Chapters:

What are the benefits of NCERT Solutions for Chapter 12 Heron's Formula Class 9 NCERT Solutions?

How many exercises in chapter 12 heron's formula, find the length of a diagonal of a square whose side is 2 cm., find the height of an equilateral triangle whose side is 2 cm., contact form.

• NCERT Solutions
• NCERT Class 9
• NCERT 9 Maths
• Chapter 12: Herons Formula

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

Ncert solutions class 9 maths chapter 12 – download free pdf.

* According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 10.

NCERT Solutions for Class 9 Maths Chapter 12 – Heron’s Formula is provided here. Heron’s formula is a fundamental concept that finds significance in countless areas and is included in the CBSE Syllabus of Class 9 Maths. Therefore, it is imperative to have a clear grasp of the concept. And one of the best ways to do just that is by referring to the NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula. These solutions are designed by knowledgeable teachers with years of experience in accordance with the latest update on the CBSE syllabus 2023-24. The NCERT Solutions for Class 9 aim at equipping the students with detailed and step-wise explanations for all the answers to the questions given in the exercises of this Chapter.

Download Exclusively Curated Chapter Notes for Class 9 Maths Chapter – 12 Heron’s Formula

Download most important questions for class 9 maths chapter – 12 heron’s formula.

Hence, one of the best guides you could adapt to your study needs is NCERT Solutions . Relevant topics are presented in an easy-to-understand format, barring the use of any complicated jargon. Furthermore, its content is updated as per the last CBSE syllabus and its guidelines.

• 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 Introduction to Probability

Download PDF of NCERT Solutions Class 9 Maths Chapter 12 Heron’s Formula

carouselExampleControls111

List of Exercises in NCERTclass 9 Maths Chapter 12:

• Exercise 12.1 Solutions – 6 Questions
• Exercise 12.2 Solutions – 9 Questions

Access Answers of NCERT Class 9 Maths Chapter 12 – Heron’s Formula

Exercise: 12.1 (Page No: 202)

1. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?

Side of the signal board = a

Perimeter of the signal board = 3a = 180 cm

∴ a = 60 cm

Semi perimeter of the signal board (s) = 3a/2

By using Heron’s formula,

Area of the triangular signal board will be =

2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m (see Fig. 12.9). The advertisements yield an earning of ₹5000 per m 2 per year. A company hired one of its walls for 3 months. How much rent did it pay?

The sides of the triangle ABC are 122 m, 22 m and 120 m respectively.

Now, the perimeter will be (122+22+120) = 264 m

Also, the semi perimeter (s) = 264/2 = 132 m

Using Heron’s formula,

Area of the triangle =

We know that the rent of advertising per year = ₹ 5000 per m 2

∴ The rent of one wall for 3 months = Rs. (1320×5000×3)/12 = Rs. 1650000

3. There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see Fig. 12.10 ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

It is given that the sides of the wall as 15 m, 11 m and 6 m.

So, the semi perimeter of triangular wall (s) = (15+11+6)/2 m = 16 m

Area of the message =

= √[16(16-15)(16-11) (16-6)] m 2

= √[16×1×5×10] m 2 = √800 m 2

4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42cm.

Assume the third side of the triangle to be “x”.

Now, the three sides of the triangle are 18 cm, 10 cm, and “x” cm

It is given that the perimeter of the triangle = 42cm

So, x = 42-(18+10) cm = 14 cm

∴ The semi perimeter of triangle = 42/2 = 21 cm

Area of the triangle,

= √[21(21-18)(21-10)(21-14)] cm 2

= √[21×3×11×7] m 2

= 21√11 cm 2

5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.

The ratio of the sides of the triangle are given as 12 : 17 : 25

Now, let the common ratio between the sides of the triangle be “x”

∴ The sides are 12x, 17x and 25x

It is also given that the perimeter of the triangle = 540 cm

12x+17x+25x = 540 cm

54x = 540cm

Now, the sides of triangle are 120 cm, 170 cm, 250 cm.

So, the semi perimeter of the triangle (s) = 540/2 = 270 cm

Area of the triangle

= 9000 cm 2

6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

First, let the third side be x.

It is given that the length of the equal sides is 12 cm and its perimeter is 30 cm.

So, 30 = 12+12+x

∴ The length of the third side = 6 cm

Thus, the semi perimeter of the isosceles triangle (s) = 30/2 cm = 15 cm

= √[15(15-12)(15-12)(15-6)] cm 2

= √[15×3×3×9] cm 2

= 9√15 cm 2

Exercise: 12.2 (Page No: 206)

1. A park, in the shape of a quadrilateral ABCD, has C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

First, construct a quadrilateral ABCD and join BD.

We know that

C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m

The diagram is:

Now, apply Pythagoras theorem in ΔBCD

BD 2 = BC 2 +CD 2

⇒ BD 2 = 12 2 +5 2

⇒ BD 2 = 169

⇒ BD = 13 m

Now, the area of ΔBCD = (½ ×12×5) = 30 m 2

The semi perimeter of ΔABD

(s) = (perimeter/2)

= (8+9+13)/2 m

= 30/2 m = 15 m

Area of ΔABD

= 6√35 m 2 = 35.5 m 2 (approximately)

∴ The area of quadrilateral ABCD = Area of ΔBCD+Area of ΔABD

= 30 m 2 +35.5m 2 = 65.5 m 2

2. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

First, construct a diagram with the given parameter.

Now, apply Pythagorean theorem in ΔABC,

AC 2 = AB 2 +BC 2

⇒ 5 2 = 3 2 +4 2

Thus, it can be concluded that ΔABC is a right angled at B.

So, area of ΔBCD = (½ ×3×4) = 6 cm 2

The semi perimeter of ΔACD (s) = (perimeter/2) = (5+5+4)/2 cm = 14/2 cm = 7 m

Now, using Heron’s formula,

Area of ΔACD

= 2√21 cm 2 = 9.17 cm 2 (approximately)

Area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD = 6 cm 2 +9.17 cm 2 = 15.17 cm 2

3. Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.

For the triangle I section:

It is an isosceles triangle and the sides are 5 cm, 1 cm and 5 cm

Perimeter = 5+5+1 = 11 cm

So, semi perimeter = 11/2 cm = 5.5 cm

= √[5.5(5.5- 5)(5.5-5)(5.5-1)] cm 2

= √[5.5×0.5×0.5×4.5] cm 2

= 0.75√11 cm 2

= 0.75 × 3.317cm 2

= 2.488cm 2 (approx)

For the quadrilateral II section:

This quadrilateral is a rectangle with length and breadth as 6.5 cm and 1 cm respectively.

∴ Area = 6.5×1 cm 2 =6.5 cm 2

For the quadrilateral III section:

It is a trapezoid with 2 sides as 1 cm each and the third side as 2 cm.

Area of the trapezoid = Area of the parallelogram + Area of the equilateral triangle

The perpendicular height of the parallelogram will be

And, the area of the equilateral triangle will be (√3/4×a 2 ) = 0.43

∴ Area of the trapezoid = 0.86+0.43 = 1.3 cm 2 (approximately).

For triangle IV and V:

These triangles are 2 congruent right angled triangles having the base as 6 cm and height 1.5 cm

Area triangles IV and V = 2×(½×6×1.5) cm 2 = 9 cm 2

So, the total area of the paper used = (2.488+6.5+1.3+9) cm 2 = 19.3 cm 2

4. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.

It is given that the parallelogram and triangle have equal areas.

The sides of the triangle are given as 26 cm, 28 cm and 30 cm.

So, the perimeter = 26+28+30 = 84 cm

And its semi perimeter = 84/2 cm = 42 cm

Now, by using Heron’s formula, area of the triangle =

= √[42(42-26)(42-28)(42-30)] cm 2

= √[42×16×14×12] cm 2

Now, let the height of parallelogram be h.

As the area of parallelogram = area of the triangle,

28 cm× h = 336 cm 2

∴ h = 336/28 cm

So, the height of the parallelogram is 12 cm.

5. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

Draw a rhombus-shaped field first with the vertices as ABCD. The diagonal AC divides the rhombus into two congruent triangles which are having equal areas. The diagram is as follows.

Consider the triangle BCD,

Its semi-perimeter = (48 + 30 + 30)/2 m = 54 m

Area of the ΔBCD =

∴ Area of field = 2 × area of the ΔBCD = (2 × 432) m 2 = 864 m 2

Thus, the area of the grass field that each cow will be getting = (864/18) m 2 = 48 m 2

6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig.12.16), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

For each triangular piece, The semi perimeter will be

s = (50+50+20)/2 cm = 120/2 cm = 60cm

Area of the triangular piece

= √[60(60-50)(60-50)(60-20)] cm 2

= √[60×10×10×40] cm 2

= 200√6 cm 2

∴ The area of all the triangular pieces = 5 × 200√6 cm 2 = 1000√6 cm 2

7. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much paper of each shade has been used in it?

As the kite is in the shape of a square, its area will be

A = (½)×(diagonal) 2

Area of the kite = (½)×32×32 = 512 cm 2.

The area of shade I = Area of shade II

512/2 cm 2 = 256 cm 2

So, the total area of the paper that is required in each shade = 256 cm 2

For the triangle section (III),

The sides are given as 6 cm, 6 cm and 8 cm

Now, the semi perimeter of this isosceles triangle = (6+6+8)/2 cm = 10 cm

By using Heron’s formula, the area of the III triangular piece will be

= √[10(10-6)(10-6)(10-8)] cm 2

= √(10×4 ×4×2) cm 2

= 8√5 cm 2 = 17.92 cm 2 (approx.)

8. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm 2 .

The semi perimeter of the each triangular shape = (28+9+35)/2 cm = 36 cm

The area of each triangular shape will be

= 36√6 cm 2 = 88.2 cm 2

Now, the total area of 16 tiles = 16×88.2 cm 2 = 1411.2 cm 2

It is given that the polishing cost of tiles = 50 paise/cm 2

∴ The total polishing cost of the tiles = Rs. (1411.2×0.5) = Rs. 705.6

9. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

First, draw a line segment BE parallel to the line AD. Then, from B, draw a perpendicular on the line segment CD.

Now, it can be seen that the quadrilateral ABED is a parallelogram. So,

AB = ED = 10 m

AD = BE = 13 m

EC = 25-ED = 25-10 = 15 m

Now, consider the triangle BEC,

Its semi perimeter (s) = (13+14+15)/2 = 21 m

By using Heron’s formula,

Area of ΔBEC =

We also know that the area of ΔBEC = (½)×CE×BF

84 cm 2 = (½)×15×BF

BF = (168/15) cm = 11.2 cm

So, the total area of ABED will be BF×DE i.e. 11.2×10 = 112 m 2

∴ Area of the field = 84+112 = 196 m 2

Chapter 12 Heron’s Formula belongs to Unit 5: Mensuration. This unit carries a total of 13 marks out of 100. Therefore, this is an important chapter and should be studied thoroughly. The important topics that are covered in this chapter are:

• Area of a Triangle – by Heron’s Formula
• Application of Heron’s Formula in Finding Areas of Quadrilaterals

Heron’s formula helps us to find the area of a triangle with 3 side lengths. Besides the formula, Heron also contributed in other ways – the most notable one being the inventor of the very first steam engine called the Aeolipile. However, Heron couldn’t find any practical applications for it; instead, it ended up being used as a toy and an object of curiosity for the ancient Greeks.

Explore more about Heron’s Formula and learn how to solve various kinds of problems only on NCERT Solutions For Class 9 Maths . It is also one of the best academic resources to revise for your exams.

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

• Well-structured content
• Relevant formulas are highlighted
• Easy-to-understand language and jargon-free explanations
• Designed by qualified teachers
• Latest questions with solutions from the updated syllabus
• A thorough analysis of previous year question papers
• Access to other learning resources, such as sample papers and more

The expert faculty team of members have formulated the solutions in a lucid manner to improve the problem-solving abilities of the students. For a more clear idea about Heron’s Formula, students can refer to the study materials available at BYJU’S.

• RD Sharma Solutions for Class 9 Maths Chapter 12 – Heron’s Formula

Disclaimer:

Dropped Topics –  12.1 Introduction and 12.3 Application of Heron’s formula in finding areas of quadrilaterals.

Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 12

What is the main aim of ncert solutions for class 9 maths chapter 12, what are the key benefits of learning ncert solutions for class 9 maths chapter 12, why should we refer to ncert solutions for class 9 maths chapter 12.

FORMULAS Related Links

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Request OTP on Voice Call

Post My Comment

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.