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

Edited By Komal Miglani | Updated on Apr 15, 2025 01:14 AM IST

Everyone already knows how to find the area of a triangle using the general formula or the specific formula for an Equilateral triangle. But have you ever wondered how we can get the area of a triangle when it is a scalene triangle, which means the length of all three sides is different, and you have no clue how to get the altitude? In such cases, Heron’s formula becomes a lifesaver, helping us find the triangle's area. In Class 9 Maths Chapter 10 solutions, students will learn about Heron’s formula, also known as the Hero’s formula. Heron’s formulas have many real-life examples, and they help calculate the area of irregular plots.

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This Story also Contains

Herons Formula Class 9 Questions And Answers PDF Free Download
Herons Formula Class 9 Solutions - Important Formulae
NCERT Solutions for class 9 Mathematics: Chapter wise
Importance of NCERT Solutions for Class 9 Maths Chapter 10: Heron’s Formula
NCERT solutions for class 9 subject-wise
NCERT Books and NCERT Syllabus

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

The concept of Heron’s formula is not only important for class 9 board exams but also for higher-level and competitive exams. These Heron’s formula class 9 NCERT solutions follow the latest CBSE guidelines and have been prepared by Careers360 teachers who have multiple years of experience in this field. Students can try to solve the exercises of NCERT textbooks on their own before checking these well-structured solutions of Heron’s formula class 9. After finishing the exercise, students can practice these NCERT Exemplar Solutions for Heron's Formula for a deeper understanding of this concept. These Heron's Formula Notes can be used as an important revision tool.

Herons Formula Class 9 Questions And Answers PDF Free Download

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Herons Formula Class 9 Solutions - Important Formulae

Triangle:

Semi-perimeter of a triangle $= s = \frac{(a + b + c)}{2}$ ,

Where 's' is the semi-perimeter, and 'a', 'b', and 'c' are the lengths of its sides.

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

Equilateral Triangle:

For an equilateral triangle with side length 'a':

Its perimeter is given by: Perimeter = 3a units.
The altitude (height) of an equilateral triangle is equal to $\frac{\sqrt{3}}{2}$ times its side length, Altitude = $\frac{\sqrt{3}}{2} \times a$ units.
The area of an equilateral triangle is equal to $\frac{\sqrt{3}}{4}$ times the square of its side length, Area = $\frac{\sqrt{3}}{4} \times a^{2}$ units. These formulas are specific to equilateral triangles and are based on their unique properties.

Heron's Formula Class 9 NCERT Solutions (Exercise)

Class 9 Maths chapter 12 Question Answer: Exercise: 12.1
Total Questions: 6
Page number: 134-135

Q1. 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?

Answer:

Given the perimeter of an equilateral triangle is 180 cm.

So, $3 a = 180$
$\Rightarrow a = 60$ cm

Hence, the length of the side is 60 cm.

Now,

Calculating the area of the signal board by Heron's Formula:

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

Where s is the half-perimeter of the triangle and a, b, and c are the sides of the triangle.

Therefore,

$s = \frac{1}{2} \times$ Perimeter $= \frac{1}{2} \times 180 = 90$ cm

$a = b = c = 60$ cm as it is an equilateral triangle.

Substituting the values in Heron's formula, we obtain

$⟹ A = \sqrt{90 (90 - 60) (90 - 60) (90 - 60)} = 900 \sqrt{3}$ cm2

Q2. 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 Rs. 5000 per m 2 per year. A company hired one of its walls for 3 months. How much rent did it pay?

1640600558297
Answer:

From the figure,

The sides of the triangle are:

$a = 122 m, b = 120 m a n d c = 22 m$

The semi-perimeter, s, will be:

$s = \frac{a + b + c}{2} = \frac{122 + 120 + 22}{2} = \frac{264}{2} = 132$ m

Therefore, the area of the triangular side wall will be calculated by Heron's Formula,

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

$= \sqrt{132 (132 - 122) (132 - 120) (132 - 22)}$

$= \sqrt{132 (10) (12) (110)}$

$= \sqrt{(12 \times 11) (10) (12) (11 \times 10)} = 1320$ m2

Given the rent for 1 year (i.e., 12 months) per meter square is Rs. 5000.

Rent for 3 months per meter square will be:

Rs. $5000 \times \frac{3}{12}$

Therefore, for 3 months for 1320 m2 :

Rs. $5000 \times \frac{3}{12} \times 1320 =$ Rs. $16, 50, 000$

Q3. There is a slide in a park. One of its side walls has been painted in some colour with the 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.

1640600584782
Answer:

Given the sides of the triangle are:

$a = 15 m, b = 11 m a n d c = 6 m .$

So, the semi-perimeter of the triangle will be:

$s = \frac{a + b + c}{2} = \frac{15 + 11 + 6}{2} = \frac{32}{2} = 16 m$

Therefore, Heron's formula will be:

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

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

$= \sqrt{16 (1) (5) (10)}$

$= \sqrt{(4 \times 4) (1) (5) (5 \times 2)}$

$= 4 \times 5 \sqrt{2} = 20 \sqrt{2} m^{2}$

Hence, the area painted in colour is $20 \sqrt{2}$ m2.

Q4. Find the area of a triangle, two sides of which are 18 cm and 10 cm, and the perimeter is 42 cm.

Answer:

Given the perimeter of the triangle is $42$ cm
and the sides length $a = 18$ cm and $b = 10$ cm

So, $a + b + c = 42$

$\Rightarrow c = 42 - 18 - 10 = 14$ cm

So, the semi-perimeter of the triangle will be:

$s = \frac{P}{2} = \frac{42}{2} = 21$ cm

Therefore, the area given by Heron's Formula will be,

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

$= \sqrt{21 (21 - 18) (21 - 10) (21 - 14)}$

$= \sqrt{(7 \times 3) (3) (11) (7)}$

$= 21 \sqrt{11}$ cm2

Hence, the area of the triangle is $21 \sqrt{11}$ cm2.

Q5. Sides of a triangle are in the ratio of 12 : 17 : 25, and its perimeter is 540 cm. Find its area.

Answer:

Given the sides of a triangle are in the ratio of $12 : 17 : 25$ and its perimeter is $540 c m$

Let us consider the length of one side of the triangle to be $a = 12 x$

Then, the remaining two sides are $b = 17 x$ and $c = 25 x .$

So, by the given perimeter, we can find the value of x:

Perimeter $= a + b + c = 12 x + 17 x + 25 x = 540$ cm

$⟹ 54 x = 540$ cm

$⟹ x = 10$

So, the sides of the triangle are:

$a = 12 \times 10 = 120$ cm

$b = 17 \times 10 = 170$ cm

$c = 25 \times 10 = 250$ cm

So, the semi-perimeter of the triangle is given by

$s = \frac{540 c m}{2} = 270$ cm

Therefore, using Heron's Formula, the area of the triangle is given by:

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

$= \sqrt{270 (270 - 120) (270 - 170) (270 - 250)}$

$= \sqrt{270 (150) (100) (20)}$

$= \sqrt{81000000} = 9000$ cm2

Hence, the area of the triangle is $9000$ cm2.

Q6. An isosceles triangle has a perimeter of 30 cm, and each of the equal sides is 12 cm. Find the area of the triangle.

Answer:

The perimeter of an isosceles triangle is 30 cm (Given).

The length of the sides, which are equal, is 12 cm.

Let the third side length be 'a' cm.

Then, the Perimeter $= a + b + c$

$\Rightarrow 30 = a + 12 + 12$

$\Rightarrow a = 6 c m$

So, the semi-perimeter of the triangle is given by,

$s = \frac{1}{2} \times$ Perimeter $= \frac{1}{2} \times 30 = 15$ cm

Therefore, using Heron's Formula, calculating the area of the triangle

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

$= \sqrt{15 (15 - 6) (15 - 12) (15 - 12)}$

$= \sqrt{15 (9) (3) (3)}$

$= 9 \sqrt{15}$ cm2

Hence, the area of the triangle is $9 \sqrt{15}$ cm2.

NCERT Solutions for class 9 Mathematics: Chapter wise

NCERT Solutions for Class 9 Mathematics Chapter 1: Number Systems

NCERT Solutions for Class 9 Mathematics Chapter 2: Polynomials

NCERT Solutions for Class 9 Mathematics Chapter 3: Coordinate Geometry

NCERT Solutions for Class 9 Mathematics Chapter 4: Linear Equations in Two Variables

NCERT Solutions for Class 9 Mathematics Chapter 5: Introduction to Euclid’s Geometry

NCERT Solutions for Class 9 Mathematics Chapter 6: Lines and Angles

NCERT Solutions for Class 9 Mathematics Chapter 7: Triangles

NCERT Solutions for Class 9 Mathematics Chapter 8: Quadrilaterals

NCERT Solutions for Class 9 Mathematics Chapter 9: Circles

NCERT Solutions for Class 9 Mathematics Chapter 10: Heron's Formula

NCERT Solutions for Class 9 Mathematics Chapter 11: Surface Areas and Volumes

NCERT Solutions for Class 9 Mathematics Chapter 12: Statistics

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Importance of NCERT Solutions for Class 9 Maths Chapter 10: Heron’s Formula

NCERT Class 9 Maths chapter 10 solutions have grave importance in students' journeys towards excellence in this concept. Here are some important points of why students need these solutions.

These solutions have been updated by following the latest CBSE syllabus and are very well structured.
Every step of the solution has been explained so that students can understand why that happens.
Practicing these solutions will help students to grasp this concept well and reduce errors during the exam.
This solution article contains important links, like NCERT exemplar solutions, NCERT notes, Reference books and the 2025-26 Maths syllabus.
Experts who have multiple years of experience in these topics have solved these problems so that students can learn these concepts easily.

NCERT solutions for class 9 subject-wise

Students can also check these subject-wise solutions. These solutions have explained every step and are written in very easy language.

NCERT Solutions for Class 9 Maths

NCERT Solutions for Class 9 Science

NCERT Books and NCERT Syllabus

The following links can be used to find the latest CBSE syllabus and a reference math book. These are very useful study materials to do well in the exam.

Frequently Asked Questions (FAQs)

1. What are the important topics in chapter Heron's Formula

Area of a Triangle – by Heron’s Formula and Application of Heron’s Formula in finding Areas of Quadrilaterals are two important topics of this chapter. Students can prioritize important topics from the NCERT syllabus and study accordingly to score well in exams. for ease you can study heron's formula class 9 pdf both online and offline mode.

2. What are the key benefits of learning NCERT Solutions for class 9 chapter 12 maths?

Here are the rephrased key benefits of NCERT Solutions for Class 9 Maths Chapter 12:

The solutions for each exercise within the chapter are easily accessible for students to refer to.
The solutions are designed with graphs and illustrations that aid in providing a clear understanding of the mathematical concepts.
The solutions are meticulously prepared by the expert team at Careers360, with a strong emphasis on accuracy.

3. How does the NCERT solutions are helpful ?

NCERT solutions are helpful for the students if they are no able to solve NCERT problems on their own. Also, these solutions are provided in a very detailed manner which will give them conceptual clarity.

4. Where can I find the complete solutions of NCERT for class 9 maths ?

Here you will get the detailed NCERT solutions for class 9 maths by clicking on the link. Practicing these solutions will give you in-depth understanding of concepts which help you to score good marks in the exam.

Also Read

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

NCERT Solutions for Class 9 Maths Chapter 9 Circles

NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

NCERT Solutions for Class 9 Maths Chapter 7 Triangles

NCERT Solutions for Class 9 Maths Chapter 6 Lines And Angles

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

NCERT Exemplar Class 9 Maths Solutions Chapter 10 Circles

NCERT Exemplar Class 9 Maths Solutions Chapter 9 Areas of Parallelograms and Triangles

NCERT Exemplar Class 9 Science Solutions Chapter 9 Force and Laws of Motion

NCERT Exemplar Class 9 Science Solutions chapter 8 Motion

NCERT Exemplar Class 9 Solutions - Subject Wise Problems with Solutions

NCERT Exemplar Class 9 Science Solutions Chapter 11 Work and Energy

NCERT Books for Class 9 Maths 2025-26; Download Free PDF

NCERT Books for Class 9 All Subjects - Download Free PDF

NCERT Books for Class 9 Social Science 2025-26: Download PDF

NCERT Books for Class 9 Hindi 2025-26: Download PDF

NCERT Books for Class 9 Science 2025-26; Download PDF (All Chapters)

NCERT Class 9 Maths Chapter 6 Notes Lines and Angles - Download PDF

Surface Areas And Volumes Class 9th Notes - Free NCERT Class 9 Maths Chapter 13 Notes - Download PDF

NCERT Class 9 Maths Notes - Download Chapter Wise PDFs

Circles Class 9th Notes - Free NCERT Class 9 Maths Chapter 10 Notes - Download PDF

Triangles Class 9th Notes - Free NCERT Class 9th Maths Chapter 7 Notes - Download PDF

NCERT Class 9 Maths Chapter 1 Notes Number System - Download Free PDF

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