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

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

Edited By Komal Miglani | Updated on May 02, 2025 01:09 PM 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 Heron’s formula. Heron’s formulas have many real-life examples, and they help calculate the area of irregular plots. The NCERT Solutions for this chapter simplify the process through practical questions and straightforward explanations.

This Story also Contains
  1. Heron's Formula Class 9 Questions And Answers PDF Free Download
  2. Herons Formula Class 9 Solutions - Important Formulae
  3. NCERT Solutions for Class 9 Maths Chapter 10: Exercise Questions
  4. Class 9 Maths NCERT Chapter 10: Extra Question
  5. Approach to Solve Questions of Heron's Formula Class 9
  6. NCERT Solutions for Class 9 Mathematics: Chapter-wise
  7. NCERT solutions for class 9, Subject-Wise
  8. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 9 Maths Chapter 10 Heron's Formula
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 the NCERT textbooks on their own before checking these well-structured solutions of Heron’s formula class 9. The latest NCERT Solutions for Class 9 Maths covers various triangle types and extends the concept to real-world problems involving quadrilaterals. Along with this, the NCERT Solutions for Class 9 materials help students apply the formula accurately, especially when dealing with application-based or word problems in exams.

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Heron's 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=(a+b+c)2,

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

Area =s(sa)(sb)(sc)

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 32 times its side length, Altitude = 32×a units.

  • The area of an equilateral triangle is equal to 34 times the square of its side length, Area = 34×a2 units. These formulas are specific to equilateral triangles and are based on their unique properties.

NCERT Solutions for Class 9 Maths Chapter 10: Exercise Questions

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

Question 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?

Answer:

Given: The Perimeter of the equilateral triangle is 180 cm.

Therefore, each side of the triangle is:

a = 1803 = 60 cm

Now, calculate the semi-perimeter:

s = Perimeter2=1802 = 90 cm

Use Heron’s Formula to find the area, that is:

Area = s(sa)(sb)(sc)

Thus,

Area = 90(9060)(9060)(9060)=90×30×30×30

Area = 90×27000=2430000=9003 cm2.

Question 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 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,

Let the sides of the triangle be:

a = 122m, b = 120m and c = 22m

Therefore, the semi-perimeter, s, will be

s=a+b+c2=122+120+222=2642=132m

Thus, Area ( using Heron’s formula ):

A=s(sa)(sb)(sc)

=132(132122)(132120)(13222) m2

=132(10)(12)(110) m2

=(12×11)(10)(12)(11×10) m2=1320 m2

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

Thus, rent for 3 months per square meter will be:

Rs. 5000×312

Therefore, for 3 months for 1320 m2:

Rs. 5000×312×1320=Rs. 16,50,000.

Question 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.

1640600584782

Answer:

We are given the sides of the triangle, which are:

a = 15 m, b = 11m and c = 6m

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

s=a+b+c2=15+11+62=322=16m

Therefore, the area painted in colour is:

A=s(sa)(sb)(sc)

=16(1615)(1611)(166)

=16(1)(5)(10)

=(4×4)(1)(5)(5×2)

=4×52=202 m2

Question 4: 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 42cm and the lengths of two sides, a = 18 cm and b = 10 cm.

So, a+b+c=42cm

Or, c=421810=14cm

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

s=Perimeter2=42cm2=21cm

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

A=s(sa)(sb)(sc)

=21(2118)(2110)(2114)

=(7×3)(3)(11)(7)

=2111 cm2

Question 5: 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 cm.

Let, the length of one side of the triangle to be a=12x

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

Thus, by the given perimeter,

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

54x=540cm

x=10

Therefore, the sides of the triangle are:

a=12×10=120cm

b=17×10=170cm

c=25×10=250cm

The semi-perimeter of the triangle is:

s=540cm2=270cm

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

A=s(sa)(sb)(sc)

=270(270120)(270170)(270250)

=270(150)(100)(20)

=81000000=9000cm2

Question 6: 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:

Given, the perimeter of an isosceles triangle is 30 cm and the length of the sides which are equal is 12 cm.

Let the third side length be 'a cm'.

Then, Perimeter=a+b+c

30=a+12+12

a=6cm

So, the semi-perimeter of the triangle is:

s=Perimeter2=12×30cm=15cm

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

A=s(sa)(sb)(sc)

=15(156)(1512)(1512)

=15(9)(3)(3)

=915 cm2


Class 9 Maths NCERT Chapter 10: Extra Question

Question: Find the area of a triangle whose sides are 9 cm, 10 cm, and 17 cm using Heron’s formula.

Answer:

Let the sides be a = 9 cm, b = 10 cm, c = 17 cm

s=a+b+c2=9+10+172=362=18cm

Area=s(sa)(sb)(sc)=18(189)(1810)(1817)

=18×9×8×1=1296=36cm2

Approach to Solve Questions of Heron's Formula Class 9

1. Memorise the Heron’s Formula: Students need to remember the standard formula, which states Area equals the square root of semi-perimeter times its difference with each side. The formula uses s to denote the semi-perimeter of a triangle.

2. Calculate semi-perimeter accurately: The calculation of s=a+b+c2 requires utmost attention because even minor errors can create problems throughout the problem-solving process.

3. Check if a triangle is valid: The triangle inequality rule should be used to verify triangle feasibility before you implement the formula.

4. Work with different triangle types: Apply Heron’s formula on various triangle shapes, including scalene, isosceles and right-angled to enhance your understanding of the concept.

5. Special attention must be given when simplifying square roots: Preparing the value inside the square root for calculation before extracting the root will reduce errors, particularly when working with irrational numbers.

6. Extend to real-world problems: When solving problems about land plots or uneven quadrilaterals, apply Heron’s rule by dividing the shape into two triangles.

NCERT Solutions for Class 9 Mathematics: Chapter-wise

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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 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:

  1. The solutions for each exercise within the chapter are easily accessible for students to refer to.

  2. The solutions are designed with graphs and illustrations that aid in providing a clear understanding of the mathematical concepts.

  3. 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.

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