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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 Jul 02, 2025 04:36 PM IST

In the world of triangles, Heron's Formula is the golden key to the unknown space within the triangle(s). Everyone knows their own ways of finding the area of a triangle as intended, either through the general area triangle formula or the specially intended formula for an Equilateral triangle. Did you ever stop and ask yourself how you would find the area of a triangle if it were of a scalene nature (different lengths for all three sides), and you had zero idea how to find an altitude? That's where we turn to Heron's formula, and it helps us find the triangle's area. Heron's formula can be used in the actual world, and they can help us to figure out the area of irregular plots. This NCERT solutions chapter effectively breaks it down with real examples of questions and makes it understandable.

This Story also Contains
  1. Heron's Formula Class 9 Questions And Answers PDF Free Download
  2. NCERT Solutions for Class 9 Maths Chapter 10: Exercise Questions
  3. Class 9 Maths NCERT Chapter 10: Extra Question
  4. Heron's Formula Class 9 Chapter 10: Topics
  5. Heron's Formula Class 9 Solutions - Important Formulae
  6. Approach to Solve Questions of Heron's Formula Class 9
  7. NCERT Solutions for Class 9 Mathematics: Chapter-wise
NCERT Solutions for Class 9 Maths Chapter 10 Heron's Formula
NCERT Solutions for Class 9 Maths Chapter 10 Heron's Formula

Heron’s formula is the poet's way of determining the area of a triangle. Heron's formula is not just important for grade 9 board exams, but also for higher grade tests and competitive exams as well. These Heron's formula class 9 NCERT solutions align with the current guidelines by CBSE for study, and they have been developed by Careers360 teachers with many years of experience in this area. Students can make an effort to solve the exercises in the NCERT textbooks on their own before validating their answers with the well-constructed solutions in Heron's formula class 9. The most current NCERT solutions for Class 9 Maths not only deal with different types of triangles, but they also take the concept further to quadrilateral problem situations in the real world. An added benefit of the NCERT Solutions for Class 9 materials is to help students apply the formula correctly, especially under test conditions, although the application-based or word problems can also become test questions.

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Heron's Formula Class 9 Questions And Answers PDF Free Download

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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= Perimeter 2=1802=90 cm
Use Heron's Formula to find the area, that is:

 Area =s(sa)(sb)(sc)

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

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=16 m

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 m2msup

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

So, a+b+c=42cm
c=421810=14 cm
Thus, the semi-perimeter of the triangle will be:

s= Perimeter 2=42 cm2=21 cm
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 cm2msup

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=540 cmx=10

Therefore, the sides of the triangle are:

a=12×10=120 cmb=17×10=170 cm
c=25×10=250 cm

The semi-perimeter of the triangle is:

s=540 cm2=270 cm
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=9000 cm2msup

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+12a=6 cm
So, the semi-perimeter of the triangle is:

s= Perimeter 2=12×30 cm=15 cm
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 cm2msup

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=18 cm Area =s(sa)(sb)(sc)=18(189)(1810)(1817)=18×9×8×1=1296=36 cm2msup

Heron's Formula Class 9 Chapter 10: Topics

The topics discussed in the NCERT Solutions for class 9, chapter 10, Heron's Formula, are:

  • Area of a Triangle — by Heron’s Formula
  • Summary

Heron's 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 to32 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.

Approach to Solve Questions of Heron's Formula Class 9

These steps will assist students in confidently handling NCERT Class 9 Heron's Formula questions.

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

We at Careers360 compiled all the NCERT class 9 Maths solutions in one place for easy student reference. The following links will allow you to access them.

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 students if they are not able to solve the NCERT problems on their own. These solutions also provide detailed explanations, giving students conceptual clarity.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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