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

Last updated date: 13th Sep 2024
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## 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.

Table of Content
1. NCERT Solutions for Maths Chapter 10 Class 9 Heron’s Formula - FREE PDF Download
2. Glance on Maths Chapter 10 Class 9 - Heron's Formula
3. Access Exercise Wise NCERT Solutions for Chapter 10 Maths Class 9
4. Exercises Under NCERT Solutions for Class 9 Maths Chapter 10 Heron's Formula
5. Access NCERT Solutions Maths Chapter 10 - Heron’s formula
6. Overview of Deleted Syllabus for CBSE Class 9 Maths Heron's Formula
7. Other Study Material for CBSE Class 9 Maths Chapter 10
8. Chapter-Specific NCERT Solutions for Class 9 Maths
FAQs

## 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.

• 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 NCERT Solutions of Class 9 Maths Heron's Formula Exercise 10.1
Competitive Exams after 12th Science

## 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?

Ans:

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:

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

Where,

$a$ , $b$ 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:

$P=122+22+120$

$P=264\ \text{m}$

We know that,

$2s=$ Perimeter of the triangle,

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

$s=132\ \text{m}$

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

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

Where,

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

$s=\dfrac{a+b+c}{2}$

So, in this question,

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

$s=132\ \text{m}$

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:

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

So,

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,

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

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

$A=\sqrt800 m^2$

$A=20\sqrt2 m^2$

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:

$P=42cm$

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

$a=18cm$

$b=10cm$

$c=xcm$

Perimeter of the triangle = sum of all sides

$18+10+x=42$

$\Rightarrow 28+x=42$

$\Rightarrow x=14$

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

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

Where,

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

$s=\dfrac{a+b+c}{2}$

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

$\Rightarrow s=21cm$

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

Area of given triangle:

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

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

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

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:

$12\times 10=120cm$

$17\times 10=170cm$

$25\times 10=250cm$

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

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

Where,

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

$s=\dfrac{a+b+c}{2}$

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

$\Rightarrow s=270cm$

Area of given triangle:

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

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

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

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.

It is given that,

Perimeter of triangle, $P=30cm$

Perimeter of triangle = Sum of the sides

$12+12+x=30$

$\Rightarrow x=6cm$

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

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

Where,

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

$s=\dfrac{a+b+c}{2}$

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

$\Rightarrow s=15cm$

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)

## Conclusion

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 Class 9 Heron's Formula Important Questions 2 Class 9 Heron's Formula Revision Notes 3 Class 9 Heron's Formula Important Formulas 4 Class 9 Heron's Formula RD Sharma Solutions

## 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.

## 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.