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NCERT Solutions for Class 9 Maths Chapter 10 - Heron’s Formula Exercise 10.1

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NCERT Solutions for Class 9 Maths Chapter 10 Heron’s Formula Exercise 10.1 - FREE PDF Download

NCERT Solutions for Class 9 Maths Chapter 10: Heron's Formula by Vedantu introduces students to the useful method for finding the area of a triangle when the lengths of all three sides are known. One of the easiest ways to grasp the concept is by practising the numerical problems given in Ex 10.1 class 9 maths NCERT Solutions with the help of Vedantu’s free PDF study material. Vedantu’s solutions provide clear explanations and detailed steps to help you understand and apply Heron's Formula correctly. By practising these problems, you'll gain confidence in solving triangle area problems and improve your overall understanding of geometry.

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Table of Content
1. NCERT Solutions for Class 9 Maths Chapter 10 Heron’s Formula Exercise 10.1 - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 10 Exercise 10.1 Class 9 | Vedantu
3. Formulas Used in Class 9 Chapter 10 Exercise 10.1
4. Access NCERT Solutions for Maths Class 9 Chapter 10 Heron’s Formula Exercise 10.1
    4.1Exercise 10.1
5. Conclusion:
6. CBSE Class 9 Maths Chapter 10 Other Study Materials
7. Chapter-Specific NCERT Solutions for Class 9 Maths
8. Important Study Materials for CBSE Class 9 Maths
FAQs


Glance on NCERT Solutions Maths Chapter 10 Exercise 10.1 Class 9 | Vedantu

  • NCERT Solutions Maths Chapter 10 Exercise 10.1 Class 9 includes Area of triangle and calculating area of triangle by Heron’s Formula.

  • Heron’s Formula is used to calculate the area of a triangle when all the three sides of a triangle are known.

  • Formula to calculate area of a triangle is $\sqrt{s\left ( s-a \right )\left ( s-b \right )\left ( s-c \right )}$

  • Where, s is a semi-perimeter and a,b,c are sides of a triangle.

  • Calculating the semi-perimeter by using the formula $s=\frac{a+b+c}{2}$

  • To find area of different types of triangles is also possible by using Heron’s formula

  • In class 9 Heron's Formula 10.1 there are 6 fully solved questions that helps you to understand the Heron’s Formula and how to use Heron’s Formula


Formulas Used in Class 9 Chapter 10 Exercise 10.1

  • Heron's Formula Class 9 Solutions are based on Area of a triangle =  $\sqrt{s\left ( s-a \right )\left ( s-b \right )\left ( s-c \right )}$

  • Formula to calculate semi-perimeter(s) = $s=\frac{a+b+c}{2}$

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NCERT Solutions for Class 9 Maths Chapter 10 - Heron’s Formula Exercise 10.1
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Access NCERT Solutions for Maths Class 9 Chapter 10 Heron’s Formula Exercise 10.1

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\[\mathbf{180}\text{ }\mathbf{cm}\] , what will be the area of the signal board?

Ans:

Side of traffic signal board$=a$ 

Perimeter of traffic signal board $=3\left( a \right)$ 

 By Heron’s formula, 

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

Area of given triangle $=\sqrt{\frac{3}{2}\left( \frac{3}{2}a-a \right)\left( \frac{3}{2}a-a \right)\left( \frac{3}{2}a-a \right)}$ 

$=\sqrt{\frac{3}{2}a\left( \frac{a}{2} \right)\left( \frac{a}{2} \right)\left( \frac{a}{2} \right)}$ 

$=\frac{\sqrt{3}}{2}{{a}^{2}}$ 

Area of given triangle$=\frac{\sqrt{3}}{2}{{a}^{2}}$ …..(1)

Perimeter of traffic signal board $=180cm$ 

Side of traffic signal board $\left( a \right)=\left( \frac{180}{3} \right)cm=60cm$

Using equation (1), area of traffic signal $=\frac{\sqrt{3}}{2}\left( 60\,c{{m}^{2}} \right)$ 

$=\left( \frac{3600}{4}\sqrt{3} \right)c{{m}^{2}}=900\sqrt{3}c{{m}^{2}}$ 


2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are\[122\text{ }m,\text{ }22\text{ }m\] , and \[\mathbf{120}\text{ }\mathbf{m}\] (see the given figure). The advertisements yield an earning of \[\mathbf{Rs}.\text{ }\mathbf{5000}\text{ }\mathbf{per}\text{ }\mathbf{m2}\] per year. A company hired one of its walls for \[\mathbf{3}\] months. How much rent did it pay?


Flyover


Flyover


Ans:

The sides of the triangle (i.e.,\[a,\text{ }b,\text{ }c\] ) are of \[122\text{ }m,\text{ }22\text{ }m,\] and \[120\text{ }m\] respectively.

Perimeter of triangle$=\left( 122+22+120 \right)m$ 

$2s=264m$ 

$s=132m$ 

By Heron’s formula,

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

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

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

Rent of $1\,{{m}^{2}}$ area per year$=Rs.5000$ 

Rent of $1\,{{m}^{2}}$ area per month$=Rs.\frac{5000}{12}$ 

Rent of \[1320\text{ }{{m}^{2}}\] area for \[3\]  months $=\frac{5000}{12}\left( 3 \right)\left( 1320 \right)=1650000$

Therefore, the company had to pay $Rs.1650000$  


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.


A black and blue triangle with text

Description automatically generated


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)} $

$A = \sqrt 8 00{m^2}$

$A = 20\sqrt 2 {m^2}$.


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

Ans:

Let the third side of the triangle be\[x\] .

Perimeter of the given triangle \[=\text{ }42\text{ }cm\text{ }\]

\[18\text{ }cm+10\text{ }cm+x=42\] 

$x=14cm$ 

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

By Heron’s formula, Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{21\left( 21-18 \right)\left( 21-10 \right)\left( 21-14 \right)}c{{m}^{2}}$ 

$=\sqrt{21\left( 3 \right)\left( 11 \right)\left( 7 \right)}$

$=21\sqrt{11}c{{m}^{2}}$ 


5. Sides of a triangle are in the ratio of \[\mathbf{12}:\mathbf{17}:\mathbf{25}\] and its perimeter is\[\mathbf{540}\text{ }\mathbf{cm}\] . 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 \[12x,\text{ }17x,\] and\[25x\] .

Perimeter of this triangle \[=540\text{ }cm\] 

\[\Rightarrow 12x+17x+25x=540\text{ }cm\] 

\[\Rightarrow 54x=540\text{ }cm\] 

\[\Rightarrow x=10cm\]

Sides of the triangle will be\[120\text{ }cm,\text{ }170\text{ }cm,\text{ }250\text{ }cm\].

$s=\frac{perimeter\,of\,triangle}{2}=\frac{540\,cm}{2}=270cm$ 

By Heron’s formula, Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{270\left( 270-120 \right)\left( 270-170 \right)\left( 270-250 \right)}c{{m}^{2}}$ 

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

$=9000c{{m}^{2}}$ 

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


6. An isosceles triangle has perimeter \[\mathbf{30}\text{ }\mathbf{cm}\] and each of the equal sides is\[\mathbf{12}\text{ }\mathbf{cm}\] . Find the area of the triangle.

Ans:

Let the third side of this triangle be\[x\] .

Perimeter of triangle \[=\text{ }30\text{ }cm\] 

\[\Rightarrow 12\text{ }cm+12\text{ }cm+x=30\text{ }cm\] 

\[\Rightarrow x=6\text{ }cm\] 

$s=\frac{perimeter\,of\,triangle}{2}=\frac{30\,cm}{2}=15cm$ 

By Heron’s formula, Area of triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$

Area of given triangle $=\sqrt{15\left( 15-12 \right)\left( 15-12 \right)\left( \left( 15-6 \right) \right)}c{{m}^{2}}$ 

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

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


Conclusion:

NCERT Solutions for Class 9 Maths Ch 10 Ex 10.1 Heron’s Formula by Vedantu are vital for mastering the technique of finding the area of triangles when the lengths of all three sides are known. Heron's Formula is a practical tool that simplifies this process without needing to know the height of the triangle. Focus on accurately calculating the semi-perimeter and correctly applying the formula to find the area. By diligently working through Vedantu’s step-by-step solutions, you can ensure a thorough understanding and ability to tackle similar problems in your exams.


CBSE Class 9 Maths Chapter 10 Other Study Materials



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.



Important Study Materials for CBSE Class 9 Maths

FAQs on NCERT Solutions for Class 9 Maths Chapter 10 - Heron’s Formula Exercise 10.1

1.  What are the steps that are used to calculate the Heron’s Formula?

Step 1: Calculate the semi-perimeter 𝑠 by the formula $s=\frac{a+b+c}{2}$

Step 2: Note the values of a,b,c nothing but sides of a given triangle

Step 3: Substitute the values of a,b,c and s in the Heron’s Formula

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

2. Is There any Alternate Way to Calculate the Area of a Triangle?

There are actually a good number of alternate ways to calculate the area of a triangle other than the one involving the semi perimeter of the triangle. In the following, we mention only three of them only for the sake of your knowledge. Use of Heron’s formula is most recommendable as it is easiest and commonly used and also, included in the syllabus.

Area = (a × h)/2

Area = a x b x sin (C)/2

Area = a2 x sin (B) x sin (C)/ (2 x sin (B + C))

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

NCERT solutions on chapter 12 Heron’s Formula consists of two exercises 12.1 and 12.2, Exercise 12.1 has 6 questions and Exercise 12.2 has 9 questions in total.  To gain more insights into this chapter, students are advised to make use of the NCERT solutions provided by the Vedantu to help them prepare for their exams. These questions are created by the subject experts. Download the PDF on the Vedantu website or the app for free of cost.

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

By practicing Heron's formula, students will be able to grasp the concept and answer better even if the questions are twisted in the exams. Vedantu offers NCERT solutions with a series of questions and answers prepared for students to practice and revise. To get well versed in this chapter, students can make use of the NCERT solution provided by Vedantu. These questions are solved by highly educated subject experts to help the students clarify their doubts.

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

“Practice makes a man perfect”.  With that quote being said, you do need to practice all the questions provided in the NCERT solution on Heron’s Formula. Working on all the problems given will help you understand the concept better. Practicing the given problems will be a great revision for the exams. To get free access to the NCERT solutions, students need to download the PDF either by clicking here or they can also download the Vedantu app.  

6. Can you please brief me on Chapter 9, Maths Class 9?

Chapter 9, Maths Class 9 focuses on Areas of Parallelograms and triangles. Vedantu offers chapter-wise NCERT solutions for the students to get well versed in the concepts. With the NCERT solutions provided by the Vedantu, students can practise the concepts in this chapter and excel in the exams. These solutions are created by a qualified group of experts. These solutions are free of cost. Students can download them and can access these solutions anytime they want to. 

7. How can I understand chapter 9, class 9 Maths?

We all know that only when you understand the topic, you will be able to answer the questions even if it is twisted. Vedantu helps you get well versed in the concepts with free online live classes and NCERT solutions. Vedantu with their PDF of important questions will assist the students to get through all the chapters. These solutions are created by the subject experts at Vedantu. To download the important questions for this chapter, visit the page-NCERT solutions to get access to the pdf for free.

8. Can Heron's Formula be used for right-angled triangles?

Yes, but it's often easier to use the standard formula for right-angled triangles A = $\frac{1}{2} \times base \times height$. However, Heron's Formula works for right-angled triangles as well.