NCERT Solutions Class 9 Maths Chapter 10 Herons Formula

NCERT Solutions for Class 9 Math Chapter 10 focuses on Heron's Formula, an essential tool for calculating the area of a triangle when the lengths of all three sides are known. Understanding this formula is important for students, as it has practical applications in geometry and various real-life situations. 

With the step-by-step explanation, NCERT Solutions help students to learn all the concepts thoroughly and build confidence in using various formulas. The solutions are designed to simplify complex topics and make learning the concepts easier. Additionally, these resources serve different learning styles, ensuring that every student can find the support they need. 

By practising with these solutions, students will be better prepared for exams and advanced math topics. Discover how to better understand Heron's Formula and develop your mathematical problem-solving abilities by going through the thorough exercises and notes given in this article in PDF format.

1.0Download NCERT Solutions for Class 9 Maths Chapter 10 : Free PDF

Triangles are fundamental shapes in mathematics, used in various fields like physics and geography. You can easily find the area of a triangle using Heron's formula. Understanding Heron's formula will give you a better grasp of triangle-related topics and their practical applications. To help you understand better, we have provided NCERT solutions in PDF format, which you can download and study at your convenience.

2.0Importance of Class 9 Herons Formulas and Its Application in Real-Life

Heron's formula can give you the area whether you need to find the area of a triangular piece of land, though. This is where Heron's formula becomes helpful. Heron's formula is the ideal method for calculating the area of real-world triangles. Since a triangle is a closed shape with three sides, you can with use this formula when you know the lengths of all three sides. Whether the triangle is scalene, isosceles, or equilateral, Heron's formula can give you the area. So, you can see how important it is to know about Heron’s formula. 

Class 9 Maths Chapter 10 Heron's Formulas Subtopics

Before moving forward to understand what are the NCERT solutions, we have provided you the subtopics that have been covered under the topic Heron’s formula so that you must know what you are going to learn in this chapter:

3.0What are NCERT solutions? Brief Overview

NCERT Solutions helps students learn important math concepts in a clear and organised way. They highlight key formulas and use simple language without complicated terms. 

Qualified teachers create these solutions, making sure the content matches the latest syllabus and includes recent questions. Students can also review past and exam papers to prepare better. They can also find helpful resources, like sample papers. 

The expert team designs these solutions to improve students' problem-solving skills. For a better understanding of topics like Heron's Formula, students can check study materials at ALLEN. 

4.0Practice Problems Covered in NCERT Solutions Class 9 Herons Formulas Chapter 10

Below we have provided the types and number of questions present in the chapter 10 class 9 maths solutions pdf. This exercise covered all the topics in this chapter:

5.0NCERT Questions with Solutions for Class 9 Maths Chapter 10 - Detailed Solutions

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 cm , what will be the area of the signal board? Sol. The equilateral triangle each side $=\mathrm{a}$ Its semi perimeter $=\frac{a+a+a}{2}=\frac{3}{2}a$ By Heron's formula, the area of the triangle $=\sqrt{\frac{3}{2}a\times \left(\frac{3}{2}a-a\right)\times \left(\frac{3}{2}a-a\right)\times \left(\frac{3}{2}a-a\right)}$ $=\frac{\sqrt{3}}{4}{\mathrm{a}}^2$ When perimeter of the triangle is 180 cm , we have $3\mathrm{a}=180\mathrm{cm}$ i.e., $\mathrm{a}=60\mathrm{cm}$. Then the area of the triangle $=\frac{\sqrt{3}}{4}(60{)}^2{\mathrm{cm}}^2=900\sqrt{3}{\mathrm{cm}}^2$
2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $122\mathrm{m},22\mathrm{m}$ and 120 m (see Fig.). The advertisements yield an earning of Rs. 5000 per ${\mathrm{m}}^2$ per year. A company hired one of its walls for 3 months. How much rent did it pay?
   
   Sol. Sides of the two equal triangular walls below the bridge are $122\mathrm{m},22\mathrm{m}$ and 120 m . $\mathrm{s}=\frac{122\mathrm{m}+22\mathrm{m}+120\mathrm{m}}{2}=132\mathrm{m}$ Area of one triangular wall $=\sqrt{132\times (132-122)\times (132-22)\times (132-120)}{\mathrm{m}}^2$ $=\sqrt{132\times 10\times 110\times 12}{\mathrm{m}}^2=1320{\mathrm{m}}^2$ Company hired only one wall for 3 months. Thus, earning from advertisements for 3 months at the rate of Rs. 5000 per ${\mathrm{m}}^2$ per year. $=$ Rs. $5000\times \frac{3}{12}\times 1320=$ Rs. $16,50,000$
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.). If the sides of the wall are $15\mathrm{m},11\mathrm{m}$ and 6 m , find the area painted in colour.
   
   Sol. The sides of the triangular wall be 15 m , 11 m and 6 m . $\mathrm{s}=\frac{15\mathrm{m}+11\mathrm{m}+6\mathrm{m}}{2}=16\mathrm{m}$ Area of the wall $=\sqrt{16\times (16-15)\times (16-11)\times (16-6)}{\mathrm{m}}^2$ $=\sqrt{16\times 1\times 5\times 10}{\mathrm{m}}^2$ $=20\sqrt{2}{\mathrm{m}}^2$
4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm . Sol. $\mathrm{a}=18\mathrm{cm}$, $\mathrm{b}=10\mathrm{cm}$, Perimeter $=42\mathrm{cm}$ we have $\mathrm{a}+\mathrm{b}+\mathrm{c}=42$ $\Rightarrow c=14\mathrm{cm}$ $\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}=\frac{42}{2}=21\mathrm{cm}$ Area of $\Delta =\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{21(21-18)(21-10)(21-14)}$ $=\sqrt{21(3)(11)(7)}=3\times 7\times \sqrt{11}$ $=21\sqrt{11}{\mathrm{cm}}^2$
5. Sides of a triangle are in the ratio of 12 : $17:25$ and its perimeter is 540 cm . Find its area. Sol. Let the sides of triangle be $12\mathrm{k},17\mathrm{k},25\mathrm{k}$ Perimeter $=12\mathrm{k}+17\mathrm{k}+25\mathrm{k}=54\mathrm{k}$. $\Rightarrow 54\mathrm{k}=540$ $\mathrm{k}=10$ $\Rightarrow \mathrm{a}=12\times \mathrm{k}=12\times 10=120$ $\mathrm{b}=17\times \mathrm{k}=17\times 10=170$ c $=25\times \mathrm{k}=25\times 10=250$ $\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}=\frac{540}{2}=270$ $\therefore$ Area $=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{270(270-120)(270-170)(270-250)}$ $==\sqrt{5\times 3\times 3\times 2\times 2\times 7}$ $=3\times 30\times 5\times 20{\mathrm{cm}}^2=9000{\mathrm{cm}}^2$
6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm . Find the area of the triangle. Sol.
   
   $\mathrm{a}=12\mathrm{cm}$ $\mathrm{b}=12\mathrm{cm}$ Perimeter $=30\mathrm{cm}$ $\Rightarrow c=30-24=6\mathrm{cm}$ $\mathrm{s}=\frac{30}{2}=15\mathrm{cm}$ $\therefore$ Area $=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{15(15-12)(15-12)(15-6)}$ $=\sqrt{\mathrm{15.3.3.9}}=9\sqrt{15}{\mathrm{cm}}^2$

6.0Benefits of Using NCERT Solutions

Here are the key benefits of NCERT Solutions for Class 9 herons formula:

• Easy to Access: Students can effortlessly access these detailed solutions for each exercise, making it convenient to study and review them whenever required.
• Use of Visual Tools: The solutions include informative graphs and illustrations that improve understanding, allowing students to visualise complex concepts and grasp the material more effectively.
• Prepared by Expert Preparation: The experienced team at ALLEN crafts these comprehensive solutions, ensuring a high level of accuracy and educational value. This helps students build a solid foundation in mathematics.