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.

1.0NCERT Solution Class 9 Maths Chapter 10 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:

5.0Benefits 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.

6.0NCERT Question With Solution

7.0Exercise: 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 $=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{15(15-12)(15-12)(15-6)}$ $=\sqrt{\mathrm{15.3.3.9}}=9\sqrt{15}{\mathrm{cm}}^2$

8.0Exercise: 10.2

1. A park, in the shape of a quadrilateral ABCD , has $\angle \mathrm{C}=90^{\circ },\mathrm{AB}=9\mathrm{m},\mathrm{BC}=12\mathrm{m}$, $\mathrm{CD}=5\mathrm{m}$ and $\mathrm{AD}=8\mathrm{m}$. How much area does it occupy? Sol. Join the diagonal AD of the quadrilateral ABCD.
   
   ${\mathrm{BD}}^2={\mathrm{BC}}^2+{\mathrm{CD}}^2$ $=(12{)}^2+(5{)}^2=169$ $\Rightarrow \mathrm{BD}=\sqrt{169}=13\mathrm{m}$ Area of right angle $△BCD$ $=\frac{1}{2}\times \mathrm{BC}\times \mathrm{CD}=\frac{1}{2}\times 12\times 5{\mathrm{m}}^2=30{\mathrm{m}}^2$ Now, the sides of the $△\mathrm{ABD}$ are $9\mathrm{m},13\mathrm{m}$ and 8 m . semi-perimeter of $△ABD$ $\mathrm{s}=\frac{9\mathrm{m}+13\mathrm{m}+8\mathrm{m}}{2}=15\mathrm{m}$. The area of the $△\mathrm{ABD}$ $=\sqrt{15\times (15-9)\times (15-13)\times (15-8)}{\mathrm{m}}^2$ $=\sqrt{15\times 6\times 2\times 7}{\mathrm{m}}^2$ $=\sqrt{5\times 3\times 3\times 2\times 2\times 7}$ $=6\times 5.916{\mathrm{m}}^2$ (approx.) $=35.496{\mathrm{m}}^2$ (approx.) $=35.5{\mathrm{m}}^2$ (approx.) Thus area of the quadrilateral ABCD $=\mathrm{ar}(△\mathrm{BCD})+\mathrm{ar}(△\mathrm{ABD})$ $=30{\mathrm{m}}^2+35.5{\mathrm{m}}^2$ (approx.) $=65.5{\mathrm{m}}^2$ (approx.)
2. Find the area of quadrilateral ABCD in which $\mathrm{AB}=3\mathrm{cm},\mathrm{BC}=4\mathrm{cm},\mathrm{CD}=4\mathrm{cm}$, DA $=5\mathrm{cm}$ and $\mathrm{AC}=5\mathrm{cm}$
   
   Sol. $\mathrm{a}=4$ $\mathrm{b}=5$ $\mathrm{c}=3$ $\therefore {\mathrm{a}}^2+{\mathrm{c}}^2={\mathrm{b}}^2$ $\Rightarrow \angle \mathrm{B}=90^{\circ }$ Area of $△\mathrm{ABC}=\frac{1}{2}\times$ Base $\times$ Height $=\frac{1}{2}\times 3\times 4=6{\mathrm{cm}}^2$ For $△\mathrm{ACD}\mathrm{a}=4,\mathrm{b}=5$ and $\mathrm{c}=5$ $\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}=\frac{14}{2}=7\mathrm{cm}$ $\therefore$ Area of $△\mathrm{ACD}$ $=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{7(7-4)(7-5)(7-5)}$ $=\sqrt{7\times 3\times 2\times 2}$ $=2\sqrt{21}{\mathrm{cm}}^2$ area of quadrilateral $\mathrm{ABCD}=6+2\sqrt{21}{\mathrm{cm}}^2$ $\approx 15.2{\mathrm{cm}}^2$
3. Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.
   
   Sol. It is triangular part and its sides are 5 cm , $5\mathrm{cm},1\mathrm{cm}$. Here, semi perimeter of the triangle $\mathrm{s}=\frac{5\mathrm{cm}+5\mathrm{cm}+1\mathrm{cm}}{2}=\frac{11}{2}\mathrm{cm}$ Area of part I in figure $=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{\frac{11}{2}\times \frac{1}{2}\times \frac{1}{2}\times \frac{9}{2}}{\mathrm{cm}}^2$ $=\frac{3}{4}\sqrt{11}{\mathrm{cm}}^2=\frac{3}{4}\times 3.31{\mathrm{cm}}^2$ $=2.482$ (approx.) Area of part II in figure $=6.5\times 1{\mathrm{cm}}^2=6.5{\mathrm{cm}}^2$ Area of part III in figure
   
   Area of $△\mathrm{BEC}=\frac{\sqrt{3}}{4}(1{)}^2{\mathrm{cm}}^2=\frac{\sqrt{3}}{4}{\mathrm{cm}}^2$ Let $h$ be the height of the $△BEC$ i.e., height of the trapezium. $\frac{1}{2}\times \mathrm{BE}\times \mathrm{h}=\frac{\sqrt{3}}{4}$ $\begin{array}{ll}\Rightarrow & \frac{1}{2}\times 1\times \mathrm{h}=\\ \Rightarrow & \mathrm{h}=\frac{\sqrt{3}}{4}\mathrm{cm}\end{array}$ Area of Trapezium ABCD $=\frac{1}{2}$ (sum of parallel sides) (height) $=\frac{1}{2}(1+2)\left(\frac{\sqrt{3}}{2}\right){\mathrm{cm}}^2$ $=\frac{3}{4}\sqrt{3}{\mathrm{cm}}^2=\frac{3}{4}=\times 1.732$ (approx.) $=1.299$ (approx.) $=1.3{\mathrm{cm}}^2$ (approx.) Area of part IV in figure $=\frac{1}{2}\times 1.5\times 6{\mathrm{cm}}^2=4.5{\mathrm{cm}}^2$ Area of part V in figure $=\frac{1}{2}\times 1.5\times 6{\mathrm{cm}}^2=4.5{\mathrm{cm}}^2$ Total area of the paper used $=\mathrm{part}(\mathrm{I}+\mathrm{II}+\mathrm{III}+\mathrm{IV}+\mathrm{V})$ $=2.482{\mathrm{cm}}^2+6.5{\mathrm{cm}}^2+1.3{\mathrm{cm}}^2+4.5{\mathrm{cm}}^2$ $+4.5{\mathrm{cm}}^2$ $=(10.282+9){\mathrm{cm}}^2$ (approx.) $=19.282{\mathrm{cm}}^2$ (approx.) $=19.3{\mathrm{cm}}^2$ (approx.)
4. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are $26\mathrm{cm},28\mathrm{cm}$ and 30 cm , and the parallelogram stands on the base 28 cm , find the height of the parallelogram.
   
   Sol. Semi perimeter of $△\mathrm{ABC}$ $\mathrm{s}=\frac{26+28+30}{2}=42\mathrm{cm}$ $\mathrm{s}-\mathrm{a}=42-26=16\mathrm{cm}$ $\mathrm{s}-\mathrm{b}=42-28=14\mathrm{cm}$ $\mathrm{s}-\mathrm{c}=42-30=12\mathrm{cm}$ Area of $△ABC=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{42\times 16\times 14\times 12}{\mathrm{cm}}^2$ $=\sqrt{14\times 3\times 4\times 4\times 14\times 4\times 3}{\mathrm{cm}}^2$ $=14\times 4\times 3\times 2{\mathrm{cm}}^2=336{\mathrm{cm}}^2$ $\therefore$ Area of parallelogram $=$ Area of triangle [Given] $\mathrm{h}\times \mathrm{AB}=336$ $\mathrm{h}\times 28=336{\mathrm{cm}}^2$ $\mathrm{h}=\frac{336}{28}=12\mathrm{cm}$
5. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m , how much area of grass field will each cow be getting?
   
   Sol. In $△\mathrm{ABC}$ Semi perimeter $=\frac{AB+BC+AC}{2}$ $=\frac{30+30+48}{2}=54\mathrm{m}$ Using Heron's formula: Area of $△ABC=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{54(54-30)(54-30)(54-48)}$ $=\sqrt{54\times 24\times 24\times 6}$ $=432{\mathrm{m}}^2$ $\therefore$ Area of field $=2\times$ area of $△ABC$ $=2\times 432=864{\mathrm{m}}^2$ Thus, the area of gross field that each cow be getting $=\frac{864}{18}=48{\mathrm{m}}^2$
6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig.) each piece measuring $20\mathrm{cm},50\mathrm{cm}$ and 50 cm . How much cloth of each colour is required for the umbrella?
   
   Sol. Sides of a triangular piece of coloured cloth are $20\mathrm{cm},50\mathrm{cm}$ and 50 cm . Its semi perimeter $=\frac{20\mathrm{cm}+50\mathrm{cm}+50\mathrm{cm}}{2}$ $=60\mathrm{cm}$ Then, the area of one triangular piece $=\sqrt{60\times 10\times 10\times 40}{\mathrm{cm}}^2=200\sqrt{6}{\mathrm{cm}}^2$ There are 5 triangular pieces one colour and 5 of the other colour. Then total area of cloth of each colour (two colours) $=5\times 200\sqrt{6}{\mathrm{cm}}^2$ $=1000\sqrt{6}{\mathrm{cm}}^2$
7. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. How much paper of each shade has been used in it ?
   
   Sol. Area of paper shade $I=\frac{1}{2}\left[\frac{1}{2}\times 32\times 32\right]$ $=256{\mathrm{cm}}^2$ Area of paper shade $\mathrm{II}=256{\mathrm{cm}}^2$ Area of paper of shade III $\mathrm{a}=8\mathrm{cm};\mathrm{b}=6\mathrm{cm};\mathrm{c}=6\mathrm{cm}$ $\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}=\frac{8+6+6}{2}=10\mathrm{cm}$ $\therefore$ Area of paper of shade III $=\sqrt{s(s-a)(s-b)(s-c)}$ $=\sqrt{10(10-8)(10-6)(10-6)}$ $=\sqrt{10\cdot 2\cdot 4\cdot 4}=8\sqrt{5}=17.89{\mathrm{cm}}^2$
8. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being $9\mathrm{cm},28\mathrm{cm}$ and 35 cm (See Fig.). Find the cost of polishing the tiles at the rate of 50 paise $/{\mathrm{cm}}^2$.
   
   Sol. Sides of a triangular tile are $9\mathrm{cm},28\mathrm{cm}$ and 35 cm . Its semi perimeter will be $\mathrm{s}=\frac{9\mathrm{cm}+28\mathrm{cm}+35\mathrm{cm}}{2}=36\mathrm{cm}$ Area of one tile $=\sqrt{36\times 27\times 8\times 1}{\mathrm{cm}}^2$ $=36\sqrt{6}{\mathrm{cm}}^2$ Total area of 16 tiles $=16\times 36\sqrt{6}{\mathrm{cm}}^2$ $=576\sqrt{6}{\mathrm{cm}}^2=576\times 2.45{\mathrm{cm}}^2$ (approx.) $=1411.20{\mathrm{cm}}^2$ (approx.) Total cost of polishing at the rate of 50 p per ${\mathrm{cm}}^2$ $=$ Rs. $1411.20\times \frac{1}{2}{\mathrm{cm}}^2=$ Rs. 705.60
9. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m . The non-parallel sides are 14 m and 13 m . Find the area of the field.
   
   Sol. Through C draw CE | DA Draw CF $\perp$ AB In $△BCE$, we have $\mathrm{s}=\frac{15+14+13}{2}=21\mathrm{m}$ $\mathrm{s}-\mathrm{a}=21-15=6\mathrm{m}$ $\mathrm{s}-\mathrm{b}=21-14=7\mathrm{m}$ $\mathrm{s}-\mathrm{c}=21-13=8\mathrm{m}$ Area of $△\mathrm{BCE}=\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}$ $=\sqrt{21\times 6\times 7\times 8}{\mathrm{m}}^2$ $=7\times 4\times 3=84{\mathrm{m}}^2$ Now, area of $△\mathrm{BCE}=84{\mathrm{m}}^2$ $\Rightarrow \frac{1}{2}\times$ base $\times$ height $=84{\mathrm{m}}^2$ $\Rightarrow \frac{1}{2}\times 15\times \mathrm{h}=84$ $\Rightarrow \mathrm{h}=\frac{84\times 2}{15}\mathrm{m}$ $\Rightarrow \mathrm{h}=$ distance between parallel sides of trapezium $=\frac{168}{15}\mathrm{m}$ Area of parallelogram, AECD $=$ base $\times$ height $=10\times \frac{168}{15}=56\times 2=112{\mathrm{m}}^2$ $\therefore$ Area of trapezium ABCD = Area of parallelogram AECD + Area of $\Delta BCE=112+84=196{\mathrm{m}}^2$