**NCERT Solutions Class 9 Maths Chapter 12 Herons Formula** – Here are all the NCERT solutions for Class 9 Maths Chapter 12. This solution contains questions, answers, images, explanations of the complete Chapter 12 titled Herons Formula of Maths taught in class 9. If you are a student of class 9 who is using NCERT Textbook to study Maths, then you must come across Chapter 12 Herons Formula. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 9 Maths Chapter 12 Herons Formula in one place.

## NCERT Solutions Class 9 Maths Chapter 12 Herons Formula

Here on **AglaSem Schools**, you can access to **NCERT Book Solutions** in free pdf for Maths for Class 9 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.

For a better understanding of this chapter, you should also see summary of Chapter 12 Herons Formula , Maths, Class 9.

Class | 9 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 12 |

Chapter Name |
Herons Formula |

### NCERT Solutions Class 9 Maths chapter 12 Herons Formula

Class 9, Maths chapter 12, Herons Formula solutions are given below in PDF format. You can view them online or download PDF file for future use.

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### Question & Answer

Q.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?

Ans :Side of traffic signal board = a Perimeter of traffic signal board \(2 s=3 a \Rightarrow s=\frac{3}{2} a\) By Heron's formula, Area of given triangle \(=\sqrt{s(s-a)(s-b)(s-c)}\) \(\begin{aligned} \text { Area of given triangle } &=\sqrt{\frac{3}{2} a\left(\frac{3}{2} a-a\right)\left(\frac{3}{2} a-a\right)\left(\frac{3}{2} a-a\right)} \\ &=\sqrt{\left(\frac{3}{2} a\right)\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)} \\ &=\frac{\sqrt{3}}{4} a^{2} \end{aligned}\)....(1) Perimeter of traffic signal board = 180 cm Side of traffic signal board \((a)=\left(\frac{180}{3}\right) \mathrm{cm}=60 \mathrm{cm}\) \(\begin{array}{l}{\text { Using equation }(1), \text { area of traffic signal board }=\frac{\sqrt{3}}{4}(60 \mathrm{cm})^{2}} \\ {=\left(\frac{3600}{4} \sqrt{3}\right) \mathrm{cm}^{2}=900 \sqrt{3} \mathrm{cm}^{2}}\end{array}\)

Q.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). The advertisements yield an earning of 5000 per \(m^{2}\) per year. A company hired one of its walls for 3 months. How much rent did it pay?

Ans :The sides of the triangle (i.e., a, b, c) are of 122 m, 22 m, and 120 m respectively. Perimeter of triangle = (122 + 22 + 120) m 2s = 264 m s = 132 m By Heron's formula, Area of triangle \(=\sqrt{s(s-a)(s-b)(s-c)}\) \(\begin{aligned} \text { Area of given triangle } &=[\sqrt{132(132-122)(132-22)(132-120)}] \mathrm{m}^{2} \\ &=[\sqrt{132(10)(110)(12)}] \mathrm{m}^{2}=1320 \mathrm{m}^{2} \end{aligned}\) 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 \(m^{2}\) area per 3 month = \(\operatorname{Rs}\left(\frac{5000}{12} \times 3 \times 1320\right)\) = Rs (5000 x 330) = Rs 1650000 Therefore, the company had to pay Rs 1650000.

Q.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 m, 11 m and 6 m, find the area painted in colour.

Ans :Sides of the triangular wall are 15 m, 11m, an 6m. Semi perimeter of triangular wall (s) = (15 + 11 + 6) / 2 m = 16 m Using Heron’s Formula, \( \begin{aligned} \text { Area of the message }=& \sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})} \\ &=\sqrt{16(16-15)(16-11)(16-6)} \mathrm{m}^{2} \\ &=\sqrt{16 \times 1 \times 5 \times 10} \mathrm{m}^{2}=\sqrt{800} \mathrm{m}^{2} \\ &=20 \sqrt{2} \mathrm{m}^{2} \end{aligned} \)

Q.4:Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.

Ans :Let the third side of the triangle be x. Perimeter of the given triangle 42 cm 18 cm + 10 cm + x = 42 x = 14 cm \(s=\frac{\text { Perimeter }}{2}=\frac{42 \mathrm{cm}}{2}=21 \mathrm{cm}\) By Heron's formula, Area of a triangle =\(=\sqrt{s(s-a)(s-b)(s-c)}\) \(\begin{aligned} \text { Area of the given triangle } &=(\sqrt{21(21-18)(21-10)(21-14)}) \mathrm{cm}^{2} \\ &=(\sqrt{21(3)(11)(7)}) \mathrm{cm}^{2} \\ &=21 \sqrt{11} \mathrm{cm}^{2} \end{aligned}\)

Q.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. Perimeter of this triangle = 540 cm 12x + 17x + 25x = 540 cm 54x = 540 cm x = 10 cm Sides of the triangle will be 120 cm, 170 cm, and 250 cm. \(s=\frac{\text { Perimeter of triangle }}{2}=\frac{540 \mathrm{cm}}{2}=270 \mathrm{cm}\) By Heron's formula, \(\begin{aligned} \text { Area of triangle } &=\sqrt{s(s-a)(s-b)(s-c)} \\ &=[\sqrt{270(270-120)(270-170)(270-250)}] \mathrm{cm}^{2} \\ &=[\sqrt{270 \times 150 \times 100 \times 20}] \mathrm{cm}^{2} \end{aligned}\) Therefore, the area Of this triangle is 9000 \(cm^{2}\) .

## NCERT / CBSE Book for Class 9 Maths

You can download the NCERT Book for Class 9 Maths in PDF format for free. Otherwise you can also buy it easily online.

- Click here for NCERT Book for Class 9 Maths
- Click here to buy NCERT Book for Class 9 Maths

### All NCERT Solutions Class 9

- NCERT Solutions for Class 9 English
- NCERT Solutions for Class 9 Hindi
- NCERT Solutions for Class 9 Maths
- NCERT Solutions for Class 9 Science
- NCERT Solutions for Class 9 Social Science
- NCERT Solutions for Class 9 Sanskrit

### All NCERT Solutions

You can also check out NCERT Solutions of other classes here. Click on the class number below to go to relevant NCERT Solutions of Class 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12.

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