NCERT Solutions for Class 9 Maths Chapter 12: Heron's Formula - Exercise 12.2

Exercise 12.2

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: Given - Traffic signal board is an equilateral triangle.

And, each side of the triangle =$a$

Also, perimeter of the triangle$\ =\ 180\ \text{cm}$

Since, the perimeter of triangle is given by $3a$.

$ \therefore \ \ a+a+a=\ 180 \\ $

$\ \ \ \ \ \ \ \Rightarrow \ 3a=\ 180 \\ $

$  \ \ \ \ \ \ \ \Rightarrow \ a\ =\ \frac{180}{3} \\ $

$\ \ \ \ \ \ \Rightarrow \ a\ =\ 60\ \ \text{cm} \\ $

$\therefore$ Each side of triangle =$60\ \text{cm}$

Now, semi-perimeter of triangle

\[\Rightarrow s=\frac{\text{perimeter}}{\text{2}}\ \frac{180}{2}\ =\ 90\ \text{cm}\]

∴ Area of the triangular signal board (Heron’s Formula) $\text{=}\,\sqrt{s(s-a)(s-b)(s-c)}$

But

\[\ a\ =\ b\ =\ c=\ 60\ \text{cm}\]

$=(s-a)\sqrt{s(s-a)} \\ $

$  =\left( 90-60 \right)\sqrt{90(90-60)}\ \text{c}{{\text{m}}^{\text{2}}} \\ $

 $ =\ 30\ \sqrt{90\times 30\ }\text{c}{{\text{m}}^{\text{2}}} \\ $

 $ =\ 30\ \sqrt{9\times 10\times 3\times 10}\ \text{c}{{\text{m}}^{\text{2}}} \\ $

$  =\,30\times 3\times 10\sqrt{3\ }\text{c}{{\text{m}}^{\text{2}}} \\ $

$  =\ 900\sqrt{3\ }\text{c}{{\text{m}}^{\text{2}}} \\ $

Therefore, area of Signal board is $900\sqrt{3\ }\ \text{c}{{\text{m}}^{\text{2}}}$

2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are\[122\text{ m, }22\text{ m and }120\text{ m}\](see Fig.). The advertisements yield an earning of

\[\text{Rs}\ 5000\ \text{per}\ {{\text{m}}^{\text{2}}}\]per year. A company hired one of its walls for 3 months. How much rent did it pay?

Ans: To calculate rent for 3 months, first we need to find area of the wall to be painted.

So, Let’s find the area of the triangular side walls having sides-

\[a=122~\text{m},~\ \text{b}\ =\ 120~\text{m}\ \ and\,\ \text{c}\ =\ 22~\text{m}\]

So, Semi-perimeter of triangular wall =

\[s=\frac{\text{a}\ +\ \text{b}\ +\ \text{c}}{2}\ =\ \frac{122+120+22}{2}\ =\ 132~\text{m}\]

By Heron’s Formula area of triangular side wall =

\[\sqrt{s(s-a)(s-b)(s-c)}\]

$ =\sqrt{132(132-122)(132-120)(132-22)}\ {{\text{m}}^{2}} \\ $

$ =\sqrt{132\times 10\times 12\times 110}~\ {{\text{m}}^{2}} \\$ 

$ =\sqrt{12\times 11\times 10\times 12\times 10\times 11}\ {{\text{m}}^{2}} \\ $

$ =12\times 11\times 10\ {{\text{m}}^{2}} \\ $

$ =1320~\ {{\text{m}}^{2}} \\ $

Given,

Rent for $1\ {{\text{m}}^{\text{2}}}$of wall for $1$ year (12 months) =  $\text{Rs}\,5000.$

$\therefore$ Rent for $1\ {{\text{m}}^{\text{2}}}$of wall for $1$ month  $=\ \text{Rs}\,\frac{5000}{12}.$

$\Rightarrow$ Rent for $1320\ {{\text{m}}^{\text{2}}}$of wall for $3$ months   $=\ \frac{5000}{12}\ \ \times \ 3$  = $\text{Rs}\,16,50,000.$

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 \[15\ \text{m},\ 11\ \text{m}\ \text{and}\ 6\ \text{m}\] , find the area painted in colour.

(image will be uploaded soon)

Ans: Given sides of triangular wall are

\[\text{ }a=15\ \text{m},\text{ }b=11\text{ m and }c=6\text{ m}\]

So, semi-perimeter =

\[s\ =\ \frac{a+\ b\ +\ c}{2}\ =\ \frac{15+11+6}{2}\ =\ \frac{32}{2}\ =\ 16\ \text{m}\]

Area to be painted in colour = Area of triangle =

\[\sqrt{s(s-a)(s-b)(s-c)}\]

$ =\sqrt{(16(16-15)(16-11)(16-6)}\ {{\text{m}}^{2}} \\ $

$ =\ 4\sqrt{1\times 5\times 10}\ {{\text{m}}^{2}} \\ $

$ =\ 4\sqrt{5\times 2\times 5}\ {{\text{m}}^{2}} \\ $

$ =\ 4\times 5\sqrt{2}\ {{\text{m}}^{2}} \\ $

$=\ 20\sqrt{2}\ {{\text{m}}^{2}} \\ $

Therefore, area to be painted in colour = $20\sqrt{2}\ {{\text{m}}^{\text{2}}}$

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

Ans: Given, $a=\ 18\ \text{cm},\ b=\ 10\ \text{cm}\ ,\ c=?$
Also, Perimeter of the triangle $=\ 42\ \text{cm}$

 $ \ \ \Rightarrow \ a+\ b\ +\ c=\ 42\ \text{cm} $

$  \Rightarrow \ 18+10+c=42\ \text{cm}$

$  \ \ \ \ \ \ \Rightarrow \ 28+c\ =42\ \text{cm} $

$  \Rightarrow \ c=42-28=14\ \text{cm} $

Now, semi-perimeter $=s=\frac{\text{Perimeter}}{\text{2}}\ =\ \frac{42}{2}\ =\ 21\ \text{cm}$

Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\ \sqrt{21(21-18)(21-10)(21-14)}\ \text{c}{{\text{m}}^{\text{2}}} \\ $

 $=\ \sqrt{21\times 3\times 11\times 7}\ \text{c}{{\text{m}}^{\text{2}}} \\$ 

 $ =\ \sqrt{3\times 7\times 3\times 11\times 7}\ \text{c}{{\text{m}}^{\text{2}}} \\$ 

 $ =\ 3\times 7\sqrt{11}\,\text{c}{{\text{m}}^{\text{2}}} \\ $

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

Therefore, area of triangle = $21\sqrt{11}\ \text{c}{{\text{m}}^{\text{2}}}$

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

Ans: Let the sides of the triangle be $a=12x\ \text{cm},\ b\ =\ 17x\ \text{cm}\,\ \text{and}\ c=\ 25x\ \text{cm}$

Given, Perimeter of the triangle  $=\ 540\ \text{cm}$

$\therefore \ 12x+17x+25x=\ 540 \\ $

$ \Rightarrow 54x\ =\ 540 \\ $

 $ \Rightarrow \ x=\ \frac{540}{54}\  \\$ 

 $ \Rightarrow \ x=10  $

$\text{Now, }s=\frac{a+b+c}{2}\ =\ \frac{120+170+250}{2}\ =\ \frac{540}{2}\ =270\ \text{cm}$

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

$=\sqrt{270(270-120)(270-170)(270-250)}\ \text{c}{{\text{m}}^{\text{2}}} \\ $

$\text{=}\ \sqrt{270(150)(100)(20)}\ \text{c}{{\text{m}}^{\text{2}}} \\ $

$ =\ 10\sqrt{3\times 9\times 10\times 3\times 5\times 10\times 5\times 4}\ \text{c}{{\text{m}}^{\text{2}}} \\ $

$ =\ 10\times 3\times 3\times 10\times 5\times 2\ \text{c}{{\text{m}}^{\text{2}}} \\ $

$ =\ 9000\ \text{c}{{\text{m}}^{\text{2}}} \\ $

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

6. An isosceles triangle has perimeter$\text{30 cm}$and each of the equal sides is

\[12\ \text{cm}\]. Find the area of the triangle.

Ans: $ \text{Here,}\ a=b=12\ \text{cm} \\ $

$ \text{given,  }a+b+c=30 \\ $

$ \ \ \ \Rightarrow \ 12+12+c=30 \\ $

$\ \ \ \ \ \ \ \ \ \Rightarrow 24+c=30 \\ $

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow c=30-24 \\ $

$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow c=6 \\ $

Now, semi-perimeter =\[s=\ \frac{30}{2}\ =\ 15\ \text{cm}\]

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

Therefore, Area of triangle is $9\sqrt{15\ }\text{c}{{\text{m}}^{\text{2}}}$

Class 9th Maths NCERT Solutions Chapter 12.2 Free PDF

The Class 9 Maths Chapter 12.2 Solutions are available for the students in a PDF format which can be downloaded for free from the official website of Vedantu. It is easy to get a soft copy or hard copy from the website. Either students or parents can take a printout and store it for future revision references. These PDFs are also helpful for students to attend different olympiads, entrance tests, competitive exams, etc.

NCERT Solutions for Class 9 Maths

• Chapter 1 - Number System

• Chapter 2 - Polynomials 

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introductions to Euclids Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelogram and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Herons formula

• Chapter 13 - Surface area and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

Along with this, students can also download additional study materials provided by Vedantu, for Chapter 12 of CBSE Class 9 Maths Solutions –

• Chapter 12: Important Questions

• Chapter 12: Important Formulas

• Chapter 12: Revision Notes

• Chapter 12: NCERT Exemplar Solutions

• Chapter 12: RD Sharma Solutions

NCERT Class 9 Maths Chapter 12 Exercise 12.2 Question 1

The first question of exercise 12.2 of class 9 maths is about a park. In the PDF, the shape of the park is like a quadrilateral, and the length of each side is also given. Also, one angle is the acute angle by considering this information; students may solve by splitting the quadrilateral into two triangles. By practising and understanding the pdf, students were already well known about the Pythagoras theorem and heron formula. I'm playing these two formulae; students can easily find out the area of the park.

Class 9 Maths Chapter 12 Exercise 12.2 Question 2

In the second question, similar to the first one, the length of different sides of a quadrilateral was given, and then the students are asked to find out the area. As the students can know the heron's formula and the concept of quadrilaterals, with the help of Class 9 Maths Chapter 12 Exercise 12.2, they can easily find the area of the quadrilateral. It is like a short answer type question which is simple and straightforward.

Exercise 12.2 Class 9 Question 3

Exercise 12.2 question number 3 is about a paper aeroplane made by a girl, Radha. Different lengths were given and asked to find the area of the paper. It is like a tricky question for the students. 9th Class Maths Chapter 12 Exercise 12.2.

Will help the students by providing various shortcuts and smart techniques to solve these kinds of problems. Hair students need to click the whole shape into multiple shapes like trapezium, triangle, rectangle, etc. Then find the areas of each shape individually. By adding all those areas, the required answer will be obtained.

NCERT Maths Class 9 Chapter 12 Exercise 12.2 Question 4

The fourth question of Exercise 12.2 Class 9 is to find the height of your parallelogram. The given data is about two figures namely-triangle and parallelogram stand on the same base. The lengths of the triangle were given. Both have equal areas. Then the students need to find out the perimeter of the triangle to find out the height of the parallelogram easily.

Maths NCERT Solutions Class 9 Chapter 12 Exercise 12.2 Question 5

The next question is a little bit logical and twisty. Students need to think wisely and understand keenly. This can be habituated for the students with proper practising and understanding of NCERT Solutions class 9 chapter 12 PDF. Even though the procedure of solving the problem is easy, understanding plays a vital role in this question.

Class 9 Maths NCERT Solutions Chapter 12 Exercise 12.2 Question 6

The sixth question is purely one of the applications of the heron's formula in your day-to-day life. Students can understand the situation properly and apply the formula for the given figure. Turn the umbrella into multiple triangles and substitute the values. It helps to solve the sum easily.

Class 9 Maths Chapter 12 Exercise 12.2 Solutions Question 7

The next question of exercise 12.2 class 9 maths is about making a paper kite. Students need to observe that it has two figures-square and triangles. Considering the given data, Apply the heron's formula and find out the quantity of paper used.

Ex 12.2 Class 9 Question 8

This is the last but one question of the whole exercise. The question seems to be generic but gives a lot of knowledge to the students. Considering the shape of the tile and the design, students need to use General mathematics applications like multiplication, edition along with the heron formula, area of a triangle, etc. to find out the cost of the painting.

Maths NCERT Solutions Class 9 Chapter 12 Exercise 12.2 Question 9

The last question is very simple and straightforward. It is about a trapezium whose lengths are given. Students need to find the area of the entire field in a step-by-step process.

Key Takeaways of NCERT Solutions Class 9 Maths Chapter 12 Exercise 12.2 Free PDF

The advantages of using it's your teeth Solutions notes for the students are:

• Students can clarify their doubts through a live chat on the official website of Vedantu.

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• The PDFs have a detailed explanation, along with previous question papers, to get enough practice.