# NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula

NCERT solutions for class 9 maths chapter 12 Heron's Formula: In the previous classes and chapters, you have learnt the formulas to calculate the area of specific kinds of triangles like- right-angled triangle, equilateral triangle, etc. In this particular chapter, you will learn a special technique to find the area of any triangle. CBSE NCERT Solutions for class 9 maths chapter 12 Heron's Formula is covering the explanation of the all questions using Heron's formula. The Formula was invented by a mathematician from Egypt Heron. Before the invention of this formula, there was no formula present which was applicable to all kinds of triangles. Solutions of NCERT for class 9 maths chapter 12 Heron's formula are designed to provide assistance while doing homework as well as for the preparation of exams. This chapter has a total of two exercises which consists of 15 questions.

NCERT solutions for class 9 maths chapter 12 Heron's Formula is covering the solutions to each and every question present in the practice exercises in detail. If three sides of triangles are a,b,c then the area in terms of a, b and c is given by the Heron's Formula. The area is given by the formula    where s is half the perimeter. That is $s=\frac{a+b+c}{2}$.  You may think about the application of Heron's Formula. This formula is used for calculating the area of irregular plots. If you have a land whose shape is not regular we will divide the plots into different triangles and then apply Heron's Formula to calculate the area. Apart from this chapter, NCERT solutions can be downloaded for other classes and subjects.

## NCERT solutions for class 9 maths chapter 12 Heron's Formula Excercise: 12.1

Given the perimeter of an equilateral triangle is 180cm.

So, $3a = 180\ cm$  or  $a = 60\ cm$.

Hence, the length of the side is 60cm.

Now,

Calculating the area of the signal board by the Heron's Formula:

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

Where, s is the half-perimeter of the triangle and a, b and c are the sides of the triangle.

Therefore,

$s = \frac{1}{2}Perimeter = \frac{1}{2}180cm = 90cm$

$a =b=c = 60cm$  as it is an equilateral triangle.

Substituting the values in the Heron's formula, we obtain

$\implies A = \sqrt{90(90-60)(90-60)(90-60)} = 900\sqrt{3}\ cm^2$.

From the figure,

The sides of the triangle are:

$a = 122m,\ b = 120m\ and\ c = 22m$

The semi perimeter, s will be

$s = \frac{a+b+c}{2} = \frac{122+120+22}{2} = \frac{264}{2} = 132m$

Therefore, the area of the triangular side wall will be calculated by the Heron's Formula,

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

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

$= \sqrt{132(10)(12)(110)}\ m^2$

$= \sqrt{(12\times11)(10)(12)(11\times10)}\ m^2 = 1320\ m^2$

Given the rent for 1 year (i.e., 12 months) per meter square is Rs. 5000.

Rent for 3 months per meter square will be:

$Rs.\ 5000\times \frac{3}{12}$

Therefore,  for 3 months for 1320 m:

$Rs.\ 5000\times \frac{3}{12}\times 1320 = Rs.\ 16,50,000.$

Given the sides of the triangle are:

$a = 15 m,\ b= 11m\ and\ c= 6m.$

So, the semi perimeter of the triangle will be:

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

Therefore, Heron's formula will be:

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

$= \sqrt{16(16-15)(16-11)(16-6)}$

$= \sqrt{16(1)(5)(10)}$

$= \sqrt{(4\times 4)(1)(5)(5\times 2 )}$

$= 4\times 5 \sqrt2 = 20\sqrt2\ m^2$

Hence, the area painted in colour is $20\sqrt2\ m^2$.

Given the perimeter of the triangle is  $42cm$ and the sides length $a= 18cm$ and $b= 10cm$

So, $a+b+c = 42cm$

Or, $c = 42 - 18-10 = 14cm$

So, the semi perimeter of the triangle will be:

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

Therefore, the area given by the Heron's Formula will be,

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

$= \sqrt{21(21-18)(21-10)(21-14)}$

$= \sqrt{(7\times3 )(3)(11)(7)}$

$= 21\sqrt{11}\ cm^2$

Hence, the area of the triangle is $21\sqrt{11}\ cm^2.$

Given the sides of a triangle are in the ratio of $12:17:25$ and its perimeter is $540cm$

Let us consider the length of one side of the triangle be $a = 12x$

Then, the remaining two sides are $b = 17x$ and $c = 25x.$

So, by the given perimeter, we can find the value of x:

$Perimeter = a+b+c = 12x+17x+25x = 540cm$

$\implies 54x = 540cm$

$\implies x = 10$

So, the sides of the triangle are:

$a = 12\times10 =120 cm$

$b = 17\times10 =170 cm$

$c = 25\times10 =250 cm$

So, the semi perimeter of the triangle is given by

$s = \frac{540cm}{2} = 270cm$

Therefore, using Heron's Formula, the area of the triangle is given by

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

$= \sqrt{270(270-120)(270-170)(270-250)}$

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

$= \sqrt{81000000} = 9000cm^2$

Hence, the area of the triangle is $9000cm^2$.

The perimeter of an isosceles triangle is 30 cm (Given).

The length of the sides which are equal is12 cm.

Let the third side length be 'a cm'.

Then, $Perimeter = a+b+c$

$\Rightarrow 30= a+12+12$

$\Rightarrow a = 6cm$

So, the semi-perimeter of the triangle is given by,

$s= \frac{1}{2}Perimeter =\frac{1}{2}\times30cm = 15cm$

Therefore, using Herons' Formula, calculating the area of the triangle

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

$= \sqrt{15(15-6)(15-12)(15-12)}$

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

$= 9\sqrt{15}\ cm^2$

Hence, the area of the triangle is $9\sqrt{15}cm^2.$

## NCERT solutions for class 9 maths chapter 12 Heron's Formula Excercise: 12.2

### Q1 A park, in the shape of a quadrilateral ABCD, has $\angle C = 90\degree$, $AB = 9 m$, $BC = 12 m$, $CD = 5 m$ and $AD = 8 m$. How much area does it occupy?

From the figure:

We have joined the BD to form two triangles so that the calculation of the area will be easy.

In triangle BCD, by Pythagoras theorem

$BD^2 = BC^2+CD^2$

$\Rightarrow BD^2 = 12^2+5^2 = 144+25 = 169$

$\Rightarrow BD = 13\ cm$

The area of triangle BCD can be calculated by,

$Area_{(BCD)} = \frac{1}{2}\times BC\times DC = \frac{1}{2}\times 12\times 5 = 30\ cm^2$

and the area of the triangle DAB can be calculated by Heron's Formula:

So, the semi-perimeter of the triangle DAB,

$s = \frac{a+b+c}{2} = \frac{9+8+13}{2} = \frac{30}{2} = 15\ cm$

Therefore, the area will be:

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

where, $a = 9 cm,\ b = 8cm\ and\ c = 13cm$.

$= \sqrt{15(15-9)(15-8)(15-13)}$

$= \sqrt{12(6)(7)(2)} = \sqrt{1260} = 35.5\ cm^2$  (Approximately)

Then, the total park area will be:

$= Area\ of\ triangle\ BCD + Area\ of\ triangle\ DAB$

$\Rightarrow$ $Total\ area\ of\ Park = 30+35.35 = 65.5\ cm^2$

Hence, the total area of the park is $65.5\ cm^2.$

From the figure:

We have joined the AC to form two triangles so that the calculation of the area will be easy.

The area of the triangle ABC can be calculated by Heron's formula:

So, the semi-perimeter, where $a = 3 cm,\ b =4cm\ and\ c = 5cm.$

$s = \frac{a+b+c}{2} = \frac{3+4+5}{2} = 6cm$

Heron's Formula for calculating the area:

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

$= \sqrt{6(6-3)(6-4)(6-5)}= \sqrt{6(3)(2)(1)} = \sqrt{36} = 6\ cm^2$

And the sides of the triangle ACD are $a' =4cm,\ b' = 5cm\ and\ c' = 5cm.$

So, the semi-perimeter of the triangle:

$s' = \frac{a'+b'+c'}{2} = \frac{4+5+5}{2} = \frac{14}{2} = 7cm$

Therefore, the area will be given by, Heron's formula

$A = \sqrt{s'(s'-a')(s'-b')(s'-c')}$.

$= \sqrt{7(7-4)(7-5)(7-5)}$

$= \sqrt{7(3)(2)(2)} = 2\sqrt{21} = 9.2\ cm^2\ \ \ \ (Approx.)$

Then, the total area of the quadrilateral will be:

$= Area\ of\ triangle\ ABC + Area\ of\ triangle\ ACD$

$\Rightarrow$ $Total\ area\ of\ quadrilateral\ ABCD = 6+9.2 = 15.2\ cm^2$

Hence, the area of the quadrilateral ABCD is $15.2 \ cm^2.$

The total area of the paper used will be the sum of the area of the sections I, II, III, IV, and V. i.e., $Total\ area = I +II+III+IV+V$

For section I:

Here, the sides are $a = 1cm\ and\ b =c = 5cm.$

So, the Semi-perimeter will be:

$s = \frac{a+b+c}{2} = \frac{5+5+1}{2} = 5.5\ cm$

Therefore, the area of section I will be given by Heron's Formula,

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

$= \sqrt{5.5(5.5-1)(5.5-5)(5.5-5)}$

$= \sqrt{5.5(4.5)(0.5)(0.5)} = \sqrt{6.1875} = 2.5\ cm^2\ \ \ \ \ \ (Approx.)$

For section II:

Here the sides of the rectangle are $l =6.5\ cm$ and $b =1 \ cm.$

Therefore, the area of the rectangle is $= l\times b = 6.5\times 1 = 6.5\ cm^2.$

For section III:

From the figure:

Drawing the parallel line AF to DC and a perpendicular line AE to BC.

$AF || DC$             ...........................by construction.

$AD || FC$            ...........................[ $\because$ ABCD is a trapezium]

Therefore, $AF = DC = 1\ cm$  and  $AD = FC = 1\ cm$

$\left [ \because Opposite\ sides\ of\ a\ parallelogram \right ]$

Therefore, $BF = BC -FC =2-1 = 1\ cm.$

$\implies$ ABF is an equilateral triangle.      $\left [ \because AB = BF =AF = 1\ cm \right ]$

Then, the area of the equilateral triangle ABF is given by:

$\implies \frac{\sqrt3}{4}a^2 = \frac{\sqrt3}{4}1^2 = \frac{\sqrt3}{4}$

$= \frac{1}{2}\times BF \times AE$

$= \frac{1}{2}\times 1cm \times AE = \frac{\sqrt3}{4}$

$\implies AE = \frac{\sqrt3}{2} = \frac{1.732}{2} = 0.866 \approx 0.9$

Hence, the area of trapezium ABCD will be:

$= \frac{1}{2}\times(AD+BC)\times AE$

$= \frac{1}{2}\times(1+2)\times 0.9$

$=1.35 =1.4\ cm^2\ \ \ \ (Approx.)$

For Section IV:

Here, the base is 1.5 cm and the height is 6 cm.

Therefore, the area of the triangle is :

$= \frac{1}{2}\times base\times height$

$= \frac{1}{2}\times 1.5\times 6 = 4.5\ cm^2$

For section V:

The base length = 1.5cm and the height is 6cm.

Therefore, the area of the triangle will be:

$= \frac{1}{2}\times 1.5\times 6 = 4.5\ cm^2$

Hence, the total area of the paper used will be:

$Total\ area = I +II+III+IV+V$

$= 2.5+6.5+1.4+4.5+4.5 = 19.4\ cm^2$

From the figure:

The sides of the triangle are $a = 26\ cm,b= 28\ cm\ and\ c =30\ cm.$

Then, calculating the area of the triangle:

So, the semi-perimeter of triangle ABE,

$s = \frac{a+b+c}{2} = \frac{28+26+30}{2} = 42\ cm.$

Therefore, its area will be given by the Heron's formula:

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

$= \sqrt{42(42-28)(42-26)(42-30)}$

$= \sqrt{42(14)(16)(12)} = \sqrt{112896} = 336\ cm^2$

Given that the area of the parallelogram is equal to the area of the triangle:

$Area\ of\ Parallelogram = Area\ of\ Triangle$

$\implies base\times corresponding\ height = 336\ cm^2$

$\implies 28\times corresponding\ height = 336\ cm^2$

$\implies height = \frac{336}{28} = 12\ cm.$

Hence, the height of the parallelogram is 12 cm.

To find the area of the rhombus:

We first join the diagonal AC of quadrilateral ABCD.  (See figure)

Here, the sides of triangle ABC are,

$a = 30\ m,\ b = 30\ m\ and\ c = 48\ m.$

So, the semi-perimeter of the triangle will be:

$s = \frac{a+b+c}{2} = \frac{30+30+48}{2} = \frac{108}{2} = 54\ m$

Therefore, the area of the triangle given by the Heron's formula,

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

$= \sqrt{54(54-30)(54-30)(54-48)}$

$= \sqrt{54(24)(24)(6)} = \sqrt{186624} = 432\ m^2$

Hence, the area of the quadrilateral will be:

$= 2\times 432\ m^2 = 864\ m^2$

Therefore, the area grazed by each cow will be given by,

$= \frac{Total\ area}{Number\ of\ cows} = \frac{864}{18} = 48\ m^2.$

The sides of the triangle are:

$a = 20\ cm,\ b = 50\ cm\ and\ c =50\ cm.$

So, the semi-perimeter of the triangle is given by,

$s = \frac{a+b+c}{2} = \frac{20+50+50}{2} = \frac{120}{2} = 60\ cm.$

Therefore, the area of the triangle can be found by using Heron's formula:

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

$= \sqrt{60(60-20)(60-50)(60-50)}$

$= \sqrt{60(40)(10)(10)} = 200\sqrt{6}\ cm^2$

Now, for the 10 triangular pieces of cloths, the area will be,

$= 10\times200\sqrt6 = 2000\sqrt6\ cm^2$

Hence, the area of cloths of each colour will be:

$= \frac{2000\sqrt6}{2} = 1000\sqrt6\ cm^2.$

From the figure:

Calculation of the area for each shade:

Here, base $BD = 32\ cm$ and the height $AO = 16\ cm.$

Therefore, the area of triangle ABD will be:

$= \frac{1}{2} \times base\times height = \frac{1}{2}\times 32\times 16$

$= 256\ cm^2$

Hence, the area of paper used in shade I is $256\ cm^2.$

Here, base $BD = 32\ cm$  and height $CO = 16\ cm.$

Therefore, the area of triangle CBD will be:

$= \frac{1}{2} \times base\times height = \frac{1}{2}\times 32\times 16$

$= 256\ cm^2$

Hence, the area of paper used in shade II is $256\ cm^2.$

Here, the sides are of lengths, $a = 6\ cm,\ b = 6\ cm\ and\ c = 8\ cm.$

So, the semi-perimeter of the triangle:

$s = \frac{a+b+c}{2} = \frac{6+6+8}{2} = \frac{20}{2} = 10\ cm.$

Therefore, the area of the triangle can be found by using Heron's formula:

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

$= \sqrt{10(10-6)(10-6)(10-8)}$

$= \sqrt{10(4)(6)(2)}$

$= 8\sqrt{5}\ cm^2$

Hence, the area of the paper used in shade III is $8\sqrt{5}\ cm^2$.

GIven the sides of the triangle are:

$a = 9\ cm,\ b = 28\ cm\ and\ c= 35\ cm.$

So, its semi-perimeter will be:

$s = \frac{a+b+c}{2} = \frac{9+28+35}{2} = 36\ cm$

Therefore, the area of the triangle using Heron's formula is given by,

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

$= \sqrt{36(36-9)(36-28)(36-35)} = \sqrt{36(27)(8)(1)}$

$= \sqrt{7776} \approx 88.2\ cm^2$

So, we have the area of each triangle tile which is $88.2\ cm^2$.

Therefore, the area of each triangular 16 tiles will be:

$= 16\times 88.2\ cm^2 = 1411.2\ cm^2$

Hence, the cost of polishing the tiles at the rate of 50 paise per cm2 will be:

$= Rs.\ 0.50\times1411.2\ cm^2 =Rs.\ 705.60$

The trapezium field is shown below in figure:

Drawing line CF parallel to AD and a line perpendicular to AB, we obtain

$CF || AD$                         ............................ $\left [ \because\ by\ construction \right ]$

$CD || AF$                        ............................  $\left [ \because ABCD\ is\ a\ trapezium \right ]$

So, $AD = CF = 13\ m$  and  $CD = AF = 10\ m$

$\left ( \because Opposite\ sides\ of\ a\ parallelogram \right )$

Therefore, $BF = AB-AF = 25-10 = 15\ m$

Now, the sides of the triangle;

$a = 13\ m,\ b =14\ m\ and\ c = 15\ m.$

So, the semi-perimeter of the triangle will be:

$s= \frac{a+b+c}{2} = \frac{13+14+15}{2} = \frac{42}{2} = 21\ m$

Therefore, the area of the triangle can be found by using Heron's Formula:

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

$= \sqrt{21(21-13)(21-14)(21-15)}$

$= \sqrt{21(8)(7)(6)}$

$= \sqrt{7056} = 84\ m^2.$

Also, the area of the triangle is given by,

$Area = \frac{1}{2}\times BF\times CG$

$\Rightarrow \frac{1}{2}\times BF\times CG = 84\ m^2$

$\Rightarrow \frac{1}{2}\times 15\times CG = 84\ m^2$

Or,

$\Rightarrow CG = \frac{84\times2}{15} = 11.2\ m$

Therefore, the area of the trapezium ABCD is:

$= \frac{1}{2} \times (AB+CD)\times CG$

$= \frac{1}{2} \times (25+10)\times 11.2$

$= 35\times5.6$

$= 196\ m^2$

Hence, the area of the trapezium field is $196\ m^2.$

## NCERT solutions for class 9 maths chapter wise

 Chapter No. Chapter Name Chapter 1 NCERT solutions for class 9 maths chapter 1 Number Systems Chapter 2 CBSE NCERT solutions for class 9 maths chapter 2 Polynomials Chapter 3 Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry Chapter 4 NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables Chapter 5 CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry Chapter 6 Solutions of NCERT class 9 maths chapter 6 Lines And Angles Chapter 7 NCERT solutions for class 9 maths chapter 7 Triangles Chapter 8 CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals Chapter 9 Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles Chapter 10 NCERT solutions for class 9 maths chapter 10 Circles Chapter 11 CBSE NCERT solutions for class 9 maths chapter 11 Constructions Chapter 12 NCERT solutions for class 9 maths chapter 12 Heron’s Formula Chapter 13 NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes Chapter 14 CBSE NCERT solutions for class 9 maths chapter 14 Statistics Chapter 15 Solutions of NCERT class 9 maths chapter 15 Probability

## How to use NCERT solutions for class 9 maths chapter 12 Heron's Formula?

• Learn some basics about the triangles.

• Observe its application on mathematical problems.

• Now, It's time to apply the application of this formula in the practice exercises' problems.

• While practicing the exercises, you can take the help of NCERT solutions for class 9 maths chapter 12 Heron's Formula.

Keep working hard and happy learning!