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 rightangled 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 . 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.
Given the perimeter of an equilateral triangle is 180cm.
So, or .
Hence, the length of the side is 60cm.
Now,
Calculating the area of the signal board by the Heron's Formula:
Where, s is the halfperimeter of the triangle and a, b and c are the sides of the triangle.
Therefore,
as it is an equilateral triangle.
Substituting the values in the Heron's formula, we obtain
.
From the figure,
The sides of the triangle are:
The semi perimeter, s will be
Therefore, the area of the triangular side wall will be calculated by the Heron's Formula,
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:
Therefore, for 3 months for 1320 m^{2 }:
Given the sides of the triangle are:
So, the semi perimeter of the triangle will be:
Therefore, Heron's formula will be:
Hence, the area painted in colour is .
Q4 Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.
Given the perimeter of the triangle is and the sides length and
So,
Or,
So, the semi perimeter of the triangle will be:
Therefore, the area given by the Heron's Formula will be,
Hence, the area of the triangle is
Q5 Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.
Given the sides of a triangle are in the ratio of and its perimeter is
Let us consider the length of one side of the triangle be
Then, the remaining two sides are and
So, by the given perimeter, we can find the value of x:
So, the sides of the triangle are:
So, the semi perimeter of the triangle is given by
Therefore, using Heron's Formula, the area of the triangle is given by
Hence, the area of the triangle is .
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,
So, the semiperimeter of the triangle is given by,
Therefore, using Herons' Formula, calculating the area of the triangle
Hence, the area of the triangle is
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
The area of triangle BCD can be calculated by,
and the area of the triangle DAB can be calculated by Heron's Formula:
So, the semiperimeter of the triangle DAB,
Therefore, the area will be:
where, .
(Approximately)
Then, the total park area will be:
Hence, the total area of the park is
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 semiperimeter, where
Heron's Formula for calculating the area:
And the sides of the triangle ACD are
So, the semiperimeter of the triangle:
Therefore, the area will be given by, Heron's formula
.
Then, the total area of the quadrilateral will be:
Hence, the area of the quadrilateral ABCD is
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.,
For section I:
Here, the sides are
So, the Semiperimeter will be:
Therefore, the area of section I will be given by Heron's Formula,
For section II:
Here the sides of the rectangle are and
Therefore, the area of the rectangle is
For section III:
From the figure:
Drawing the parallel line AF to DC and a perpendicular line AE to BC.
We have the quadrilateral ADCF,
...........................by construction.
...........................[ ABCD is a trapezium]
So, ADCF is a parallelogram.
Therefore, and
Therefore,
ABF is an equilateral triangle.
Then, the area of the equilateral triangle ABF is given by:
Hence, the area of trapezium ABCD will be:
For Section IV:
Here, the base is 1.5 cm and the height is 6 cm.
Therefore, the area of the triangle is :
For section V:
The base length = 1.5cm and the height is 6cm.
Therefore, the area of the triangle will be:
Hence, the total area of the paper used will be:
From the figure:
The sides of the triangle are
Then, calculating the area of the triangle:
So, the semiperimeter of triangle ABE,
Therefore, its area will be given by the Heron's formula:
Given that the area of the parallelogram is equal to the area of the triangle:
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,
So, the semiperimeter of the triangle will be:
Therefore, the area of the triangle given by the Heron's formula,
Hence, the area of the quadrilateral will be:
Therefore, the area grazed by each cow will be given by,
Answer:
The sides of the triangle are:
So, the semiperimeter of the triangle is given by,
Therefore, the area of the triangle can be found by using Heron's formula:
Now, for the 10 triangular pieces of cloths, the area will be,
Hence, the area of cloths of each colour will be:
From the figure:
Calculation of the area for each shade:
The shade I: Triangle ABD
Here, base and the height
Therefore, the area of triangle ABD will be:
Hence, the area of paper used in shade I is
Shade II: Triangle CBD
Here, base and height
Therefore, the area of triangle CBD will be:
Hence, the area of paper used in shade II is
Shade III: Triangle CEF
Here, the sides are of lengths,
So, the semiperimeter of the triangle:
Therefore, the area of the triangle can be found by using Heron's formula:
Hence, the area of the paper used in shade III is .
GIven the sides of the triangle are:
So, its semiperimeter will be:
Therefore, the area of the triangle using Heron's formula is given by,
So, we have the area of each triangle tile which is .
Therefore, the area of each triangular 16 tiles will be:
Hence, the cost of polishing the tiles at the rate of 50 paise per cm^{2} will be:
The trapezium field is shown below in figure:
Drawing line CF parallel to AD and a line perpendicular to AB, we obtain
Then in quadrilateral ADCF,
............................
............................
Therefore, ADCF is a parallelogram.
So, and
Therefore,
Now, the sides of the triangle;
So, the semiperimeter of the triangle will be:
Therefore, the area of the triangle can be found by using Heron's Formula:
Also, the area of the triangle is given by,
Or,
Therefore, the area of the trapezium ABCD is:
Hence, the area of the trapezium field is
Chapter No. 
Chapter Name 
Chapter 1 

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 

Chapter 7 

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 

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 
Learn some basics about the triangles.
Read the Heron's formula.
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!
Q2. Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
Q6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?
Q2. 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. 12.9). The advertisements yield an earning of Rs. 5000 per m^{2} per year. A company hired one of its walls for 3 months. How much rent did it pay?