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

Prepare for your tests in class and the annual exam easily with NCERT Solutions for CBSE Class 9 Mathematics Chapter 12 Heron’s Formula by TopperLearning subject experts. Practise different types of textbook problems and check your answers with the accurate answers in our structured Maths chapter solutions. You can go through the detailed solutions and learn to use Heron’s formula for finding the area of a given triangle and solving Maths-based real-world problems. 

In addition, understand how to apply Heron’s formula to compute the area of a quadrilateral with our online NCERT textbook solutions for quick reference. To relearn the chapter basics, check our CBSE Class 9 Maths concept videos, practice tests and more.

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Chapter 12 - Heron's Formula Exercise Ex. 12.1

Solution 1
Side of traffic signal board = a
Perimeter of traffic signal board = 3 Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula a
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
By Heron's formula
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Perimeter of traffic signal board = 180 cm
Side of traffic signal board Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Using equation (1), area of traffic of signal board Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Solution 2
We may observe that sides of triangle 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
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Rent of 1 m2 area per year = Rs.5000
    Rent of 1 m2 area per month = Rs Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
    Rent of 1320 m2 area for 3 months  Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
                            = Rs.(5000 Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula 330) = Rs.1650000
    So, company had to pay Rs.1650000.

Solution 3
We may observe that the area to be painted in colour is a triangle, having its sides as 11 m, 6 m, and 15 m.
    Perimeter of such triangle = (11 + 6 + 15) m
                                    2 s = 32 m
                                       s = 16 m
By Heron's formula
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
    So, the area painted in colour is Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula.
Solution 4
Let third side of triangle be x.
  Perimeter of given triangle = 42 cm
        18 cm + 10 cm + x = 42
                                 x = 14 cm
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
By Heron's formula
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Solution 5
Let the common ratio between the sides of given triangle be x.
So, side of 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 triangle will be 120 cm, 170 cm, and 250 cm.
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
By Heron's formula
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
    So, area of this triangle will be 9000 cm2.
Solution 6
Let third side of this triangle be x.
Perimeter of triangle = 30 cm
  12 cm + 12 cm + x = 30 cm
                           x = 6 cm
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
By Heron's formula
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

Chapter 12 - Heron's Formula Exercise Ex. 12.2

Solution 1
Let us join BD.
In Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaBCD applying Pythagoras theorem
BD2 = BC2 + CD2
       = (12)2 + (5)2
       = 144 + 25
BD2 = 169
  BD = 13 m
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Area of Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaBCD
                  Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
For Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaABD
                    Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
By Heron's formula Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Area of triangle  Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
                         Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
      Area of park = Area of Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaABD + Area of Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaBCD
                         = 35.496 + 30 m2
                         = 65.496 m2
                         = 65. 5 m2 (approximately)

Solution 2
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
For Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaABC
AC2 = AB2 + BC2
(5)2 = (3)2 + (4)2
So, Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaABC is a right angle triangle, right angled at point B.
Area of Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaABCNcert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
For Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaDAC
Perimeter = 2s = DA + AC + CD = (5 + 5 + 4) cm = 14 cm
            s = 7 cm
By Heron's formula
Area of triangle Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Area of ABCD = Area of Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaABC + Area of Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaACD
    = (6 + 9.166) cm2 = 15.166 cm2 = 15.2 cm2 (approximately)

Solution 3
                                   Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
For triangle I

This triangle is a isosceles triangle.
Perimeter = 2s = (5 + 5 + 1) cm = 11cm
           Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
For quadrilateral II
This quadrilateral is a rectangle.
Area = l Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula b = (6.5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula 1) cm2 = 6.5 cm2

For quadrilateral III
This quadrilateral is a trapezium.
Perpendicular height of parallelogram Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
                                                      Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Area = Area of parallelogram + Area of equilateral triangle
  Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula= 0.866 + 0.433 = 1.299 cm2
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Area of triangle (iv) = Area of triangle in (v)
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Total area of the paper used = 2.488 + 6.5 + 1.299 + 4.5 Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula 2
                                         = 19.287 cm2

Solution 4
For triangle
Perimeter of triangle = (26 + 28 + 30) cm = 84 cm
2s = 84 cm
s = 42 cm
By Heron's formula
Area of triangle Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Area of triangleNcert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula 
                      Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Let height of parallelogram be h
Area of parallelogram = Area of triangle
                h Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula 28 cm = 336 cm2
                            h = 12 cm
So, height of the parallelogram is 12 cm.

Solution 5
 
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Let ABCD be a rhombus shaped field.

For Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaBCD
Semi perimeter, Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Area of triangleNcert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Therefore area of Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaBCDNcert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
                                 Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Area of field = 2 Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula Area of Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaBCD
 = (2 Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula 432) m2 = 864 m2
    Area for grazing for 1 cow= Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula = 48 m2
    Each cow will be getting 48 m2 area of grass field.

Solution 6
For each triangular piece
 Semi perimeter   Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
 
By Heron's formula
Area of triangle Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
 

Since, there are 5 triangular pieces made of two different colours cloth.

So, area of each cloth required Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

                                            Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Solution 7
We know that
Area of square Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula (diagonal)2
Area of given kite  Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
 
Area of 1st shade = Area of 2nd shade
 Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

So, area of paper required in each shape = 256 cm2.

For IIIrd triangle
Semi perimeter  Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
 
By Heron's formula
 
Area of triangle Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula

 
Area of IIIrd triangle  Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
 
                               Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
 
Area of paper required for IIIrd shade  = 17.92 cm2

Solution 8
We may observe that
Semi perimeter of each triangular shaped tile
          Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
By Heron's formula
 
Area of triangle Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
 
 
Area of each tile Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
                        Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
 
                        = (36 Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula 2.45) cm2
                        =88.2 cm2
Area of 16 tiles = (16 Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula 88.2) cm2= 1411.2 cm2

Cost of polishing per cm2 area = 50 p
Cost of polishing 1411.2 cm2 area = Rs. (1411.2 Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula 0.50) = Rs.705.60
So, it will cost Rs.705.60 while polishing all the tiles.
 

Solution 9
Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
Draw a line BE parallel to AD and draw a perpendicular BF on CD.

Now we may observe that ABED is a parallelogram.
BE = AD = 13 m
ED = AB = 10 m
EC = 25 - ED = 15 m
For Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaBEC

Semi perimeter  Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
 
By Heron's formula
 
Area of triangle Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
 
Area of Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons FormulaBEC Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
 
                    Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula m2 = 84 m2
 
 
 
        Area of Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula BEC Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
 
            Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula
 
Area of ABED = BF Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula DE = 11.2 Ncert Solutions Cbse Class 9 Mathematics Chapter - Herons Formula 10
                   = 112 m2
Area of field = 84 + 112
                   = 196 m2