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# Class 9 NCERT Solutions 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.

## Heron's Formula Exercise Ex. 12.1

### Solution 1

Side of traffic signal board = a
Perimeter of traffic signal board = 3  a
By Heron's formula
Perimeter of traffic signal board = 180 cm
Side of traffic signal board
Using equation (1), area of traffic of signal board

### 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
Rent of 1 m2 area per year = Rs.5000
Rent of 1 m2 area per month = Rs
Rent of 1320 m2 area for 3 months
= Rs.(5000  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
So, the area painted in colour is .

### Solution 4

Let third side of triangle be x.
Perimeter of given triangle = 42 cm
18 cm + 10 cm + x = 42
x = 14 cm
By Heron's 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.
By Heron's 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
By Heron's formula

## Heron's Formula Exercise Ex. 12.2

### Solution 7

We know that
Area of square  (diagonal)2
Area of given kite

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

For IIIrd triangle
Semi perimeter

By Heron's formula

Area of triangle

Area of IIIrd triangle

Area of paper required for IIIrd shade  = 17.92 cm2

### Solution 1

Let us join BD.
In BCD applying Pythagoras theorem
BD2 = BC2 + CD2
= (12)2 + (5)2
= 144 + 25
BD2 = 169
BD = 13 m
Area of BCD

For ABD

By Heron's formula
Area of triangle

Area of park = Area of ABD + Area of BCD
= 35.496 + 30 m2
= 65.496 m2
= 65. 5 m2 (approximately)

### Solution 2

For ABC
AC2 = AB2 + BC2
(5)2 = (3)2 + (4)2
So, ABC is a right angle triangle, right angled at point B.
Area of ABC
For DAC
Perimeter = 2s = DA + AC + CD = (5 + 5 + 4) cm = 14 cm
s = 7 cm
By Heron's formula
Area of triangle
Area of ABCD = Area of ABC + Area of ACD
= (6 + 9.166) cm2 = 15.166 cm2 = 15.2 cm2 (approximately)

### Solution 3

For triangle I

This triangle is a isosceles triangle.
Perimeter = 2s = (5 + 5 + 1) cm = 11cm

Area = l  b = (6.5  1) cm2 = 6.5 cm2

Perpendicular height of parallelogram

Area = Area of parallelogram + Area of equilateral triangle
= 0.866 + 0.433 = 1.299 cm2
Area of triangle (iv) = Area of triangle in (v)
Total area of the paper used = 2.488 + 6.5 + 1.299 + 4.5  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
Area of triangle

Let height of parallelogram be h
Area of parallelogram = Area of triangle
h  28 cm = 336 cm2
h = 12 cm
So, height of the parallelogram is 12 cm.

### Solution 5

Let ABCD be a rhombus shaped field.

For BCD
Semi perimeter,
Area of triangle
Therefore area of BCD

Area of field = 2  Area of BCD
= (2  432) m2 = 864 m2
Area for grazing for 1 cow= = 48 m2
Each cow will be getting 48 m2 area of grass field.

### Solution 6

For each triangular piece
Semi perimeter

By Heron's formula
Area of triangle

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

So, area of each cloth required

### Solution 8

We may observe that
Semi perimeter of each triangular shaped tile

By Heron's formula

Area of triangle

Area of each tile

= (36  2.45) cm2
=88.2 cm2
Area of 16 tiles = (16  88.2) cm2= 1411.2 cm2

Cost of polishing per cm2 area = 50 p
Cost of polishing 1411.2 cm2 area = Rs. (1411.2  0.50) = Rs.705.60
So, it will cost Rs.705.60 while polishing all the tiles.

### Solution 9

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 BEC

Semi perimeter

By Heron's formula

Area of triangle

Area of BEC

m2 = 84 m2

Area of  BEC

Area of ABED = BF  DE = 11.2  10
= 112 m2
Area of field = 84 + 112
= 196 m2