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.

Ex. 12.1

Ex. 12.2

Heron's Formula Exercise Ex. 12.1

Solution 1

Side of traffic signal board = a
Perimeter of traffic signal board = 3

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

Area of 1st shade = Area of 2nd shade

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

For III^rd triangle

Semi perimeter

By Heron's formula

Area of triangle

Area of IIIrd triangle

Area of paper required for III^rd 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

For quadrilateral II
This quadrilateral is a rectangle.
Area = l

b = (6.5

1) cm² = 6.5 cm²

For quadrilateral III
This quadrilateral is a trapezium.
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