# 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

Perimeter of traffic signal board = 3 a

Side of traffic signal board

### Solution 2

Perimeter of triangle = (122 + 22 + 120) m

2s = 264 m

s = 132 m

By Heron's formula

^{2}area per year = Rs.5000

Rent of 1 m

^{2}area per month = Rs

^{2}area for 3 months

= Rs.(5000 330) = Rs.1650000

So, company had to pay Rs.1650000.

### Solution 3

Perimeter of such triangle = (11 + 6 + 15) m

2 s = 32 m

s = 16 m

By Heron's formula

### Solution 4

Perimeter of given triangle = 42 cm

18 cm + 10 cm + x = 42

x = 14 cm

### Solution 5

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.

^{2}.

### Solution 6

Perimeter of triangle = 30 cm

12 cm + 12 cm + x = 30 cm

x = 6 cm

## Heron's Formula Exercise Ex. 12.2

### Solution 7

Area of square (diagonal)2

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

**For III**

^{rd}triangle^{rd}shade = 17.92 cm

^{2}

### Solution 1

In BCD applying Pythagoras theorem

BD

^{2}= BC

^{2}+ CD

^{2}

= (12)

^{2}+ (5)

^{2}

= 144 + 25

BD

^{2}= 169

BD = 13 m

= 35.496 + 30 m

^{2}

^{2}

^{2}(approximately)

### Solution 2

AC

^{2}= AB

^{2}+ BC

^{2}

(5)

^{2}= (3)

^{2}+ (4)

^{2}

So, ABC is a right angle triangle, right angled at point B.

Area of ABC

Perimeter = 2s = DA + AC + CD = (5 + 5 + 4) cm = 14 cm

s = 7 cm

By Heron's formula

Area of triangle

= (6 + 9.166) cm

^{2}= 15.166 cm

^{2}= 15.2 cm

^{2}(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) cm

^{2}= 6.5 cm

^{2}

**For quadrilateral III**

This quadrilateral is a trapezium.

Perpendicular height of parallelogram

= 0.866 + 0.433 = 1.299 cm2

= 19.287 cm

^{2}

### Solution 4

Perimeter of triangle = (26 + 28 + 30) cm = 84 cm

2s = 84 cm

s = 42 cm

By Heron's formula

Area of triangle

Area of parallelogram = Area of triangle

h 28 cm = 336 cm

^{2}

h = 12 cm

So, height of the parallelogram is 12 cm.

### Solution 5

For BCD

Semi perimeter,

= (2 432) m

^{2}= 864 m

^{2}

Area for grazing for 1 cow= = 48 m

^{2}

Each cow will be getting 48 m

^{2}area of grass field.

### Solution 6

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

### Solution 8

Semi perimeter of each triangular shaped tile

^{2}

=88.2 cm

^{2}

Area of 16 tiles = (16 88.2) cm

^{2}= 1411.2 cm

^{2}

Cost of polishing per cm

^{2}area = 50 p

Cost of polishing 1411.2 cm

^{2}area = Rs. (1411.2 0.50) = Rs.705.60

So, it will cost Rs.705.60 while polishing all the tiles.

### Solution 9

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

^{2}= 84 m

^{2}

= 112 m

^{2}

Area of field = 84 + 112

= 196 m

^{2}