# NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula

## NCERT Solutions Class 9 Maths Chapter 12 โ Download Free PDF

**NCERT Solutions for Class 9 Maths Chapter 12 โ Heronโs Formula** is provided here. Heronโs formula is a fundamental concept that finds significance in countless areas. Therefore, it is imperative to have a clear grasp of the concept. And one of the best ways to do just that is by referring the **NCERT Solutions** for Class 9 Maths Chapter 12 Heronโs Formula. These solutions are designed by knowledgeable teachers with years of experience. The NCERT Solutions for Class 9 aim at equipping the students with detailed and step-wise explanations for all the answers to the questions given in the exercises of this Chapter.

Hence, one of the best guides you could adapt for your study needs is NCERT Solutions. Relevant topics are presented in an easy to understand format, barring the use of any complicated jargons. Furthermore, its content is updated as per prescribed CBSE syllabus.Chapter 12-Heronโs Formula

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**List of Exercises in NCERTclass 9 Maths Chapter 12:**

### Access Answers of NCERT Class 9 Maths Chapter 12 โ Heronโs Formula

**Exercise: 12.1 (Page No: 202)**

**1. A traffic signal board, indicating โSCHOOL AHEADโ, is an equilateral triangle with side โaโ. Find the area of the signal board, using Heronโs formula. If its perimeter is 180 cm, what will be the area of the signal board?**

**Solution:**

Given,

Side of the signal board = a

Perimeter of the signal board = 3a = 180 cm

โด a = 60 cm

Semi perimeter of the signal board (s) = 3a/2

By using Heronโs formula,

Area of the triangular signal board will be =

**2. 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 โน5000 per m ^{2} per year. A company hired one of its walls for 3 months. How much rent did it pay?**

**Solution:**

The sides of the triangle ABC are 122 m, 22 m and 120 m respectively.

Now, the perimeter will be (122+22+120) = 264 m

Also, the semi perimeter (s) = 264/2 = 132 m

Using Heronโs formula,

Area of the triangle =

=1320 m^{2}

We know that the rent of advertising per year = โน 5000 per m^{2}

โด The rent of one wall for 3 months = Rs. (1320ร5000ร3)/12 = Rs. 1650000

**3. There is a slide in a park. One of its side walls has been painted in some colour with a message โKEEP THE PARK GREEN AND CLEANโ (see Fig. 12.10 ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.**

**Solution:**

It is given that the sides of the wall as 15 m, 11 m and 6 m.

So, the semi perimeter of triangular wall (s) = (15+11+6)/2 m = 16 m

Using Heronโs formula,

Area of the message =

= โ[16(16-15)(16-11) (16-6)] m^{2}

= โ[16ร1ร5ร10] m^{2 }= โ800 m^{2}

= 20โ2 m^{2}

**4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42cm.**

**Solution:**

Assume the third side of the triangle to be โxโ.

Now, the three sides of the triangle are 18 cm, 10 cm, and โxโ cm

It is given that the perimeter of the triangle = 42cm

So, x = 42-(18+10) cm = 14 cm

โด The semi perimeter of triangle = 42/2 = 21 cm

Using Heronโs formula,

Area of the triangle,

=

= โ[21(21-18)(21-10)(21-14)] cm^{2}

= โ[21ร3ร11ร7] m^{2}

= 21โ11 cm^{2}

**5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.**

**Solution:**

The ratio of the sides of the triangle are given as 12 : 17 : 25

Now, let the common ratio between the sides of the triangle be โxโ

โด The sides are 12x, 17x and 25x

It is also given that the perimeter of the triangle = 540 cm

12x+17x+25x = 540 cm

54x = 540cm

So, x = 10

Now, the sides of triangle are 120 cm, 170 cm, 250 cm.

So, the semi perimeter of the triangle (s) = 540/2 = 270 cm

Using Heronโs formula,

Area of the triangle

= 9000 cm^{2}

**6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.**

**Solution:**

First, let the third side be x.

It is given that the length of the equal sides is 12 cm and its perimeter is 30 cm.

So, 30 = 12+12+x

โด The length of the third side = 6 cm

Thus, the semi perimeter of the isosceles triangle (s) = 30/2 cm = 15 cm

Using Heronโs formula,

Area of the triangle

=

= โ[15(15-12)(15-12)(15-6)] cm^{2}

= โ[15ร3ร3ร9] cm^{2}

= 9โ15 cm^{2}

**Exercise: 12.2 (Page No: 206)**

**1. A park, in the shape of a quadrilateral ABCD, has C = 90ยฐ, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?**

**Solution:**

First, construct a quadrilateral ABCD and join BD.

We know that

C = 90ยฐ, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m

The diagram is:

Now, apply Pythagoras theorem in ฮBCD

BD^{2} = BC^{2 }+CD^{2}

โ BD^{2} = 12^{2}+5^{2}

โ BD^{2} = 169

โ BD = 13 m

Now, the area of ฮBCD = (ยฝ ร12ร5) = 30 m^{2}

The semi perimeter of ฮABD

(s) = (perimeter/2)

= (8+9+13)/2 m

= 30/2 m = 15 m

Using Heronโs formula,

Area of ฮABD

= 6โ35 m^{2 }= 35.5 m^{2} (approximately)

โด The area of quadrilateral ABCD = Area of ฮBCD+Area of ฮABD

= 30 m^{2}+35.5m^{2} = 65.5 m^{2}

**2. 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.**

**Solution:**

First, construct a diagram with the given parameter.

Now, apply Pythagorean theorem in ฮABC,

AC^{2} = AB^{2}+BC^{2}

โ 5^{2} = 3^{2}+4^{2}

โ 25 = 25

Thus, it can be concluded that ฮABC is a right angled at B.

So, area of ฮBCD = (ยฝ ร3ร4) = 6 cm^{2}

The semi perimeter of ฮACD (s) = (perimeter/2) = (5+5+4)/2 cm = 14/2 cm = 7 m

Now, using Heronโs formula,

Area of ฮABD

= 2โ21 cm^{2 }= 9.17 cm^{2} (approximately)

Area of quadrilateral ABCD = Area of ฮABC + Area of ฮABD = 6 cm^{2 }+9.17 cm^{2} = 15.17 cm^{2}

**3. Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.**

**Solution:**

**For the triangle I section:**

It is an isosceles triangle and the sides are 5 cm, 1 cm and 5 cm

Perimeter = 5+5+1 = 11 cm

So, semi perimeter = 11/2 cm = 5.5 cm

Using Heronโs formula,

Area = โ[s(s-a)(s-b)(s-c)]

= โ[5.5(5.5- 5)(5.5-5)(5.5-1)] cm^{2}

= โ[5.5ร0.5ร0.5ร4.5] cm^{2}

= 0.75โ11 cm^{2}

= 0.75 ร 3.317cm^{2}

= 2.488cm^{2} (approx)

**For the quadrilateral II section:**

This quadrilateral is a rectangle with length and breadth as 6.5 cm and 1 cm respectively.

โด Area = 6.5ร1 cm^{2}=6.5 cm^{2}

**For the quadrilateral III section:**

It is a trapezoid with 2 sides as 1 cm each and the third side as 2 cm.

Area of the trapezoid = Area of the parallelogram + Area of the equilateral triangle

The perpendicular height of the parallelogram will be

= 0.86 cm

And, the area of the equilateral triangle will be (โ3/4รa^{2}) = 0.43

โด Area of the trapezoid = 0.86+0.43 = 1.3 cm^{2 }(approximately).

**For triangle IV and V:**

These triangles are 2 congruent right angled triangles having the base as 6 cm and height 1.5 cm

Area triangles IV and V = 2ร(ยฝร6ร1.5) cm^{2 }= 9 cm^{2}

So, the total area of the paper used = (2.488+6.5+1.3+9) cm^{2 }= 19.3 cm^{2}

**4. A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.**

**Solution:**

Given,

It is given that the parallelogram and triangle have equal areas.

The sides of the triangle are given as 26 cm, 28 cm and 30 cm.

So, the perimeter = 26+28+30 = 84 cm

And its semi perimeter = 84/2 cm = 42 cm

Now, by using Heronโs formula, area of the triangle =

= โ[42(42-26)(42-28)(42-30)] cm^{2}

= โ[42ร16ร14ร12] cm^{2}

= 336 cm^{2}

Now, let the height of parallelogram be h.

As the area of parallelogram = area of the triangle,

28 cmร h = 336 cm^{2}

โด h = 336/28 cm

So, the height of the parallelogram is 12 cm.

**5. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?**

**Solution:**

Draw a rhombus-shaped field first with the vertices as ABCD. The diagonal AC divides the rhombus into two congruent triangles which are having equal areas. The diagram is as follows.

Consider the triangle BCD,

Its semi-perimeter = (48 + 30 + 30)/2 m = 54 m

Using Heronโs formula,

Area of the ฮBCD =

= 432 m^{2}

โด Area of field = 2 ร area of the ฮBCD = (2 ร 432) m^{2 }= 864 m^{2}

Thus, the area of the grass field that each cow will be getting = (864/18) m^{2 }= 48 m^{2}

**6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig.12.16), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?**

**Solution:**

For each triangular piece, The semi perimeter will be

s = (50+50+20)/2 cm = 120/2 cm = 60cm

Using Heronโs formula,

Area of the triangular piece

=

= โ[60(60-50)(60-50)(60-20)] cm^{2}

= โ[60ร10ร10ร40] cm^{2}

= 200โ6 cm^{2}

โด The area of all the triangular pieces = 5 ร 200โ6 cm^{2 }= 1000โ6 cm^{2}

**7. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much paper of each shade has been used in it?**

**Solution:**

As the kite is in the shape of a square, its area will be

A = (ยฝ)ร(diagonal)^{2}

Area of the kite = (ยฝ)ร32ร32 = 512 cm^{2.}

The area of shade I = Area of shade II

512/2 cm^{2 }= 256 cm^{2}

So, the total area of the paper that is required in each shade = 256 cm^{2}

For the triangle section (III),

The sides are given as 6 cm, 6 cm and 8 cm

Now, the semi perimeter of this isosceles triangle = (6+6+8)/2 cm = 10 cm

By using Heronโs formula, the area of the III triangular piece will be

=

= โ[10(10-6)(10-6)(10-8)] cm^{2}

= โ(10ร4 ร4ร2) cm^{2}

= 8โ5 cm^{2} = 17.92 cm^{2} (approx.)

**8. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm ^{2} .**

**Solution:**

The semi perimeter of the each triangular shape = (28+9+35)/2 cm = 36 cm

By using Heronโs formula,

The area of each triangular shape will be

= 36โ6 cm^{2 }= 88.2 cm^{2}

Now, the total area of 16 tiles = 16ร88.2 cm^{2 }= 1411.2 cm^{2}

It is given that the polishing cost of tiles = 50 paise/cm^{2}

โด The total polishing cost of the tiles = Rs. (1411.2ร0.5) = Rs. 705.6

**9. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.**

**Solution:**

First, draw a line segment BE parallel to the line AD. Then, from B, draw a perpendicular on the line segment CD.

Now, it can be seen that the quadrilateral ABED is a parallelogram. So,

AB = ED = 10 m

AD = BE = 13 m

EC = 25-ED = 25-10 = 15 m

Now, consider the triangle BEC,

Its semi perimeter (s) = (13+14+15)/2 = 21 m

By using Heronโs formula,

Area of ฮBEC =

= 84 m^{2}

We also know that the area of ฮBEC = (ยฝ)รCEรBF

84 cm^{2 }= (ยฝ)ร15รBF

BF = (168/15) cm = 11.2 cm

So, the total area of ABED will be BFรDE i.e. 11.2ร10 = 112 m^{2}

โด Area of the field = 84+112 = 196 m^{2}

Chapter 12 Heronโs Formula belongs to Unit 5: Mensuration. This unit carries a total of 13 marks out of 100. Therefore, this is an important chapter and should be studied thoroughly. The important topics that are covered under this chapter are:

- Area of a Triangle โ by Heronโs Formula
- Application of Heronโs Formula in finding Areas of Quadrilaterals

Heronโs formula helps us to find the area of a triangle with 3 side lengths. Besides the formula, Heron also contributed in other ways โ the most notable one being the inventor of the very first steam engine called the Aeolipile. However, Heron couldnโt find any practical applications for it, instead, it ended up being used as a toy and an object of curiosity for the ancient Greeks.

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