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

This article provides you with detailed NCERT Solutions for Class 9 Maths Chapter 12-Heron’s Formula. They have been solved by a team of experienced professionals at GFG, to make sure that every student can learn how to solve these questions with the easiest approach. This article provides solutions to all the problems asked in Class 9 Maths Chapter 12-Heron’s Formula of the NCERT textbook in a step-by-step manner.

Heron’s formula was given by the great mathematician cum engineer Heron of Alexandria, hence the name Heron’s Formula. Class 9 Maths Chapter 12-Heron’s Formula is the foundational chapter for getting a practical and smooth understanding of the concepts related to Heron’s Formula.

NCERT Solutions Class 9 Maths Chapter 12 Heron’s Formula Exercises:
NCERT Solutions Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.1 – 6 Questions & Solutions (2 Short Answers, 2 Long Answers, 2 Very Long Answers)
NCERT Solutions Class 9 Maths Chapter 12 Heron’s Formula Exercise 12.2 – 9 Questions & Solutions (4 Long Answers, 5 Very Long Answers)

Solutions to all the exercises in the chapter 12 Heron’s Formula of NCERT textbook have been covered in the NCERT Solutions for Class 9 Maths. NCERT Solutions Class 9 Maths Chapter 12 Heron’s Formula on this page are regularly revised to check errors and updated keeping the latest CBSE Syllabus 2023-2024 in mind.

NCERT Solutions Class 9 Maths Chapter 12 Heron’s Formula: Exercise 12.1

Question 1. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?

Solution:

As it is mentioned here that, ΔABC is an equilateral triangle having side length = a.

So, here

AB =BC = AC = a

Perimeter of Equilateral triangle = 3× (Length of a side)

 = 3×a = 3a

and perimeter = 180 cm (given)

So, 3a = 180

a = 60 cm.

Hence, length of each side is 60 cm.

Now, Area of △ABC can be calculated by Heron’s Formula, where

AB = a = 60 cm

BC = b = 60 cm

AC = c = 60 cm

Semi Perimeter (s) = (a+b+c)/2

s = 180/2

s = 90 cm

ar(△ABC) = √s(s-a)(s-b)(s-c)

= √90(90-60)(90-60)(90-60)

= √90×(30)×(30)×(30)

= 900√3 cm²

Hence, the area of the signal board = 900√3 cm²

Question 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.). The advertisements yield an earning of ₹ 5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Solution:

Here, Area of △ABC can be calculated by Heron’s Formula, where

AB = a = 120 m

BC = b = 22 m

AC = c = 122 m

Semi Perimeter (s) = (a+b+c)/2

s = (120+22+122)/2

s = 132 m

ar(△ABC) = √s(s-a)(s-b)(s-c)

= √132(132-120)(132-22)(132-122)

= √132×(12)×(110)×(10)

= 1320 m²

As it is given that,

For 1 year we cost

1 m² = ₹ 5000 

So, for 3 months,

1 m² = ₹ 5000 × (1/4)

For area of walls 1320 m² = 5000×(1/4)×(1320)

= ₹ 16,50,000

Hence, ₹ 16,50,000 much rent company will pay for 3 months.

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

Solution:

Here, Area of △ABC can be calculated by Heron’s Formula, where

AB = a = 11 m

BC = b = 6 m

AC = c = 15 m

Semi Perimeter (s) = (a+b+c)/2

s = (11+6+15)/2

s = 16 m

ar(△ABC) = √s(s-a)(s-b)(s-c)

= √16(16-11)(16-6)(16-15)

= √16×(5)×(10)×(1)

= 20√2 m²

Hence, the area painted in colour is 20√2 m²

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

Solution:

Here, length of two sides are given as 18 cm and 10 cm respectively.

and, perimeter = 42 cm.

Hence, length of third side = (Perimeter)-(length of two side)

= 42-(18+10)

AC = 14 cm

Here, Area of △ABC can be calculated by Heron’s Formula, where

AB = a = 18 cm

BC = b = 10 cm

AC = c = 14 cm

Semi Perimeter (s) = (a+b+c)/2

s = (18+10+14)/2

s = 21 cm

ar(△ABC) = √s(s-a)(s-b)(s-c)

= √21(21-18)(21-10)(21-14)

= √21×(3)×(11)×(7)

= 21√11 cm²

Hence, the area of triangle is 21√11 cm²

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

Solution:

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

Lets consider the common ratio between the sides of the triangle be p

Then, the sides are 12p, 17p and 25p

The perimeter of the triangle = 540 cm  (Given)

12p+17p+25p = 540 cm

54p = 540cm

So, p = 10

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

Here, Area of △ABC can be calculated by Heron’s Formula, where

AB = a = 250 cm

BC = b = 120 cm

AC = c = 170 cm

Semi Perimeter (s) = (a+b+c)/2

s = (250+120+170)/2

s = 270 cm

ar(△ABC) = √s(s-a)(s-b)(s-c)

= √270(270-250)(270-120)(270-170)

= √270×(20)×(150)×(100)

= 9000 cm²

Hence, the area of triangle is 9000 cm² .

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

Solution:

Here, length of two equal sides of isosceles triangle are given as 12 cm.

and, perimeter = 30 cm.

Hence, length of third side = (Perimeter)-(length of two side)

= 30-(12+12)

AC = 6 cm

Here, Area of △ABC can be calculated by Heron’s Formula, where

AB = a = 12 cm

BC = b = 12 cm

AC = c = 6 cm

Semi Perimeter (s) = (a+b+c)/2

s = (12+12+6)/2

s = 15 cm

ar(△ABC) = √s(s-a)(s-b)(s-c)

= √15(15-12)(15-12)(15-6)

= √15×(3)×(3)×(9)

= 9√15 cm²

Hence, the area of triangle is 9√15 cm²

NCERT Solutions Class 9 Maths Chapter 12 Heron’s Formula: Exercise 12.2

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

Solution:

Given, a quadrilateral ABCD where ∠C = 90º.

Construction: Join diagonal BD.

As we can see that, △DCB is right-angled at C

Hence, BC is the base and CD is height of △DCB, so

ar(△DCB) =  × Base × Height

=  × 12 × 5

= 30 m²……………………………………(1)

As △DCB is right angle triangle we can calculate third side by Pythagoras theorem

BD² = CB² + CD²

BD² = 12² + 5²

BD = √(144+25)

BD = √169

BD =13 m

Now, Area of △DAB can be calculated by Heron’s Formula, where 

AB = a = 9 m

AD = b = 8 m

BD = c = 13 m 

Semi Perimeter (s) = 

s = 

s = 15 m

ar(△DAB) = √s(s-a)(s-b)(s-c)

= √15(15-9)(15-8)(15-13)

= √15×(6)×(7)×(2)

= 6√35 m²……………………………………..(2)

From (1) and (2), we can conclude that,

ar(ABCD) = ar(△DCB)+ar(△DAB)

= (30 + 6√35)

= (30 + 35.5)

≈ 65.5 m² (approx.)

Question 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:

Here, we can notice that in △ABC

AC² = AB² + BC²

5² = 3² + 4²  

25 = 25

LHS = RHS

As this triangle is satisfying the Pythagoras theorem, Therefore, △ABC is a right angle triangle, 90° at B.

Hence, BC is the base and AB is height of △ABC. so

So, ar(△ABC) =  × Base × Height

=  × 4 × 3

= 6 cm²……………………………….(1)

Now, Area of △DAC can be calculated by Heron’s Formula, where

AD = a = 5 cm

DC = b = 4 cm

AC = c = 5 cm

Semi Perimeter (s) = 

s = 

s = 7 cm

ar(△DAC) = √s(s-a)(s-b)(s-c)

= √7(7-5)(7-4)(7-5)

= √7×(2)×(3)×(2)

= 2√21 cm²……………………………………..(2)

From (1) and (2), we can conclude that,

ar(ABCD) = ar(△ABC)+ar(△DAC)

= (6 + 2√21)

= (6 + 9.2(approx.))

≈ 15.2 cm² (approx.)

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

Solution:

Total area of the paper used = (Area I + Area II + Area III + Area IV + Area V)

Area I: Triangle

Now, Area of triangle can be calculated by Heron’s Formula, where

a = 5 cm

b = 5 cm

c = 1 cm

Semi Perimeter (s) = 

s = 

s = 5.5 cm

ar(△) = √s(s-a)(s-b)(s-c)

= √5.5(5.5-5)(5.5-5)(5.5-1)

= √5.5×(0.5)×(0.5)×(4.5)

= 0.5×0.5×3√11

= 0.75√11

 ≈ 2.5 cm² ……………………………………..(1)

Area II: Rectangle

Area of Rectangle = length×Breadth

= 1 × 6.5 = 6.5 cm²……………………………….(2)

Area III: Trapezium = Area of parallelogram EFAO + △ AFD

OD = 2cm

AD = OD-OA = 2-1 = 1 cm

Hence, △ AFD is equilateral.

PD =  AD =  cm

△ PFD is right angled at P, Pythagoras theorem is applicable 

1²=h² +()²

h = √(1-)

h = √ cm

Area III: 

= (Base × Height) + ( × Base × Height)

= (1 × ) + ( × 1 × )

= 

= = 1.29 cm²…………………………………(3)

Area IV and V: 2 times Triangle

ar(△) =  × Base × Height

=  × 6 × 1.5

= 4.5 cm²

Area IV + Area V = 2×ar(△)

= 2×4.5

= 9 cm² …………………………………(4)

Hence, Total area of the paper used = (Area I + Area II + Area III + Area IV + Area V)

= (1) + (2) + (3) + (4)

= 2.5 + 6.5 + 1.29 +9

= 19.29 cm²

Question 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:

Now, Area of △AEB can be calculated by Heron’s Formula, where

AE = a = 28 cm

EB = b = 30 cm

AB = c = 26 cm

Semi Perimeter (s) = 

s = (28+30+26)/2

s = 42 cm

ar(△AEB) = √s(s-a)(s-b)(s-c)

= √42(42-28)(42-30)(42-26)

= √42×(14)×(12)×(16)

= 336 cm²

As it is given, ar(△AEB) = ar(parallelogram ABCD)

336 = Base × Height

336 = 28 × h

h = 

h = 12 cm

Hence, the height of the parallelogram = 12 cm

Question 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:

ABCD is a rhombus having diagonal AC = 48 cm

side AB=BC=CD=AD=30 cm

Diagonal of Rhombus divides the area into two equal parts.

Now, ar(△ABC) can be calculated by Heron’s Formula, where

AB = a = 30 m

BC = b = 30 m

AC = c = 48 m

Semi Perimeter (s) = 

s = 

s = 54 m

ar(△ABC) = √s(s-a)(s-b)(s-c)

= √54(54-30)(54-30)(54-48)

= √54×(24)×(24)×(6)

= 432 m²

Hence, Area of rhombus = 2 × (ar(△))

= 2 × 432 = 864 m²

Area for 18 cows = 864 m²

Area for each cow = 864 / 18 = 48 m²

Question 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:

Let’s consider for each triangle.

Now, for each ar(△)can be calculated by Heron’s Formula, where

a = 50 cm

b = 50 cm

c = 20 cm

Semi Perimeter (s) = 

s = 

s = 60 cm

ar(△) = √s(s-a)(s-b)(s-c)

= √60(60-50)(60-50)(60-20)

= √60×(10)×(10)×(40)

= 200√6 cm²

Hence, the Total Area = 5×200√6

= 1000√6 cm²

Question 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. How much paper of each shade has been used in it? 

Solution:

As the area of kite is in the square, it area will be

Area of kite = ×(diagonal)²

= ×32×32

= 512 cm²

Diagonal divides area into equal areas.

Area of kite = Area I + Area II

512 = 2 × Area I

Area I =Area II = 256 cm²……………………………..(1)

Area III: Area of triangle

Now, for each ar(△)can be calculated by Heron’s Formula, where

a = 6 cm

b = 6 cm

c = 8 cm

Semi Perimeter (s) = 

s = 

s = 10 cm

ar(△) = √s(s-a)(s-b)(s-c)

= √10(10-6)(10-6)(10-8)

= √10×(4)×(4)×(2)

= 8√5 cm²………………………………(2)

Question 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). Find the cost of polishing the tiles at the rate of 50p per cm².

Solution:

Area for each triangle will be:

Now, for each ar(△)can be calculated by Heron’s Formula, where

a = 9 cm

b = 28 cm

c = 35 cm

Semi Perimeter (s) = 

s = 

s = 36 cm

ar(△) = √s(s-a)(s-b)(s-c)

= √36(36-9)(36-28)(36-35)

= √36×(27)×(8)×(1)

= 36√6 cm²

As there are 16 tiles, so total area = 16 × 36√6

= 1410.906 cm²

As 1 cm² = 50 p = ₹ 0.5

1410.906 cm² = ₹ 0.5 ×1410.906

= ₹ 705.45

Question 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:

AB = 25 m

EB = AB-AE = 25-10 = 15 m

Now, for ar(△ECB) can be calculated by Heron’s Formula, where

a = 13 m

b = 14 m

c = 15 m

Semi Perimeter (s) = 

s = 

s = 21 m

ar(△ECB) = √s(s-a)(s-b)(s-c)

= √21(21-3)(21-14)(21-15)

= √21×(18)×(7)×(6)

= 84 m² ………………………….(1)

ar(△ECB) =  × Base × Height

84 m² =  × 15 × h

h =  m

h = 11.2 m

Total Area = Area of parallelogram AECD + ar(△ECB)

= (Base × Height) + 84m² 

= 10 × 11.2 + 84

Total Area = 196 m²

Key Features of NCERT Solutions Class 9 Maths Chapter 12 Heron’s Formula :

• These NCERT solutions are developed by the GFG team, with a focus on students’ benefit.
• These solutions are entirely accurate and can be used by students to prepare for their board exams. 
• Each solution is presented in a step-by-step format with comprehensive explanations of the intermediate steps.

NCERT Solutions Class 9 Maths Chapter 12 Heron’s Formula- FAQs

1. Why is it important to learn Heron’s Formula?

Learning Heron’s formula is important because it provides a method to calculate the area of a triangle when the lengths of its sides are known. This formula is particularly useful when the triangle’s height or base is not easily determined. By understanding and applying Heron’s formula, students can accurately find the area of triangles, even in cases where other methods may be challenging or impractical.

2. What topics are covered in NCERT Solutions Class 9 Maths Chapter 12 Heron’s Formula?

NCERT Solutions for Class 9 Maths Chapter 12-Heron’s Formula includes Heron’s formula and its applications.

3. How can NCERT Solutions Class 9 Maths Chapter 12 Heron’s Formula help me?

NCERT Solutions for Class 9 Maths Chapter 12-Heron’s Formula can help you solve the NCERT exercise without any limitations. If you are stuck on a problem, you can find its solution in these solutions and free yourself from the frustration of being stuck on some question.

4. How many exercises are there in NCERT Solutions Class 9 Maths Chapter 12 Heron’s Formula?

There are 2 exercises in the Class 9 Maths Chapter 12-Heron’s Formula which covers all the important topics and sub-topics.

5. Where can I find NCERT Solutions Class 9 Maths Chapter 12 Heron’s Formula?

You can find these NCERT Solutions in this article created by our team of experts at GeeksforGeeks.