NCERT Solutions for Class 9 Maths Chapter 10 – Heron’s Formula

Heron’s Formula is an important and scoring chapter in Class 9 Mathematics that provides a method for finding the area of a triangle when the lengths of all three sides are known. This chapter covers key topics such as Heron’s formula, semi-perimeter of a triangle, and its application to find the area of various types of triangles. This formula is extremely useful in school exams and competitive exams like JEE and NEET.

NCERT Solutions for Class 9 Maths Chapter 10 – Heron’s Formula are prepared strictly according to the latest CBSE syllabus and exam pattern. The solutions are written in simple, step-by-step language with clear explanations, solved examples, and tips to apply Heron’s formula effectively.

NCERT Solutions for Class 9 Maths Chapter 10 – Heron’s Formula

Heron\'s Formula

Medium

Q.

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?

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Heron\'s Formula

Difficult

Q.

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 following figure.). 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?
 

           

 

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Heron\'s Formula

Difficult

Q.

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

  

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Heron\'s Formula

Difficult

Q.

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

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Heron\'s Formula

Difficult

Q.

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

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Heron\'s Formula

Medium

Q.

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

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Heron\'s Formula

Medium

Q.

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?

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Heron\'s Formula

Medium

Q.

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.

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Heron\'s Formula

Medium

Q.

Radha made a picture of an aeroplane with coloured paper as shown in the following figure. Find the total area of the paper used.
 

   

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Heron\'s Formula

Medium

Q.

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.

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Heron\'s Formula

Medium

Q.

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?

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Heron\'s Formula

Medium

Q.

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?

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Heron\'s Formula

Medium

Q.

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

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Heron\'s Formula

Medium

Q.

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

  

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Heron\'s Formula

Medium

Q.

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.

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Q1) The perimeter of a triangle is 60 cm. Its sides are in the ratio 3:4:5. Find the area of the triangle.

Answer:
Let the sides of the triangle be

$3x, 4x, 5x$

3x,4x,5x.

The perimeter is given as 60 cm, so:

$$3x + 4x + 5x = 60$$

3x+4x+5x=60

$$12x = 60$$

12x=60

$$x = \frac{60}{12} = 5$$

x=1260​=5

Thus, the sides of the triangle are:

$$3x = 15 \, \text{cm}, \quad 4x = 20 \, \text{cm}, \quad 5x = 25 \, \text{cm}.$$

3x=15cm,4x=20cm,5x=25cm.

Now, we use Heron’s formula to find the area of the triangle.

The semi-perimeter

$s$

s is given by:

$$s = \frac{15 + 20 + 25}{2} = 30 \, \text{cm}.$$

s=215+20+25​=30cm.

Using Heron’s formula:

$$A = \sqrt{s(s-a)(s-b)(s-c)}$$

A=s(s−a)(s−b)(s−c)​

where

$a = 15$

a=15,

$b = 20$

b=20, and

$c = 25$

c=25.

Substitute the values:

$$A = \sqrt{30(30-15)(30-20)(30-25)}$$

A=30(30−15)(30−20)(30−25)​

$$A = \sqrt{30 \times 15 \times 10 \times 5}$$

A=30×15×10×5​

$$A = \sqrt{22500}$$

A=22500​

$$A = 150 \, \text{cm}^2.$$

A=150cm2.

Thus, the area of the triangle is 150 cm².

Q2) The area of an equilateral triangle is 16√3 cm². Find its perimeter.

Answer:
Let the side of the equilateral triangle be

$a$

a.

The area

$A$

A of an equilateral triangle is given by the formula:

$$A = \frac{\sqrt{3}}{4} a^2.$$

A=43​​a2.

We are given that the area is

$16\sqrt{3} \, \text{cm}^2$

163​cm2, so:

$$\frac{\sqrt{3}}{4} a^2 = 16\sqrt{3}.$$

43​​a2=163​.

Simplify the equation:

$$a^2 = \frac{16\sqrt{3} \times 4}{\sqrt{3}} = 64.$$

a2=3​163​×4​=64.

$$a = \sqrt{64} = 8 \, \text{cm}.$$

a=64​=8cm.

Now, the perimeter

$P$

P of an equilateral triangle is given by:

$$P = 3a = 3 \times 8 = 24 \, \text{cm}.$$

P=3a=3×8=24cm.

Thus, the perimeter of the equilateral triangle is 24 cm.

Q3) A right-angled triangle has legs 6 cm and 8 cm. Find its area.

Answer:
In a right-angled triangle, the area

$A$

A is given by:

$$A = \frac{1}{2} \times \text{base} \times \text{height}.$$

A=21​×base×height.

Here, the legs are 6 cm and 8 cm, so:

$$A = \frac{1}{2} \times 6 \times 8 = \frac{1}{2} \times 48 = 24 \, \text{cm}^2.$$

A=21​×6×8=21​×48=24cm2.

Thus, the area of the right-angled triangle is 24 cm².

Q4) Find the area of a triangle whose sides are 7 cm, 8 cm, and 9 cm.

Answer:
Let the sides of the triangle be

$a = 7 \, \text{cm}, b = 8 \, \text{cm}, c = 9 \, \text{cm}$

a=7cm,b=8cm,c=9cm.

We can use Heron’s formula to find the area of the triangle.

First, calculate the semi-perimeter

$s$

s:

$$s = \frac{7 + 8 + 9}{2} = 12 \, \text{cm}.$$

s=27+8+9​=12cm.

Now, use Heron’s formula:

$$A = \sqrt{s(s-a)(s-b)(s-c)}.$$

A=s(s−a)(s−b)(s−c)​.

Substitute the values:

$$A = \sqrt{12(12-7)(12-8)(12-9)}$$

A=12(12−7)(12−8)(12−9)​

$$A = \sqrt{12 \times 5 \times 4 \times 3} = \sqrt{720}.$$

A=12×5×4×3​=720​.

$$A = 26.83 \, \text{cm}^2.$$

A=26.83cm2.

Thus, the area of the triangle is approximately 26.83 cm².

FAQs: Class 9 Maths Chapter 10 – Heron’s Formula

Q1. Is Heron’s Formula important for Class 9 exams?
Yes, it is a high-weightage chapter with numericals based on finding the area of a triangle.

Q2. What is Heron’s formula?
Heron’s formula for the area of a triangle is:
Area = √[s(s-a)(s-b)(s-c)], where s is the semi-perimeter and a, b, c are the sides of the triangle.

Q3. Are numericals asked from this chapter?
Yes, area of triangles using Heron’s formula is a common question in exams.

Q4. Are there any important tips for solving problems in this chapter?
Yes, always calculate the semi-perimeter (s) first before applying Heron’s formula. Double-check the side lengths to avoid calculation errors.

Q5. How do NCERT Solutions help?
They provide step-by-step solutions with detailed explanations and help in mastering the application of Heron’s formula.