NCERT Solutions for Class 9 Maths Chapter 12 - Heron's Formula

Question

Our NCERT Solutions Class 9 Maths Chapter 12 Heron's Formula is one of the best ways to study for the CBSE term II examination. We have discussed all kinds of concepts and topics in such a way that it is easy to understand for the students. NCERT solutions provided here are based on the latest syllabus proposed by the CBSE. Students who are looking to clear the basic concepts of NCERT chapters must prepare for competitive exams like JEE Mains, JEE Advance, NTSE, Olympiad, or any similar exams and must opt for our solutions.

NCERT Solutions for Class 9 Maths Chapter 12 - Heron's formula is a fundamental concept that has applications in a wide range of fields, and it is covered in the first term of the CBSE Class 9 Maths syllabus. As a result, it is critical to have a firm grasp of the concept. One of the best ways to achieve this is to consult the NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula. These solutions are created by knowledgeable teachers with years of experience in accordance with the most recent CBSE term-wise syllabus update 2022-23. The NCERT Solutions for Class 9 aim to provide students with detailed and step-by-step explanations for all of the answers to the questions in this Chapter's exercises.

Heron’s Formula is used to find the area of a triangle. This formula is helpful where it is not possible to find the height of the triangle easily.

NCERT Solutions is one of the best study guides you could use. Relevant topics are presented in an easy-to-understand format, with no complicated jargon used. Furthermore, its content is kept up to date with the most recent CBSE term-wise syllabus and guidelines.

Answer 1

NCERT Solutions for Class 9 Maths Chapter 12 - Heron's Formula

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:

Let each side of the equilateral triangle be a.

Semi-perimeter of the triangle,

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:

Let the sides of the triangular will be

a = 122m, b = 12cm, c = 22m

Semi-perimeter, \(s = \frac {a + b + c}2\)

\((\frac{122+ 120 + 22}4)m = \frac{264}2 m = 132 m\)

The area of the triangular side wall

Rent for 1 year (i.e. 12 months) per m² = Rs. 5000

∴ Rent for 3 months per m² = Rs. 5000 x \(\frac3{12}\)

= Rent for 3 months for 1320 m² = Rs. 5000 x \(\frac3{12}\) x 1320 = Rs. 16,50,000.

3. 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” (see Fig). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Solution:

Let the sides of the wall be

a = 15m, b = 11m, c = 6m

Semi-perimeter,

Thus, the required area painted in colour

= 20√2 m²

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

Solution:

Let the sides of the triangle be a = 18 cm, b = 10 cm and c = x cm

Since, perimeter of the triangle = 42 cm

∴ 18cm + 10 cm + x cm = 42

x = [42 – (18 + 10)cm = 14cm

Now, semi-perimeter, s = \(\frac{42}2\) cm = 21 cm

Thus, the required area of the triangle = \(21\sqrt{11}\) cm²

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

Solution:

Let the sides of the triangle be

a = 12x cm, b = 17x cm, c = 25x cm

Perimeter of the triangle = 540 cm

Now, 12x + 17x + 25x = 540

⇒ 54x = 54 ⇒ x = 10

∴ a = (12 x 10)cm = 120cm,

b = (17 x 10) cm = 170 cm

and c = (25 x 10)cm = 250 cm

Now, semi-perimeter, s = \(\frac{540}2\)cm = 270 cm

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

Solution:

Let the sides of an isosceles triangle be

a = 12cm, b = 12cm, c = x cm

Since, perimeter of the triangle = 30 cm

∴ 12cm + 12cm + x cm = 30 cm

⇒ x = (30 – 24) = 6

Now, semi-perimeter, s = \(\frac{30}2\)cm =15 cm

Thus, the required area of the triangle = 9√15 cm²

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

Solution:

Given, a quadrilateral ABCD with ZC = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m.

Let us join B and D, such that ABCD is a right angled triangle.

Now, to find the area of ∆ABD, we need the length of BD.

In right-angled ∆BCD, by Pythagoras theorem

BD² = 50² + CD²

⇒ BD² = 12² + 5²

⇒ BD² = 144 + 25 = 169

⇒ BD = 13 m

Now, for ∆ABD, we have

a = AB = 9 m, b = AD = 8 m, c = BD = 13 m

∴ Area of quadrilateral ABCD = area of ∆BCD + area of ∆ABD

= 30 m² + 35.5 m²

= 65.5 m² (approx.)

8. 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:

Given a quadrilateral ABCD with AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

For ∆ABC, a = AB = 3 cm, b = BC = 4 cm and c = AC = 5 cm

Now, area of quadrilateral ABCD = area of ∆ABC + area of ∆ACD

= 6 cm² + 9.2 cm²

= 15.2 cm² (approx.)

Answer 2

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

Solution:

For surface I:

It is an isosceles triangle whose sides are a = 5 cm, b = 5 cm, c = 1 cm

= (0.75 x 3.3) cm²

= 2.475 cm² (approx.)

For surface II:

It is a rectangle with length 6.5 cm and breadth 1 cm.

∴ Area of surface II = Length x Breadth

= (6.5 x 1) cm² = 6.5 cm²

For surface III:

It is a trapezium whose parallel sides are 1 cm and 2 cm as shown in the figure given below:

For surface IV and V:

Surface V is a right-angled triangle with base 6cm arid height 1.5 cm.

Also, area of surface IV = area of surface V

= 12 x base x height

= (12 x 6 x 15) cm² = 4.5 cm²

Thus, the total area of the paper used = (area of surface I) + (area of surface II) + (area of surface III) + (area of surface IV) + (area of surface V)

= [2.475 + 6.5 + 1.3 + 4.5 + 4.5] cm²

= 19.275 cm²

= 19.3 cm² (approx.)

10. 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:

For the given triangle, we have a = 28 cm, b = 30 cm, c = 26 cm

Area of the given parallelogram = Area of the given triangle

∴ Area of the parallelogram = 336 cm²

⇒ base x height = 336

⇒ 28 x h = 336, where ‘h’ be the height of the parallelogram.

⇒ h = \(\frac{336}{28}\) = 12

Thus, the required height of the parallelogram = 12 cm

11. 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:

Here, each side of the rhombus = 30 m.

Let ABCD be the given rhombus and the diagonal, BD = 48 m

Sides ∆ABC are a = AB = 30m, b = AD = 30m, c = BD = 48m

Since, a diagonal divides the rhombus into two congruent triangles.

∴ Area of triangle II = 432 m²

Now, total area of the rhombus = Area of triangle I + Area of triangle II

= 432 m² + 432 m² = 864 m²

Area of grass for 18 cows to graze = 864 m²

⇒ Area of grass for 1 cow to graze = \(\frac{864}{18}\) m²

= 48 m²

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

Solution:

Let the sides of each triangular piece be

a = 20 cm, b = 50 cm, c = 50 cm

13. 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?

Solution:

Each shade of paper is divided into 3 triangles i.e., I, II, III

8 cm

For triangle I:

ABCD is a square [Given]

∵ Diagonals of a square are equal and bisect each other.

∴ AC = BD = 32 cm

Height of AABD = OA = (\(\frac12\) x 32 )cm

= 16 cm

Area of triangle I = (\(\frac12\) x 32 x 16 ) cm²

= 256cm²

For triangle II:

Since, diagonal of a square divides it into two congruent triangles.

So, area of triangle II = area of triangle I

∴ Area of triangle II = 256 cm²

For triangle III:

The sides are given as a = 8 cm, b = 6 cm and c = 6 cm

Thus, the area of different shades are:

Area of shade I = 256 cm²

Area of shade II = 256 cm²

and area of shade III = 17.92 cm²

Answer 3

14. 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 figure). Find the cost of polishing the tiles at the rate of 50 paise per cm.

Solution:

Let the sides of the triangle be a = 9 cm, b = 28 cm, c = 35 cm

Total area of all the 16 triangles = (16 x 88.2) cm² = 1411.2 cm² (approx.)

Cost of polishing the tiles = Rs. 0.5 per cm²

∴ Cost of polishing all the tiles = Rs. (0.5 x 1411.2) = Rs. 705.60 (approx.)

15. 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:

Let the given field is in the form of a trapezium ABCD such that parallel sides are AB = 10 m and DC = 25 m

Non-parallel sides are AD = 13 m and BC = 14 m.

We draw BE || AD, such that BE = 13 m.

The given field is divided into two shapes (i) ∆BCE, (ii) parallelogram ABED For ∆BCE:

Sides of the triangle are a = 13 m, b = 14 m, c = 15 m

For parallelogram ABED:

Let the height of the ∆BCE corresponding to the side EC be h m.

Area of a triangle = \(\frac12\) x base x height

∴ \(\frac12\) x 15 x h = 84

⇒ 10 + \(\frac{82\times 2}{15}\) = \(\frac{56}5\)

Now, area of a parallelogram = base x height

= (10 x \(\frac{56}5\)) = (2 x 56) m² = 112 m²

So, area of the field

= area of ∆BCE + area of parallelogram ABED

= 84 m² + 112 m² = 196 m²