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

NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula.

Board	CBSE
Textbook	NCERT
Class	Class 9
Subject	Maths
Chapter	Chapter 12
Chapter Name	Heron’s Formula
Exercise	Ex 12.1, Ex 12.2.
Number of Questions Solved	15
Category	NCERT Solutions

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

Chapter 12 Heron’s Formula Ex 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 he the area of the signal board?
Solution:
We know that, an equilateral triangle has equal sides. So, all sides are equal to a.

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 Rs. 5000 per m² per year. A company hired one of its walls for 3 months. How much rent did it pay?

Solution:
The lengths of the sides of the triangular walls are 122 m, 22 m and 120 m.


Question 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 figure). If the sides of the wall are 15m, 11 m and 6m, find the area painted in colour.

Solution:

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:
Let the sides of a triangle, a = 18 cm, b = 10 cm and c
We have, perimeter = 42 cm
⇒  a + b + c = 42
⇒ 18 +10 + c = 42
⇒  c = (42 28) cm
⇒ c = 14 cm
Now a + b + c

Question 5.
Sides of a triangle are in the ratio of 12:17: 25 and its perimeter is 540 cm. Find its area.
Solution:
Given, sides of a triangle are in the ratio 12 :17 :25 and perimeter = 540 cm .
Let the sides of a triangle be a a = 12x, b = 17x and c = 25x.
Perimeter of a triangle = a + b + c
540 = 12x + 17x + 25x
⇒ 540 = 54x
⇒ x = 
∴ x = 10

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, each equal side of an isosceles triangle is 12 cm.
And perimeter of the given triangle is 30 cm
∴ Third side of the triangle = 30 -12 -12 = 6 cm

Chapter 12 Heron’s Formula Ex 12.2

Question 1.
A park in the shape of a quadrilateral ABCD has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Solution:
Given, quadrilateral ABCD in which AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m and ∠C = 90°. Now, join the diagonal BD which divides quadrilateral ABCD into two triangles.

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:
Area of quadrilateral ABCD = Area of Δ ABC + Area of Δ ACD
In Δ ABC,
We have, AB = 3 cm, BC = 4 cm, CA = 5 cm
Therefore, AB² + BC² = 3² + 4² = 9 +16 = 25 = (5)² = CA²

Question 3.
Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.
Solution:
For region I (triangle),
Let a = 5 cm, b = 5 cm and c = 1 cm
Then, semi-perimeter of I triangle




Similarly area of region IV (right angled triangle )
= 4.5 cm² [Since, it has same dimensions as region V]
∴ Total area of paper used = Area of region I + Area of region II + Area of region III +Area of region IV + Area of region V
= 2.49 + 6.5 +1. 30 + 4.5 + 4.5= 19.29 = 19.3 cm² (approx.)

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:
Let ABC be a triangle with sides
AB = 26 cm, BC = 28 cm, CA = 30 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:
Given, ABCD is a rhombus shaped field whose each side is 30 m and longer diagonal AC is 48 m.

Clearly, the diagonal AC divides the rhombus into two triangles, Δ ABC and Δ ADC which are congruent.
Area of Δ ABC = Area of Δ ADC
Also, Δ ABC and Δ ADC have equal perimeters.
Now, semi-perimeter of Δ ABC,

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

Solution:
Let given sides of one triangular piece of cloth be a = 20 cm, b = 50 cm and c = 50 cm.
Then, semi-perimeter of one triangular piece.

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 figure. How much paper of each shade has been used in it?
Solution:

Let the kite is made square ABCD and an isosceles ADEF.
Given, sides of ADEF are DE = DF = 6 cm and EF = 8 cm.
Also, diagonal of square ABCD = 32 cm
We know that, diagonals of a square bisect each other at right angle.
OA = OB = OC = OD = = 16 cm
Now, area of region I = 2 x Area of AAOB = 2 x x OA x OB
[∵ area of right angled triangle = x base x height]

Hence, area of paper of I colour used in making kite = 256 cm² Area of paper of II colour used in making kite = 256 cm² and area of paper of III colour used in making kite = 17.92 cm²

Question 8.
A floral design on a floor is made up of 16 tiles, which are triangular, the sides of the triangle being 28 cm, 9 cm and 35 cm, Find the cost of polishing the tiles at the rate of 50 paise per cm².

Solution:
Given, length of the side of the triangular tile are 28 cm, 9 cm and 35 cm.
Let a = 28 cm, b = 9 cm and c = 35 cm
Then, semi-perimeter of one triangular tile,

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

Let ABCD be the given field in the form of a trapezium in which AB = 25 m and CD = 10 m, BC = 13 m, AD = 14 m and DC || AB.
Through C, draw CE||DA and let it meets AB at E.
Let CP h m be the height of the trapezium.
Now, DC || AE     [∵ DC 11 AB given]
and  CE||DA       [by construction]
Thus, AECD is a parallelogram.
⇒ AE = DC = 10 m and CE = DA = 14 m
In A CEB, we have, CB = 13m, CE = 14 m
and BE = AB – AE = 25 -10 = 15 m
Let a = 14 m, b = 13 m and c = 15 m

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