CLASS IX (NCERT) Heron's Formula

Class 9 Chapter 12 Heron’s Formula Ex 12.1

July 2, 2021
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Heron’s Formula

  Class 9 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

Each side of the equilateral triangle be a.                            (Given)

Derivation of Area of Equilateral TriangleSemi-perimeter of the triangle

AB = a     BC = b       AC = c

S=\frac{a+b+c}{2}

\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}           (Heron’s Formula)

\sqrt{S\left ( S-a \right )\left ( S-a \right )\left ( S-a \right )}

a = b = c= a

S=\frac{a+a+a}{2}

S=\frac{3a}{2}

Area of triangle = \sqrt{\tfrac{3}2{a}\left ( \tfrac{3}2{a-a} \right )\left ( \tfrac{3}2{a-a} \right )\left ( \tfrac{3}2{a-a} \right )}

=   \sqrt{\left ( \tfrac{3}2{a} \right )\left ( \frac{3a-2a}{2} \right )\left ( \frac{3a-2a}{2} \right )\left ( \frac{3a-2a}{2} \right )}

=  \sqrt{\frac{3}2{a}\left ( \frac{a}2{} \right )\left ( \frac{a}{2} \right )\left ( \frac{a}{2} \right )}

= \sqrt{3\times \frac{a}{2}\times \frac{a}{2}\times \frac{a}{2}\times \frac{a}{2}}                   ( Now make  Pair of a/2)

∴               =  \frac{a}{2}\times \frac{a}{2}\sqrt{3}

Area of equilateral triangle = \frac{\sqrt{3}}{4}a^{2}

Perimeter of triangle = 180

a + b + c = 180

a + a + a = 180

3a = 180

a=60

The area of the signal board= \frac{\sqrt{3}}{4}\times 60\times 60

=900\sqrt{3}cm^{2}

 

Class 9 Exercise 12.1 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 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?
NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula Ex 12.1 Q2

Solution

a = 122       b = 120        c = 22

S = \frac{a+b+c}{2}        (Semi-Perimeter)

S = \frac{122+120+22}{2}

S = \frac{164}{2}

S = 132

The area of the triangular side wall = \sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}           (Heron’s Formula)

=\sqrt{132\left ( 132-122 \right )\left ( 132-120 \right )\left ( 132-22 \right )}

=\sqrt{132\times 10\times 12\times 110}

=\sqrt{2\times 2\times 3\times 11\times 2\times 5\times 2\times 2\times 3\times 2\times 5\times 11 }

\sqrt{2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 5\times 5\times 11\times 11 }     (Now, arrange and make pair)

=  2\times 3\times 11\times 2\times 2\times 5

= 120\times 11

= 1320 m^{2}

Rent for 3 months for 1320 m2  = 1320\times \frac{5000}{12}\times 3       ( rent for one month \frac{5000}{12} )

= 330\times 5000

= ₹ 1650000

 

Class 9 Exercise 12.1 Question 3.
There is a slide in a park. One of its side Company hired one of its walls for 3 months. 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 15 m, 11 m and 6m, find the area painted in colour.
NCERT Solutions for Class 9 Maths Chapter 12 Heron's Formula Ex 12.1 Q3
Solution:

Let the sides of the wall be

a = 11               b = 15             c = 6

S=\frac{a+b+c}{2}        (Semi-Perimeter)

=\frac{11+15+6}{2}

S=\frac{32}{2}

S=16

Area of The Triangular Surface of the wall =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}                 (Heron’s Formula)

=\sqrt{16\left ( 16-11 \right )\left ( 16-15 \right )\left ( 16-6 \right )}

=\sqrt{16\times 5\times 1\times 10}

=\sqrt{4\times 4\times 5\times 5\times 2}

=4\times 5\sqrt{2}

Thus, the required area painted in colour =20\sqrt{2}m^{^{2}}

 

Class 9 Exercise 12.1 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:

a= 18cm        b = 10cm      Perimeter Of Triangle = 42cm (Given)

We know, Perimeter Of Triangle = a + b+ c

42  =  18 + 10 + c

c   =  42-28

c   =  14

S = \frac{a+b+c}{2}        (Semi-Perimeter)

S = \frac{18+10+14}{2} = \frac{42}{2}

S = 21 cm

The required area of the triangle  =  \sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}      (Heron’s Formula)

=  \sqrt{21\left ( 21-18 \right )\left ( 21-10 \right )\left ( 21-14 \right )}

=  \sqrt{21\times 3\times 11\times 7}

=  \sqrt{7\times 3\times 3\times 11\times 7}

=  \sqrt{7\times 7\times 3\times 3\times 11}                      (Now, arrange and make pair)

=  7\times 3\sqrt{11}

Thus, The required area of the triangle= 21\sqrt{11}cm^{2}

Class 9 Exercise 12.1 Question 5.
Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
Solution:

Perimeter of triangle = 540 cm

Let the sides of the triangle be

First side = a = 12x

Second side = b = 17x

Third side = c = 25x

Perimeter of Triangle = a + b+ c

540 = 12x+17x+25x     (Giver Perimeter = 540)

540 = 54x

x = \frac{540}{54} = 10

x = 10 cm

a = 12\times 10 = 120cm

b = 17\times 10 = 170cm

c = 25\times 10 = 250cm

a= 120cm      b= 170cm      c= 250cm

S = \frac{a+b+c}{2}              (Semi-Perimeter)

S = \frac{120+170+250}{2} = \frac{540}{2}

S = 270cm

Area of Triangle = \sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}         (Heron’s Formula)

= \sqrt{270\left ( 270-120 \right )\left ( 270-170 \right )\left ( 270-250 \right )}

= \sqrt{270\times 150\times 100\times 20}

= \sqrt{3\times 3\times 3\times 10\times 5\times 3\times 10\times 10\times 10\times 2\times 10}

= \sqrt{3\times 3\times 3\times 3\times 10\times 10\times 10\times 10\times 5\times 2\times 10}        (Now, arrange and make pair)

= 3\times 3\times 10\times 10\times 10

thus, Area of Triangle = 9000 cm^{2}

 

Class 9 Exercise 12.1 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:

a = b = 12 cm  ( Equal Sides of isosceles Triangle)

c = ?

Perimeter of Triangle = 30 cm

a + b + c = 30

12+12+c = 30

c = 30 – 24

c = 6 cm

S = \frac{a+b+c}{2}          (Semi-Perimeter)

S = \frac{12 + 12 + 6}{2} = \frac{30}{2} = 15

∴ Area of Triangle  = \sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}      (Heron’s Formula)

= \sqrt{15\left ( 15-12 \right )\left ( 15-12 \right )\left ( 15-6 \right )}

= \sqrt{15\left ( 3 \right )\left ( 3 \right )\left ( 9\right )}

= \sqrt{3\times 5\times 3\times 3\times 3\times 3}

= \sqrt{3\times 3\times 3\times 3\times 3\times 5}      (Now, arrange and make pair)

= 3\times 3\sqrt{15}

Thus, The required area of the triangle= 9\sqrt{15}cm^{2}

 

 

 

 

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