NCERT Solutions Class 9 Maths
Chapter – 12 (Heron’s Formula)
The NCERT Solutions in English Language for Class 9 Mathematics Chapter – 12 Heron’s Formula Exercise 12.1 has been provided here to help the students in solving the questions from this exercise.
Chapter 12: Heron’s Formula
Exercise – 12.1
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?
Answer – Given : Dimensions of the traffic signal board and its perimeter.
Side of the signal board = a
Perimeter of the signal board = 3a = 180 cm
∴ a = 180/3 cm = 60 cm
Semi Perimeter, s = (a + b + c)/2 = (a + a + a)/2 =
By using Heron’s formula,
√[s(s – a)(s – b)(s – c)]
= √[s(s – a)(s – a)(s – a)]
= (s – a) √[s(s – a)]
We know, s = 3a/2,
Area = (3a/2 – a)√[3a/2(3a/2 – a)]
= (a/2) √[3a/2(a/2)]
= a/2 × a/2 × √3
= (√3/4)a2
Area of the signal board = (√3/4)a2 sq. units
Substituting the value of ‘a’.
Area of the signal board = (√3/4)(60)2
= (√3/4)(3600)
= 900√3
Area of the signal board = 900√3 cm2
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. 12.9). 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?
Answer – The sides of the triangle ABC are 122 m, 22 m and 120 m respectively.
Now, the perimeter will be (122 + 22 + 120) = 264 m
Semi Perimeter, s = (a + b + c)/2
= (122 + 22 + 120)/2
= 264/2
= 132 m
Using Heron’s formula,
Area of the triangle = √s(s – a)(s – b)(s – c)
= √132(132 – 122) (132 – 22) (132 – 120)
= √132 × 10 × 110 × 12
= √11 × 12 × 10 × 10 × 11 ×12
= 11 × 12 ×10
= 1320 m2
We know that the rent of advertising per year = ₹ 5000 per m2
∴ The rent of one wall for 3 months = Rs. (1320 × 5000 × 3)/12 = 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. 12.10 ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.
Answer – It is given that the sides of the wall as 15 m, 11 m and 6 m.
So, the semi perimeter of triangular wall (s) = (15 + 11 + 6)/2 m = 16 m
Using Heron’s formula,
Area of the message = √s(s – a)(s – b)(s – c)
= √[16(16 – 15)(16 – 11)(16 – 6)] m2
= √[16 × 1 × 5 × 10] m2
= √800 m2
= 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.
Answer – The sides of triangle given: a =18 cm, b = 10 cm
Perimeter of the triangle = (a + b + c)
42 = 18 + 10 + c
42 = 28 + c
c = 42 – 28
c = 14 cm
Semi Perimeter
s = (a + b + c) = 42/2 = 21 cm
Using Heron’s formula,
Area of the triangle = √s(s – a)(s – b)(s – c)
= √[21(21 – 18)(21 – 10)(21 – 14)] cm2
= √[21 × 3 × 11 × 7] m2
= 21√11 cm2
5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.
Answer – The ratio of the sides of the triangle are given as 12 : 17 : 25
So, we can assume the length of the sides of the triangle as 12x cm, 17x cm, and 25x cm.
It is also given that the perimeter of the triangle = 540 cm
12x + 17x + 25x = 540 cm
54x = 540cm
x = 10
Therefore, the sides of the triangle:
12x = 12 × 10 = 120 cm,
17x = 17 × 10 = 170 cm,
25x = 25 × 10 = 250 cm
a = 120cm, b = 170 cm, c = 250 cm
Semi-perimeter(s) = (120 + 170 + 250)/2 = 540/2 = 270 cm
By using Heron’s formula,
Area of a triangle = √s(s – a)(s – b)(s – c)
= √270(270 – 120)(270 – 170)(270 – 250)
= √270 × 150 × 100 × 20
= √81000000
= 9000 cm2
Area of the triangle = 9000 cm2.
6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Answer – First, let the third side be x.
It is given that the length of the equal sides is 12 cm and its perimeter is 30 cm.
30 = 12 + 12 + x
∴ The length of the third side = 6 cm
Semi Perimeter (s) = (a + b + c)/2 =
s = (12 + 12 + 6)/2 = 30/2 = 15 cm
Using Heron’s formula,
Area of the triangle = √s(s – a)(s – b)(s – c)
= √[15(15 – 12)(15 – 12)(15 – 6)] cm2
= √[15 × 3 × 3 × 9] cm2
= 9√15 cm2
