NCERT solutions for class 9 maths chapter 12 Ex 12.1
In this page we have NCERT solutions for class 9 maths chapter 12 Heron Formula for
EXERCISE 12.1 . Hope you like them and do not forget to like , social share
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Chapter 12 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 be the area of the signal board? Solution
Semi-perimeter of triangle =s=(a+a+a)/2=3a/2
Using Heron's formula,
$A= \sqrt {s(s-a)(s-b)(s-c)}$
Where s is the semi-perimeter of triangle .
And, a, b and c are lengths of sides of triangle.
$A= \sqrt {s(s-a)(s-b)(s-c)}$
$A= \sqrt {(3a/2) (\frac {3a}{2}-a)(\frac {3a}{2}-b)(\frac {3a}{2}-c)}$
$A=\frac {a^2 \sqrt 3}{4}$
If perimeter of equilateral triangle is equal to 180 cm then we can easily find length of each side.
Let length of each side =a cm.
According to the given condition, we have
a+a+a=180
3a=180
a=60 cm
Putting this value in Area formula we derived above, we get
Area of triangle =900√3 cm^{2}
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 the given figure). 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
Sides of triangle a, b, c are of 122 m, 22 m, and 120 m respectively
s = (122 + 22 + 120)/2 m
s = 132 m
By Heron's formula
$A= \sqrt {s(s-a)(s-b)(s-c)}$
A=1320 m^{2}
Rent of 1 m^{2 }area per year = Rs 5000
Rent of 1 m^{2} area per month = Rs 5000/12
Rent of 1320 m^{2} area for 3 months = Rs.(5000/12) X 1320
= Rs 1650000
So, company had to pay Rs 1650000.
Question 3
There is a slide in a park. One of its side walls has been painted in some color with a message “KEEP THE PARK GREEN AND CLEAN” . If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in color.
Solution
It can be observed that the area to be painted in color is a triangle, having its sides as 11 m, 6 m, and 15 m.
s = (11 + 6 + 15)/2 m
s = 16 m
By Heron’s formula,
$A= \sqrt {s(s-a)(s-b)(s-c)}$
A=20√2 m^{2}
Therefore, the area painted in color is 20√2 m^{2}
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=18 cm
b=10 cm
Perimeter of triangle (a+b+c) =42 cm
c =42−18−10=14 cm
s =(18+10+14)/2=21 cm
Using Heron's formula to find area of triangle, we get
$A= \sqrt {s(s-a)(s-b)(s-c)}$
A=20√11 cm^{2}
Question 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 common ratio between the sides of given triangle be x.
So, side of triangle will be 12x, 17x, and 25x.
Perimeter of this triangle = 540 cm
12x + 17x + 25x = 540 cm
54x = 540 cm
x = 10 cm
Sides of triangle will be 120 cm, 170 cm, and 250 cm.
s=(a+b+c)/2=270 cm
By Heron's formula
$A= \sqrt {s(s-a)(s-b)(s-c)}$
A=9000 cm^{2}
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
Let the third side of this triangle be x.
Perimeter of triangle = 30 cm
12 cm + 12 cm + x = 30 cm
x = 6 cm
s= 30/2=15 cm
By Heron’s formula,
$A= \sqrt {s(s-a)(s-b)(s-c)}$
$A=9 \sqrt {15} \ cm^2$
Summary
NCERT solutions for class 9 maths chapter 12 Heron Formula Ex 12.1 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also Download this assignment as pdf
This chapter 12 has total 2 Exercise 12.1 and 12.2. This is the first exercise in the chapter.