NCERT Solutions Class 9 Maths Chapter 10 Heron's Formula Exercise 10.1
Introduction:
In this exercise/article we will learn about Heron's Formula. We know you have studied in earlier classes about area of triangle and perimeter but in this chapter different types of formulae you will learn in the chapter. Perimeter of the triangle = \(\displaystyle \frac{{a\,+\,b\,+\,c}}{2}\) and Area of a triangle = \(\displaystyle \sqrt{{s\,(\,s\,-\,a\,)\,(\,s\,-\,b\,)\,(\,s\,-\,c\,)}}\) .
NCERT Class 9 Maths Chapter 10 Heron's Formula :
Class 9 Maths Exercise 10.1 (Page- 134)
Q1. 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 :
According to the question,
Let the each sides of a 'SCHOOL AHEAD' = a
Here, the semi-perimeter of the triangle = \(\displaystyle \frac{{a\,+\,a\,+\,a}}{2}\)
⇒ s = \(\displaystyle \frac{{a\,+\,a\,+\,a}}{2}\)
⇒ s = \(\displaystyle \frac{{3a}}{2}\)
So, Area of a triangle = \(\displaystyle \sqrt{{s\,(\,s\,-\,a\,)\,(\,s\,-\,b\,)\,(\,s\,-\,c\,)}}\)
= \(\displaystyle \sqrt{{\frac{{3a}}{2}(\frac{{3a}}{2}-a)(\frac{{3a}}{2}-a)(\frac{{3a}}{2}-a\,)}}\)
= \(\displaystyle \sqrt{{\frac{{3a}}{2}(\frac{{3a}}{2}-\frac{a}{1})(\frac{{3a}}{2}-\frac{a}{1})(\frac{{3a}}{2}-\frac{a}{1}\,)}}\) [ Use the LCM method ]
= \(\displaystyle \sqrt{{\frac{{3a}}{2}(\frac{{3a\,-\,2a}}{2})(\frac{{3a\,-\,2a}}{2})(\frac{{3a\,-\,2a}}{2}\,)}}\)
= \(\displaystyle \sqrt{{\frac{{3a}}{2}(\frac{a}{2})(\frac{a}{2})(\frac{a}{2}\,)}}\)
= \(\displaystyle \sqrt{{\frac{{3a}}{2}\times \frac{a}{2}\times \frac{a}{2}\times \frac{a}{2}\,}}\)
= \(\displaystyle \frac{a}{2}\,\times \,\frac{a}{2}\sqrt{3}\text{ }~\text{ }\)
= \(\displaystyle \frac{{{{a}^{2}}}}{4}\,\sqrt{3}\text{ }\)
Now, perimeter of the triangle = 180 cm
Sum of the sides are equal = a + a + a
⇒ a + a + a = 180
⇒ 3a = 180
⇒ a = \(\displaystyle \frac{{180}}{3}\)
⇒ a = 60 cm
Then, Area of the triangle \(\displaystyle \frac{{{{{60}}^{2}}}}{4}\,\sqrt{3}\,\text{ }~\text{ }\)
= \(\displaystyle \frac{{3600}}{4}\sqrt{3}\text{ }\!\!~\!\!\text{ }\)
= 900\(\displaystyle \sqrt{3}\) cm2
∴ The area of the signal board = 900\(\displaystyle \sqrt{3}\) cm2 .
Q2. 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. 10.6 ). 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 :
According to the question,
Given, The sides of triangular walls
⇒ a = 122 m
⇒ b = 22 m
⇒ c = 120 m
Now, Semi-perimeter of the triangle = \(\displaystyle \frac{{a\,+\,b\,+\,c}}{2}\)
s = \(\displaystyle \frac{{122\,+\,22\,+\,120}}{2}\)
s = \(\displaystyle \frac{{264}}{2}\)
s = 132 m
Here, Area of a triangle = \(\displaystyle \sqrt{{s\,(\,s\,-\,a\,)\,(\,s\,-\,b\,)\,(\,s\,-\,c\,)}}\)
= \(\displaystyle \sqrt{{132\,(\,10\,)\,(\,110\,)\,(\,12\,)}}\)
= \(\displaystyle \sqrt{{132\,\times \,10\,\times \,110\,\times \,12}}\)
= \(\displaystyle \sqrt{{11\,\times \,12\,\times \,10\,\times \,10\,\times \,11\,\times \,12}}\)
= 11 × 12 × 10
= 1320 m2
∴ The advertisements yield an 1 years earning = Rs 5000 m2
= A company rent pay the walls for 3 months = \(\displaystyle \frac{{5000\,\times \,1320\,\times \,3}}{{12}}\) [ because 1 years = 12 months ]
= 5000 × 110 × 3
= 16,50,000 m2
∴ The rent of the walls = 16,50,000 m2 .
Q3. 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. 10.7 ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.
Solution :
According to the question,
Given, the sides of the walls
⇒ a = 15 m
⇒ b = 11 m
⇒ c = 6 m
Now, Semi-perimeter of the triangle = \(\displaystyle \frac{{a\,+\,b\,+\,c}}{2}\)
s = \(\displaystyle \frac{{15\,+\,11\,+\,6}}{2}\)
s = \(\displaystyle \frac{{32}}{2}\)
s = 16 m
Here, Area of a triangle = \(\displaystyle \sqrt{{s\,(\,s\,-\,a\,)\,(\,s\,-\,b\,)\,(\,s\,-\,c\,)}}\)
= \(\displaystyle \sqrt{{16\,(\,16\,-\,15\,)\,(\,16\,-\,11\,)\,(\,16\,-\,6\,)}}\)
= \(\displaystyle \sqrt{{16\,(\,1\,)\,(\,5\,)\,(\,10\,)}}\)
= \(\displaystyle \sqrt{{16\,\times \,1\,\times \,5\,\times \,10}}\)
= \(\displaystyle \sqrt{{4\,\times \,4\,\times \,5\,\times \,2\,\times \,5}}\)
= 4 × 5\(\displaystyle \sqrt{2}\)
= 20\(\displaystyle \sqrt{2}\) m2
∴ The area painted in the colour = 20\(\displaystyle \sqrt{2}\) m2 .
Q4. Find the area of a triangle two sides of which are 18 cm, 10 cm and the perimeter is 42 cm.
Solution :
According to the question,
Here, a = 18 cm
b = 10 cm
Let the c = x cm
Since, the perimeter of the triangle = 42
⇒ a + b + c = 42
⇒ 18 + 10 + x = 42
⇒ 28 + x = 42
⇒ x = 42 - 28
⇒ x = 14 cm
Now, Semi-perimeter of the triangle = \(\displaystyle \frac{{a\,+\,b\,+\,c}}{2}\)
s = \(\displaystyle \frac{{42}}{2}\)
s = 21 cm
So, Area of a triangle = \(\displaystyle \sqrt{{s\,(\,s\,-\,a\,)\,(\,s\,-\,b\,)\,(\,s\,-\,c\,)}}\)
= \(\displaystyle \sqrt{{21(21-18)(21-10)(21-14)}}\)
= \(\displaystyle \sqrt{{21\,(\,3\,)\,(\,11\,)\,(\,7\,)}}\)
= \(\displaystyle \sqrt{{21\,\times \,3\,\times \,11\,\times \,7}}\)
= \(\displaystyle \sqrt{{3\,\times \,7\,\times \,3\,\times \,11\,\times \,7}}\)
= 3 × 7\(\displaystyle \sqrt{{11}}\)
= 21\(\displaystyle \sqrt{{11}}\) cm2
∴ The area of a triangle = 21\(\displaystyle \sqrt{{11}}\) cm2 .
Q5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
Solution :
According to the question,
Given, The ratio of the triangle = 12 : 17 : 25
The perimeter of the triangle = 540 cm
Let, the side of a triangle = 12x , 17x , 25x
So, 12x , 17x , 25x = 540
⇒ 54x = 540
⇒ x = \(\displaystyle \frac{{540}}{{54}}\)
⇒ x = 10 cm
Now, the side of a triangle = 12 × 10 = 120 cm
⇒ 17 × 10 = 170 cm
⇒ 25 × 10 = 250 cm
So, Semi-perimeter of the triangle = \(\displaystyle \frac{{a\,+\,b\,+\,c}}{2}\)
⇒ s = \(\displaystyle \frac{{540}}{2}\)
⇒ s = 270 cm
Now, Area of a triangle = \(\displaystyle \sqrt{{s\,(\,s\,-\,a\,)\,(\,s\,-\,b\,)\,(\,s\,-\,c\,)}}\)
= \(\displaystyle \sqrt{{270\,(270-120)(\,270-170)(270-\,250)}}\)
= \(\displaystyle \sqrt{{270\,(\,150\,)\,(\,100\,)\,(\,20\,)}}\)
= \(\displaystyle \sqrt{{2\,\times \,5\,\times \,3\,\times \,3\,\times \,3\,\times \,10\,\times \,5\,\times \,3\,\times \,10\,\times \,10\,\times \,2\,\times \,10}}\)
= 2 × 3 × 3 × 5 × 10 × 10
= 9,000 cm2
∴ The area of a triangle = 9,000 cm2 .
Q6. An isosceles triangle has perimeter 30 cm and each of the equal side is 12 cm. Find the area of the triangle.
Solution :
According to the question,
Given, the perimeter of triangle = 30 cm
Here, a = 12 cm
b = 12 cm
Let the c = x
So, a + b + c = 30
⇒ 12 + 12 + x = 30
⇒ 24 + x = 30
⇒ x = 30 - 24
⇒ x = 6 cm
Now, Semi-perimeter of the triangle = \(\displaystyle \frac{{a\,+\,b\,+\,c}}{2}\)
⇒ s = \(\displaystyle \frac{{30}}{2}\)
⇒ s = 15 cm
Then, Area of a triangle = \(\displaystyle \sqrt{{s\,(\,s\,-\,a\,)\,(\,s\,-\,b\,)\,(\,s\,-\,c\,)}}\)
= \(\displaystyle \sqrt{{15(15-12)(15-12)(15-6)}}\)
= \(\displaystyle \sqrt{{15\,(\,3\,)\,(\,3\,)\,(\,9\,)}}\)
= \(\displaystyle \sqrt{{3\,\times \,5\,\times \,3\,\times \,3\,\times \,3\,\times \,3}}\)
= 3 × 3\(\displaystyle \sqrt{{3\,\times \,5}}\)
= 9\(\displaystyle \sqrt{{15}}\) cm2
∴ The area of the triangle = 9\(\displaystyle \sqrt{{15}}\) cm2 .