Chapter 10: Heron's Formula
Exercise 10.1
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:
Part 1: General Formula
Let the side of the traffic signal board be \( a \).
Perimeter of the traffic signal board \( = 3a \)
Semi-perimeter \( s = \frac{3a}{2} \)
By Heron's formula,
\[ \text{Area of triangle} = \sqrt{s(s-a)(s-b)(s-c)} \]
Since it is an equilateral triangle, \( a = b = c \). Therefore,
\[ \text{Area} = \sqrt{\frac{3a}{2} \left(\frac{3a}{2} - a\right) \left(\frac{3a}{2} - a\right) \left(\frac{3a}{2} - a\right)} \]
\[ = \sqrt{\frac{3a}{2} \left(\frac{a}{2}\right) \left(\frac{a}{2}\right) \left(\frac{a}{2}\right)} \]
\[ = \sqrt{3 \times \frac{a^4}{16}} \]
\[ = \frac{\sqrt{3}}{4} a^2 \]
Part 2: Specific Calculation
Given, Perimeter of traffic signal board \( = 180 \text{ cm} \).
\[ 3a = 180 \text{ cm} \]
\[ a = \frac{180}{3} \text{ cm} = 60 \text{ cm} \]
Using the derived formula, the area of the traffic signal board is:
\[ \text{Area} = \frac{\sqrt{3}}{4} (60)^2 \text{ cm}^2 \]
\[ = \left( \frac{3600}{4} \sqrt{3} \right) \text{ cm}^2 \]
\[ = 900\sqrt{3} \text{ cm}^2 \]
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 the figure). The advertisements yield an earning of Rs 5000 per \( m^2 \) per year. A company hired one of its walls for 3 months. How much rent did it pay?
Solution:
The sides of the triangle (i.e., \( a, b, c \)) are 122 m, 22 m, and 120 m respectively.
First, we find the perimeter and semi-perimeter (\( s \)).
\[ \text{Perimeter} = 122 + 22 + 120 = 264 \text{ m} \]
\[ s = \frac{264}{2} = 132 \text{ m} \]
By Heron's formula,
\[ \text{Area of triangle} = \sqrt{s(s-a)(s-b)(s-c)} \]
\[ = \sqrt{132(132-122)(132-22)(132-120)} \text{ m}^2 \]
\[ = \sqrt{132(10)(110)(12)} \text{ m}^2 \]
We can break the numbers into factors to simplify the square root:
\[ 132 = 12 \times 11 \]
\[ 110 = 11 \times 10 \]
Substituting these back:
\[ \text{Area} = \sqrt{(12 \times 11) \times 10 \times (11 \times 10) \times 12} \text{ m}^2 \]
\[ = \sqrt{12^2 \times 11^2 \times 10^2} \text{ m}^2 \]
\[ = 12 \times 11 \times 10 \text{ m}^2 = 1320 \text{ m}^2 \]
Rent Calculation:
Rent of \( 1 \text{ m}^2 \) area per year = Rs 5000
Rent of \( 1 \text{ m}^2 \) area per month = Rs \( \frac{5000}{12} \)
Rent for 3 months for \( 1320 \text{ m}^2 \) area:
\[ \text{Total Rent} = \text{Rate} \times \text{Area} \times \text{Time} \]
\[ = \text{Rs } \left( \frac{5000}{12} \times 1320 \times 3 \right) \]
\[ = \text{Rs } \left( 5000 \times 110 \times 3 \right) \]
\[ = \text{Rs } 1650000 \]
Therefore, the company had to pay Rs 1,650,000.
Q3
There is a slide in the park. One of its side walls has been painted in the same colour with a message "KEEP THE PARK GREEN AND CLEAN" (see the figure). If the sides of the wall are 15 m, 11 m, and 6 m, find the area painted in colour.
Solution:
The area to be painted is a triangle with sides 15 m, 11 m, and 6 m.
Perimeter \( = 15 + 11 + 6 = 32 \text{ m} \)
Semi-perimeter \( s = \frac{32}{2} = 16 \text{ m} \)
By Heron's formula,
\[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]
\[ = \sqrt{16(16-15)(16-11)(16-6)} \text{ m}^2 \]
\[ = \sqrt{16 \times 1 \times 5 \times 10} \text{ m}^2 \]
\[ = \sqrt{16 \times 50} \text{ m}^2 \]
\[ = 4 \times \sqrt{25 \times 2} \text{ m}^2 \]
\[ = 4 \times 5 \sqrt{2} \text{ m}^2 \]
\[ = 20\sqrt{2} \text{ m}^2 \]
Therefore, the area painted in colour is \( 20\sqrt{2} \text{ m}^2 \).
Q4
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 third side of the triangle be \( x \).
Given perimeter \( = 42 \text{ cm} \).
\[ 18 \text{ cm} + 10 \text{ cm} + x = 42 \text{ cm} \]
\[ 28 \text{ cm} + x = 42 \text{ cm} \]
\[ x = 14 \text{ cm} \]
Semi-perimeter \( s = \frac{42}{2} = 21 \text{ cm} \).
By Heron's formula,
\[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]
\[ = \sqrt{21(21-18)(21-10)(21-14)} \text{ cm}^2 \]
\[ = \sqrt{21(3)(11)(7)} \text{ cm}^2 \]
Grouping factors:
\[ = \sqrt{(3 \times 7) \times 3 \times 11 \times 7} \text{ cm}^2 \]
\[ = \sqrt{3^2 \times 7^2 \times 11} \text{ cm}^2 \]
\[ = 3 \times 7 \times \sqrt{11} \text{ cm}^2 \]
\[ = 21\sqrt{11} \text{ cm}^2 \]
Q5
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 be \( x \).
The sides of the triangle are \( 12x \), \( 17x \), and \( 25x \).
Given perimeter \( = 540 \text{ cm} \).
\[ 12x + 17x + 25x = 540 \]
\[ 54x = 540 \]
\[ x = 10 \]
Therefore, the sides are:
- \( a = 12 \times 10 = 120 \text{ cm} \)
- \( b = 17 \times 10 = 170 \text{ cm} \)
- \( c = 25 \times 10 = 250 \text{ cm} \)
Semi-perimeter \( s = \frac{540}{2} = 270 \text{ cm} \).
By Heron's formula,
\[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]
\[ = \sqrt{270(270-120)(270-170)(270-250)} \text{ cm}^2 \]
\[ = \sqrt{270 \times 150 \times 100 \times 20} \text{ cm}^2 \]
Simplifying:
\[ = \sqrt{27 \times 10 \times 15 \times 10 \times 100 \times 20} \]
\[ = \sqrt{81000000} \text{ cm}^2 \]
Alternatively, breaking into factors:
\[ = \sqrt{(9 \times 30) \times (5 \times 30) \times 100 \times (4 \times 5)} \]
\[ = \sqrt{9 \times 100 \times 4 \times 25 \times 30 \times 30} \]
\[ = 3 \times 10 \times 2 \times 5 \times 30 \]
\[ = 9000 \text{ cm}^2 \]
Therefore, the area of the triangle is \( 9000 \text{ cm}^2 \).
Q6
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 the triangle be \( x \).
Perimeter \( = 30 \text{ cm} \).
Equal sides are 12 cm each.
\[ 12 \text{ cm} + 12 \text{ cm} + x = 30 \text{ cm} \]
\[ 24 \text{ cm} + x = 30 \text{ cm} \]
\[ x = 6 \text{ cm} \]
Semi-perimeter \( s = \frac{30}{2} = 15 \text{ cm} \).
By Heron's formula,
\[ \text{Area} = \sqrt{s(s-a)(s-b)(s-c)} \]
\[ = \sqrt{15(15-12)(15-12)(15-6)} \text{ cm}^2 \]
\[ = \sqrt{15 \times 3 \times 3 \times 9} \text{ cm}^2 \]
\[ = \sqrt{15 \times 81} \text{ cm}^2 \]
\[ = 9\sqrt{15} \text{ cm}^2 \]