Heron’s Formula
NCERT Solutions • Class 9 Maths • Chapter 10Exercise 10.1
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?
Part 1: Area using Heron’s Formula
Semi-perimeter $s = \frac{a+a+a}{2} = \frac{3a}{2}$.
Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{3a}{2}(\frac{3a}{2}-a)(\frac{3a}{2}-a)(\frac{3a}{2}-a)}$
$= \sqrt{\frac{3a}{2}(\frac{a}{2})(\frac{a}{2})(\frac{a}{2})} = \frac{a^2}{4}\sqrt{3} =$ $\frac{\sqrt{3}}{4}a^2$.
Semi-perimeter $s = \frac{a+a+a}{2} = \frac{3a}{2}$.
Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{3a}{2}(\frac{3a}{2}-a)(\frac{3a}{2}-a)(\frac{3a}{2}-a)}$
$= \sqrt{\frac{3a}{2}(\frac{a}{2})(\frac{a}{2})(\frac{a}{2})} = \frac{a^2}{4}\sqrt{3} =$ $\frac{\sqrt{3}}{4}a^2$.
Part 2: If Perimeter = 180 cm
Side $a = \frac{180}{3} = 60 \text{ cm}$.
Area $= \frac{\sqrt{3}}{4}(60)^2 = \frac{\sqrt{3}}{4}(3600) = 900\sqrt{3}$.
Area $= 900\sqrt{3} \text{ cm}^2$.
Side $a = \frac{180}{3} = 60 \text{ cm}$.
Area $= \frac{\sqrt{3}}{4}(60)^2 = \frac{\sqrt{3}}{4}(3600) = 900\sqrt{3}$.
Area $= 900\sqrt{3} \text{ cm}^2$.
2. The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $122 \text{ m}, 22 \text{ m}$ and $120 \text{ m}$. The advertisements yield an earning of ₹ 5000 per $\text{m}^2$ per year. A company hired one of its walls for 3 months. How much rent did it pay?
Step 1: Calculate Area
$a = 122, b = 22, c = 120$.
$s = \frac{122 + 22 + 120}{2} = \frac{264}{2} = 132 \text{ m}$.
Area $= \sqrt{132(132-122)(132-22)(132-120)}$
$= \sqrt{132 \times 10 \times 110 \times 12} = \sqrt{(12 \times 11) \times 10 \times (11 \times 10) \times 12}$
$= \sqrt{12^2 \times 11^2 \times 10^2} = 12 \times 11 \times 10 = 1320 \text{ m}^2$.
$a = 122, b = 22, c = 120$.
$s = \frac{122 + 22 + 120}{2} = \frac{264}{2} = 132 \text{ m}$.
Area $= \sqrt{132(132-122)(132-22)(132-120)}$
$= \sqrt{132 \times 10 \times 110 \times 12} = \sqrt{(12 \times 11) \times 10 \times (11 \times 10) \times 12}$
$= \sqrt{12^2 \times 11^2 \times 10^2} = 12 \times 11 \times 10 = 1320 \text{ m}^2$.
Step 2: Calculate Rent
Rate = ₹ 5000 per $\text{m}^2$ per year.
Time = 3 months $= \frac{3}{12}$ year $= \frac{1}{4}$ year.
Total Rent $= \text{Area} \times \text{Rate} \times \text{Time}$
$= 1320 \times 5000 \times \frac{1}{4} = 330 \times 5000 =$ ₹ 16,50,000.
Rate = ₹ 5000 per $\text{m}^2$ per year.
Time = 3 months $= \frac{3}{12}$ year $= \frac{1}{4}$ year.
Total Rent $= \text{Area} \times \text{Rate} \times \text{Time}$
$= 1320 \times 5000 \times \frac{1}{4} = 330 \times 5000 =$ ₹ 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”. If the sides of the wall are $15 \text{ m}, 11 \text{ m}$ and $6 \text{ m}$, find the area painted in colour.
Step 1: Semi-perimeter
$a=15, b=11, c=6$.
$s = \frac{15+11+6}{2} = \frac{32}{2} = 16 \text{ m}$.
Step 2: Area
Area $= \sqrt{16(16-15)(16-11)(16-6)}$
$= \sqrt{16 \times 1 \times 5 \times 10} = \sqrt{16 \times 50} = 4\sqrt{25 \times 2}$
$= 4 \times 5\sqrt{2} =$ $20\sqrt{2} \text{ m}^2$.
$a=15, b=11, c=6$.
$s = \frac{15+11+6}{2} = \frac{32}{2} = 16 \text{ m}$.
Step 2: Area
Area $= \sqrt{16(16-15)(16-11)(16-6)}$
$= \sqrt{16 \times 1 \times 5 \times 10} = \sqrt{16 \times 50} = 4\sqrt{25 \times 2}$
$= 4 \times 5\sqrt{2} =$ $20\sqrt{2} \text{ m}^2$.
4. Find the area of a triangle two sides of which are $18 \text{ cm}$ and $10 \text{ cm}$ and the perimeter is $42 \text{ cm}$.
Step 1: Find third side (c)
Perimeter $= a + b + c = 42$.
$18 + 10 + c = 42 \Rightarrow 28 + c = 42 \Rightarrow c = 14 \text{ cm}$.
$s = \frac{42}{2} = 21 \text{ cm}$.
Step 2: Calculate Area
Area $= \sqrt{21(21-18)(21-10)(21-14)}$
$= \sqrt{21 \times 3 \times 11 \times 7} = \sqrt{(7 \times 3) \times 3 \times 11 \times 7}$
$= \sqrt{7^2 \times 3^2 \times 11} = 21\sqrt{11}$.
Area $= 21\sqrt{11} \text{ cm}^2$.
Perimeter $= a + b + c = 42$.
$18 + 10 + c = 42 \Rightarrow 28 + c = 42 \Rightarrow c = 14 \text{ cm}$.
$s = \frac{42}{2} = 21 \text{ cm}$.
Step 2: Calculate Area
Area $= \sqrt{21(21-18)(21-10)(21-14)}$
$= \sqrt{21 \times 3 \times 11 \times 7} = \sqrt{(7 \times 3) \times 3 \times 11 \times 7}$
$= \sqrt{7^2 \times 3^2 \times 11} = 21\sqrt{11}$.
Area $= 21\sqrt{11} \text{ cm}^2$.
5. Sides of a triangle are in the ratio of $12 : 17 : 25$ and its perimeter is $540 \text{ cm}$. Find its area.
Step 1: Find sides
Let sides be $12x, 17x, 25x$.
$12x + 17x + 25x = 540 \Rightarrow 54x = 540 \Rightarrow x = 10$.
Sides: $a=120, b=170, c=250$.
$s = \frac{540}{2} = 270 \text{ cm}$.
Let sides be $12x, 17x, 25x$.
$12x + 17x + 25x = 540 \Rightarrow 54x = 540 \Rightarrow x = 10$.
Sides: $a=120, b=170, c=250$.
$s = \frac{540}{2} = 270 \text{ cm}$.
Step 2: Calculate Area
Area $= \sqrt{270(270-120)(270-170)(270-250)}$
$= \sqrt{270 \times 150 \times 100 \times 20}$
$= \sqrt{(27 \times 10) \times (15 \times 10) \times 100 \times (2 \times 10)}$
$= \sqrt{9 \times 3 \times 3 \times 5 \times 2 \times 100000} = \sqrt{9 \times 9 \times 10 \times 100000}$
$= \sqrt{81000000} = 9000$.
Area $= 9000 \text{ cm}^2$.
Area $= \sqrt{270(270-120)(270-170)(270-250)}$
$= \sqrt{270 \times 150 \times 100 \times 20}$
$= \sqrt{(27 \times 10) \times (15 \times 10) \times 100 \times (2 \times 10)}$
$= \sqrt{9 \times 3 \times 3 \times 5 \times 2 \times 100000} = \sqrt{9 \times 9 \times 10 \times 100000}$
$= \sqrt{81000000} = 9000$.
Area $= 9000 \text{ cm}^2$.
6. An isosceles triangle has perimeter $30 \text{ cm}$ and each of the equal sides is $12 \text{ cm}$. Find the area of the triangle.
Step 1: Find third side (b)
$a = 12, c = 12$. Perimeter $= 30$.
$12 + b + 12 = 30 \Rightarrow b = 6 \text{ cm}$.
$s = \frac{30}{2} = 15 \text{ cm}$.
Step 2: Calculate Area
Area $= \sqrt{15(15-12)(15-6)(15-12)}$
$= \sqrt{15 \times 3 \times 9 \times 3} = \sqrt{15 \times 81}$
$= 9\sqrt{15}$.
Area $= 9\sqrt{15} \text{ cm}^2$.
$a = 12, c = 12$. Perimeter $= 30$.
$12 + b + 12 = 30 \Rightarrow b = 6 \text{ cm}$.
$s = \frac{30}{2} = 15 \text{ cm}$.
Step 2: Calculate Area
Area $= \sqrt{15(15-12)(15-6)(15-12)}$
$= \sqrt{15 \times 3 \times 9 \times 3} = \sqrt{15 \times 81}$
$= 9\sqrt{15}$.
Area $= 9\sqrt{15} \text{ cm}^2$.