Exercise 10.1 Practice
Area of Triangles using Heron's Formula
Q1: Equilateral Triangle
00:00
A traffic signal board, indicating 'SLOW DOWN', is an equilateral triangle with side '$a$'. Find the area of the signal board, using Heron's formula. If its perimeter is $180 \text{ cm}$, what will be the area of the signal board?
Using Heron's Formula for side '$a$':
Semi-perimeter $s = \frac{a+a+a}{2} = \frac{3a}{2}$.
Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{3a}{2}(\frac{3a}{2}-a)(\frac{3a}{2}-a)(\frac{3a}{2}-a)}$
$= \sqrt{\frac{3a}{2} \cdot \frac{a}{2} \cdot \frac{a}{2} \cdot \frac{a}{2}} = \sqrt{\frac{3a^4}{16}} = \frac{\sqrt{3}}{4}a^2$.
Semi-perimeter $s = \frac{a+a+a}{2} = \frac{3a}{2}$.
Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{\frac{3a}{2}(\frac{3a}{2}-a)(\frac{3a}{2}-a)(\frac{3a}{2}-a)}$
$= \sqrt{\frac{3a}{2} \cdot \frac{a}{2} \cdot \frac{a}{2} \cdot \frac{a}{2}} = \sqrt{\frac{3a^4}{16}} = \frac{\sqrt{3}}{4}a^2$.
For Perimeter = 180 cm:
$3a = 180 \Rightarrow a = 60 \text{ cm}$.
Area $= \frac{\sqrt{3}}{4}(60)^2 = \frac{\sqrt{3}}{4} \times 3600 = 900\sqrt{3} \text{ cm}^2$.
$3a = 180 \Rightarrow a = 60 \text{ cm}$.
Area $= \frac{\sqrt{3}}{4}(60)^2 = \frac{\sqrt{3}}{4} \times 3600 = 900\sqrt{3} \text{ cm}^2$.
Area $= 900\sqrt{3} \text{ cm}^2$
Q2: Advertisement Rent
00:00
The triangular side walls of a flyover have been used for advertisements. The sides of the walls are $50 \text{ m}$, $120 \text{ m}$ and $130 \text{ m}$. The advertisements yield an earning of ₹$4000$ per $\text{m}^2$ per year. A company hired one of its walls for $4$ months. How much rent did it pay?
Sides are $a=50, b=120, c=130$.
$s = \frac{50+120+130}{2} = \frac{300}{2} = 150 \text{ m}$.
$s = \frac{50+120+130}{2} = \frac{300}{2} = 150 \text{ m}$.
Area $= \sqrt{s(s-a)(s-b)(s-c)}$
$= \sqrt{150(150-50)(150-120)(150-130)}$
$= \sqrt{150 \times 100 \times 30 \times 20}$
$= \sqrt{9000000} = 3000 \text{ m}^2$.
$= \sqrt{150(150-50)(150-120)(150-130)}$
$= \sqrt{150 \times 100 \times 30 \times 20}$
$= \sqrt{9000000} = 3000 \text{ m}^2$.
Rent for 1 year = ₹$4000$ per $\text{m}^2$.
Rent for 4 months $= 3000 \times 4000 \times \frac{4}{12}$
$= 12000000 \times \frac{1}{3} =$ ₹$40,00,000$.
Rent for 4 months $= 3000 \times 4000 \times \frac{4}{12}$
$= 12000000 \times \frac{1}{3} =$ ₹$40,00,000$.
Rent = ₹$40,00,000$
Q3: Painted Area
00:00
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 $15 \text{ m}$, $20 \text{ m}$ and $25 \text{ m}$, find the area painted in colour.
Sides are $a=15, b=20, c=25$.
$s = \frac{15+20+25}{2} = \frac{60}{2} = 30 \text{ m}$.
$s = \frac{15+20+25}{2} = \frac{60}{2} = 30 \text{ m}$.
Area $= \sqrt{30(30-15)(30-20)(30-25)}$
$= \sqrt{30 \times 15 \times 10 \times 5}$
$= \sqrt{(2 \times 3 \times 5) \times (3 \times 5) \times (2 \times 5) \times 5}$
$= \sqrt{2^2 \times 3^2 \times 5^4} = 2 \times 3 \times 25 = 150 \text{ m}^2$.
$= \sqrt{30 \times 15 \times 10 \times 5}$
$= \sqrt{(2 \times 3 \times 5) \times (3 \times 5) \times (2 \times 5) \times 5}$
$= \sqrt{2^2 \times 3^2 \times 5^4} = 2 \times 3 \times 25 = 150 \text{ m}^2$.
Area $= 150 \text{ m}^2$
Q4: Find Third Side
00:00
Find the area of a triangle two sides of which are $11 \text{ cm}$ and $15 \text{ cm}$ and the perimeter is $40 \text{ cm}$.
Perimeter $= 40 \text{ cm}$.
Third side $c = 40 - (11 + 15) = 40 - 26 = 14 \text{ cm}$.
Third side $c = 40 - (11 + 15) = 40 - 26 = 14 \text{ cm}$.
$s = \frac{40}{2} = 20 \text{ cm}$.
Area $= \sqrt{20(20-11)(20-15)(20-14)}$
$= \sqrt{20 \times 9 \times 5 \times 6}$
$= \sqrt{5400} = \sqrt{900 \times 6} = 30\sqrt{6} \text{ cm}^2$.
$= \sqrt{20 \times 9 \times 5 \times 6}$
$= \sqrt{5400} = \sqrt{900 \times 6} = 30\sqrt{6} \text{ cm}^2$.
Area $= 30\sqrt{6} \text{ cm}^2$
Q5: Ratio of Sides
00:00
Sides of a triangle are in the ratio of $3 : 5 : 7$ and its perimeter is $300 \text{ m}$. Find its area.
Let the sides be $3x, 5x, 7x$.
$3x + 5x + 7x = 300 \Rightarrow 15x = 300 \Rightarrow x = 20$.
$3x + 5x + 7x = 300 \Rightarrow 15x = 300 \Rightarrow x = 20$.
Sides are:
$a = 3(20) = 60 \text{ m}$
$b = 5(20) = 100 \text{ m}$
$c = 7(20) = 140 \text{ m}$
$a = 3(20) = 60 \text{ m}$
$b = 5(20) = 100 \text{ m}$
$c = 7(20) = 140 \text{ m}$
$s = \frac{300}{2} = 150 \text{ m}$.
Area $= \sqrt{150(150-60)(150-100)(150-140)}$
$= \sqrt{150 \times 90 \times 50 \times 10}$
$= \sqrt{6750000} = \sqrt{2250000 \times 3} = 1500\sqrt{3} \text{ m}^2$.
Area $= \sqrt{150(150-60)(150-100)(150-140)}$
$= \sqrt{150 \times 90 \times 50 \times 10}$
$= \sqrt{6750000} = \sqrt{2250000 \times 3} = 1500\sqrt{3} \text{ m}^2$.
Area $= 1500\sqrt{3} \text{ m}^2$
Q6: Isosceles Triangle
00:00
An isosceles triangle has perimeter $32 \text{ cm}$ and each of the equal sides is $10 \text{ cm}$. Find the area of the triangle.
Perimeter $= 32 \text{ cm}$. Equal sides $a = b = 10 \text{ cm}$.
Third side $c = 32 - (10 + 10) = 12 \text{ cm}$.
Third side $c = 32 - (10 + 10) = 12 \text{ cm}$.
$s = \frac{32}{2} = 16 \text{ cm}$.
Area $= \sqrt{16(16-10)(16-10)(16-12)}$
$= \sqrt{16 \times 6 \times 6 \times 4}$
$= 4 \times 6 \times 2 = 48 \text{ cm}^2$.
$= \sqrt{16 \times 6 \times 6 \times 4}$
$= 4 \times 6 \times 2 = 48 \text{ cm}^2$.
Area $= 48 \text{ cm}^2$