Exercise-10.1
Maths - Exercise
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?
Given: The signal board is an equilateral triangle with perimeter \(180~\text{cm}\)
Find the length of each side
$$\begin{aligned} a &= \frac{\text{Perimeter}}{3} \\&= \frac{180}{3} \\&= 60~\text{cm} \end{aligned}$$
Find the semi-perimeter
$$\begin{aligned} s &= \frac{\text{Perimeter}}{2} \\&= \frac{180}{2} \\&= 90~\text{cm} \end{aligned}$$
Calculate the area using Heron's formula
$$ \begin{aligned} \text{Area} &= \small\sqrt{s(s-a)(s-b)(s-c)} \\ &= \scriptsize\sqrt{90 \times (90-60) \times (90-60) \times (90-60)} \\ &= \sqrt{90 \times 30 \times 30 \times 30} \\ &= \sqrt{90 \times 27,000} \\ &= \sqrt{2,430,000} \\ &= 900 \sqrt{3}~\text{cm}^2 \end{aligned} $$
Answer: The area of the signal board is $$900\sqrt{3}~\text{cm}^2$$
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 Fig. 10.6). The advertisements yield an earning of ` 5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?
Given: The sides of the triangular wall are $$\begin{aligned} a &= 122\,\text{m}\\b &= 22\,\text{m}\\ c &= 120\,\text{m} \end{aligned}$$
Calculate the semi-perimeter
$$\begin{aligned}
s &= \frac{a + b + c}{2} \\&= \frac{122 + 22 + 120}{2} \\&= \frac{264}{2} \\&= 132\,\text{m}
\end{aligned}$$
Calculate $$ \begin{aligned} s - a &= 132 - 122 = 10\,\text{m} \\ s - b &= 132 - 22 = 110\,\text{m} \\ s - c &= 132 - 120 = 12\,\text{m} \end{aligned} $$
Calculate area using Heron's formula
$$
\begin{aligned}
A &= \sqrt{s(s-a)(s-b)(s-c)} \\
&= \sqrt{132 \times 10 \times 110 \times 12} \\
&= \sqrt{132 \times 10 \times 110 \times 12} \\
&= \sqrt{1,742,400} \\
&= 1,320\,\text{m}^2
\end{aligned}
$$
Calculate advertising cost for 3 months
Yearly rate per \( = ₹5000 /\text{m}^2\)
So, cost for 3 months:
$$\begin{aligned}
\text{Cost} &= \text{Area} \times \text{Rate} \times \frac{3}{12} \\\\&= 1,320 \times 5,000 \times
\frac{3}{12}\\\\
&= 1,320 \times 5,000 \times \frac{1}{4} \\\\
&= 1,320 \times 1,250 \\
&= ₹1,650,000
\end{aligned}
$$
Answer: The company paid ₹1,650,000 for 3 months’ rent.
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.
Given: The sides of the wall are $$\begin{aligned} a &= 15\, \text{m}\\b& = 11\, \text{m}\\c &= 6\, \text{m}\end{aligned}$$
Calculate the semi-perimeter
$$\begin{aligned}
s &= \frac{a + b + c}{2} \\&= \frac{15 + 11 + 6}{2} \\&= \frac{32}{2} \\&= 16\, \text{m}
\end{aligned}$$
Calculate
$$
\begin{aligned}
s - a &= 16 - 15 = 1\\
s - b &= 16 - 11 = 5\\
s - c &= 16 - 6 = 10
\end{aligned}
$$
Find the area using Heron's formula
$$
\begin{aligned}
A &= \sqrt{s(s-a)(s-b)(s-c)}\\
&= \sqrt{16 \times 1 \times 5 \times 10}\\
&= \sqrt{16 \times 50}\\
&= \sqrt{800}\\
&= \sqrt{16 \times 50} \\
&= 4 \sqrt{50}\\
&= 4 \times \sqrt{25 \times 2} \\
&= 4 \times 5 \sqrt{2}\\
&= 20 \sqrt{2}\, \text{m}^2
\end{aligned}
$$
Answer: The area painted in colour is $$20 \sqrt{2}\, \text{m}^2$$
Q4. Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.
Given: Two sides of the triangle are $$\begin{aligned}a &= 18\,\text{cm and } \\b &= 10\,\text{cm}\end{aligned}$$ with perimeter $$P = 42\,\text{cm}$$
Find the third side
$$\begin{aligned}
c &= P - (a + b) \\&= 42 - (18 + 10) \\&= 42 - 28 \\&= 14\,\text{cm}
\end{aligned}$$
Step 2: Calculate the semi-perimeter
$$
s = \frac{P}{2} = \frac{42}{2} = 21\,\text{cm}
$$
Calculate \((s-a),\,(s-b)\text{ and } (s-c)\)
$$
\begin{aligned}
s - a &= 21 - 18 = 3\,\text{cm}\\
s - b &= 21 - 10 = 11\,\text{cm}\\
s - c &= 21 - 14 = 7\,\text{cm}
\end{aligned}
$$
Calculate the area using Heron's formula
$$
\begin{aligned}
A &= \sqrt{s(s-a)(s-b)(s-c)} \\
&= \sqrt{21 \times 3 \times 11 \times 7} \\
&= \sqrt{21 \times 3 \times 7 \times 11} \\
&= \sqrt{21 \times 21 \times 11} \\
&= 21\sqrt{11}\,\text{cm}^2
\end{aligned}
$$
Answer: The area of the triangle is $$21\sqrt{11}\,\text{cm}^2$$
Q5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.
Given: The sides of the triangle are in the ratio $$12 : 17 : 25$$ and the perimeter is $$540\,\text{cm}$$
Let the sides be $$12x, 17x, 25x$$
Sum of ratios: $$12 + 17 + 25 = 54$$
Find the value of \(x\)
$$\begin{aligned}
12x + 17x + 25x &= 54x\\54x &= 540\\
\\\implies x &= \frac{540}{54} \\&= 10
\end{aligned}$$
Find the lengths of the sides
$$
\begin{aligned}
a &= 12x = 12 \times 10 = 120\,\text{cm}\\
b &= 17x = 17 \times 10 = 170\,\text{cm}\\
c &= 25x = 25 \times 10 = 250\,\text{cm}
\end{aligned}
$$
Calculate the semi-perimeter
$$\begin{aligned}
s &= \frac{a + b + c}{2} \\&= \frac{540}{2} \\&= 270\,\text{cm}
\end{aligned}$$
Find \((s - a),\, (s - b),\, (s - c)\)
$$
\begin{aligned}
s-a &= 270 - 120 = 150\,\text{cm}\\
s-b &= 270 - 170 = 100\,\text{cm}\\
s-c &= 270 - 250 = 20\,\text{cm}
\end{aligned}
$$
Calculate the area using Heron's formula
$$
A = \sqrt{ 270 \times 150 \times 100 \times 20 }
$$
On calculating:
$$
270 \times 150 = 40,500 \\
40,500 \times 100 = 4,050,000 \\
4,050,000 \times 20 = 81,000,000 \\
A = \sqrt{81,000,000 } = 9,000\,\text{cm}^2
$$
Answer: The area of the triangle is $$9,000\,\text{cm}^2$$
Q6. An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.
Given: Perimeter of the isosceles triangle is $$30\,\text{cm},$$ and each of the equal sides is $$12\,\text{cm}$$
Find the length of the third side
$$\begin{aligned}
c &= 30 - (12 + 12) \\&= 30 - 24 \\&= 6\,\text{cm}
\end{aligned}$$
List the side lengths
$$
\begin{aligned}
a &= 12\,\text{cm} \\
b &= 12\,\text{cm} \\
c &= 6\,\text{cm}
\end{aligned}
$$
Calculate the semi-perimeter
$$\begin{aligned}
s &= \frac{a + b + c}{2} \\&= \frac{30}{2} \\&= 15\,\text{cm}
\end{aligned}$$
Calculate \((s-a),\,(s-b),\,(s-c)\)
$$
\begin{aligned}
s - a &= 15 - 12 = 3\,\text{cm}\\
s - b &= 15 - 12 = 3\,\text{cm}\\
s - c &= 15 - 6 = 9\,\text{cm}
\end{aligned}
$$
Find the area using Heron's formula
$$
\begin{aligned}
A &= \sqrt{ s (s-a)(s-b)(s-c) }\\
&= \sqrt{ 15 \times 3 \times 3 \times 9 }\\
&= \sqrt{ 15 \times 9 \times 9 }\\
&= \sqrt{ 15 \times 81 }\\
&= \sqrt{ 1,215 }\\
&= 9 \sqrt{15}\,\text{cm}^2
\end{aligned}
$$
Answer: The area of the triangle is $$9\sqrt{15}\,\text{cm}^2$$
