Exercise 10.1 (Revised) - Chapter 12 - Heron's Formula - Ncert Solutions class 9 - Maths

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Chapter 10: Heron's Formula - NCERT Solutions for Class 9 Maths

Ex 10.1 Question 1.

A traffic signal board, indicating 'SCHOOLAHEAD' is an equilateral triangle with side ' $a$ '. Find the area of the signal board, using Heron's formula. If its perimeter is $\mathbf{1 8 0} \mathbf{~ c m}$, what will be the area of the signal board?

Answer.

Let the Traffic signal board is $\triangle \mathrm{ABC}$.

According to question, Semi-perimeter of $\Delta \mathrm{ABC}(s)=\frac{a+a+a}{2}=\frac{3 a}{2}$

Using Heron's Formula, Area of triangle $\mathrm{ABC}=\sqrt{s(s-a)(s-b)(s-c)}$
$
\begin{aligned}
& =\sqrt{\frac{3 a}{2}\left(\frac{3 a}{2}-a\right)\left(\frac{3 a}{2}-a\right)\left(\frac{3 a}{2}-a\right)} \\
& =\sqrt{\frac{3 a}{2} \times \frac{a}{2} \times \frac{a}{2} \times \frac{a}{2}}=\sqrt{3\left(\frac{a}{2}\right)^4} \\
& =\frac{\sqrt{3} a^2}{4}
\end{aligned}
$

Now, Perimeter of this triangle $=180 \mathrm{~cm} \Rightarrow$ Side of triangle ${ }^{(a)}=\frac{180}{3}=60 \mathrm{~cm}$
$\Rightarrow$ Semi-perimeter of this triangle $=\frac{180}{2}=90 \mathrm{~cm}$
Using Heron's Formula, Area of this triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$
\begin{aligned}
& =\sqrt{90(90-60)(90-60)(90-60)} \\
& =\sqrt{90 \times 30 \times 30 \times 30} \\
& =30 \times 30 \sqrt{3} \\
& =900 \sqrt{3} \mathrm{~cm}^2
\end{aligned}
$
Ex 10.1 Question 2.

The triangular side walls of a flyover has been used for advertisements. The sides of the walls are $122 \mathrm{~m}, \mathbf{2 2} \mathrm{m}$ and $120 \mathrm{~m}$ (see figure). The advertisement yield an earning of Rs. $\mathbf{5 0 0 0}$ per $\mathbf{m}^2$ per year. A company hired one of its walls for 3 months, how much rent did it pay?

Answer.

Given: $a=122 \mathrm{~m}, b=22 \mathrm{~m}$ and $c=120 \mathrm{~m}$

Semi-perimeter of triangle $(s)=\frac{122+22+120}{2}=\frac{264}{2}=132 \mathrm{~m}$

Using Heron's Formula,

Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{132(122-132)(132-22)(132-120)}$
$=\sqrt{132 \times 10 \times 110 \times 12}$
$=\sqrt{11 \times 12 \times 10 \times 10 \times 11 \times 12}$
$=10 \times 11 \times 12$
$=1320 m^2$
$\because$ Rent for advertisement on wall for 1 year $=$ Rs. 5000 per $m^2$
$\therefore$ Rent for advertisement on wall for 3 months for $1320 \mathrm{~m}^2=\frac{5000}{12} \times 3 \times 1320$
$=$ Rs. 1650000

Hence rent paid by company $=$ Rs. $16,50,000$
Ex 10.1 Question 3.

There is 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 figure). If the sides of the wall are 15 $\mathrm{m}, 11 \mathrm{~m}$ and $\mathbf{6}$, find the area painted in colour.

Answer.

Since, sides of coloured triangular wall are $15 \mathrm{~m}, 11 \mathrm{~m}$ and $6 \mathrm{~m}$.
$\therefore$ Semi-perimeter of coloured triangular wall $=\frac{15+11+6}{2}=\frac{32}{2}=16 \mathrm{~m}$

Now, Using Heron's formula,

Area of coloured triangular wall $=\sqrt{s(s-a)(s-b)(s-c)}$
$
\begin{aligned}
& =\sqrt{16(16-15)(16-11)(16-6)} \\
& =\sqrt{16 \times 1 \times 5 \times 10}=20 \sqrt{2} \mathrm{~m}^2
\end{aligned}
$

Hence area painted in blue colour $=20 \sqrt{2} m^2$
Ex 10.1 Question 4.

Find the area of a triangle two sides of which are $18 \mathrm{~cm}$ and $10 \mathrm{~cm}$ and the perimeter is 42 cm.

Answer.

Given: $a=18 \mathrm{~cm}, b=10 \mathrm{~cm}$.

Since Perimeter $=42 \mathrm{~cm}$
$
\Rightarrow a+b+c=42
$

$\begin{aligned}
& \Rightarrow 18+10+c=42 \\
& \Rightarrow c=42-28=14 \mathrm{~cm} \\
& \therefore \text { Semi-perimeter of triangle }=\frac{18+10+14}{2}=21 \mathrm{~cm} \\
& \therefore \text { Area of triangle }=\sqrt{s(s-a)(s-b)(s-c)} \\
& =\sqrt{21(21-18)(21-10)(21-14)} \\
& =\sqrt{21 \times 3 \times 11 \times 7}=\sqrt{7 \times 3 \times 3 \times 11 \times 7} \\
& =21 \sqrt{11}=21 \times 3.3=69.3 \mathrm{~cm}^2
\end{aligned}$

Ex 10.1 Question 5.

Sides of a triangle are in the ratio of 12: $17: 25$ and its perimeter is $540 \mathrm{~cm}$. Find its area.

Answer.

Let the sides of the triangle be $12 x, 17 x$ and $25 x$.

Therefore, $12 x+17 x+15 x=540$
$\Rightarrow 54 x=540$
$\Rightarrow x=10$
$\therefore$ The sides are $120 \mathrm{~cm}, 170 \mathrm{~cm}$ and $250 \mathrm{~cm}$.

Semi-perimeter of triangle $(s)=\frac{120+170+250}{2}=270 \mathrm{~cm}$

Now, Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{270(270-120)(270-170)(270-250)}$
$=\sqrt{270 \times 150 \times 100 \times 20}=9000 \mathrm{~cm}^2$

Ex 10.1 Question 6.

An isosceles triangle has perimeter $30 \mathrm{~cm}$ and each of the equal sides is $12 \mathrm{~cm}$. Find the area of the triangle.

Answer.

Given: $a=12 \mathrm{~cm}, b=12 \mathrm{~cm}$
$
\begin{aligned}
& \text { Since Perimeter }=30 \mathrm{~cm} \\
& \Rightarrow a+b+c=30 \\
& \Rightarrow 12+12+c=30 \\
& \Rightarrow c=30-24=6 \mathrm{~cm}
\end{aligned}
$
$\therefore$ Semi-perimeter of triangle $=\frac{12+12+6}{2}=15 \mathrm{~cm}$
$\therefore$ Area of triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
$
\begin{aligned}
& =\sqrt{15(15-12)(15-12)(15-6)} \\
& =\sqrt{15 \times 3 \times 3 \times 9}=\sqrt{5 \times 3 \times 3 \times 3 \times 3 \times 3} \\
& =9 \sqrt{15} \mathrm{~cm}^2
\end{aligned}
$