Examples (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

Example 1 :

Find the area of a triangle, two sides of which are $8 \mathrm{~cm}$ and $11 \mathrm{~cm}$ and the perimeter is $32 \mathrm{~cm}$ (see Fig. 10.3).
Solution :

Here we have perimeter of the triangle $=32 \mathrm{~cm}, a=8 \mathrm{~cm}$ and $b=11 \mathrm{~cm}$.
Third side $c=32 \mathrm{~cm}-(8+11) \mathrm{cm}=13 \mathrm{~cm}$
So, $2 s=32$, i.e., $s=16 \mathrm{~cm}$,
$
\begin{aligned}
& s-a=(16-8) \mathrm{cm}=8 \mathrm{~cm} \\
& s-b=(16-11) \mathrm{cm}=5 \mathrm{~cm} \\
& s-c=(16-13) \mathrm{cm}=3 \mathrm{~cm} .
\end{aligned}
$

Therefore, area of the triangle $=\sqrt{s(s-a)(s-b)(s-c)}$
$
=\sqrt{16 \times 8 \times 5 \times 3} \mathrm{~cm}^2=8 \sqrt{30} \mathrm{~cm}^2
$

Example 2 :

A triangular park ABC has sides 120m, 80m and 50m (see Fig. 10.4). A gardener Dhania has to put a fence all around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of ₹ 20 per metre leaving a space $3 \mathrm{~m}$ wide for a gate on one side.
Solution :

For finding area of the park, we have
$
2 s=50 \mathrm{~m}+80 \mathrm{~m}+120 \mathrm{~m}=250 \mathrm{~m}
$

i.e., $\quad s=125 \mathrm{~m}$

Now,
$
\begin{aligned}
& s-a=(125-120) \mathrm{m}=5 \mathrm{~m}, \\
& s-b=(125-80) \mathrm{m}=45 \mathrm{~m}, \\
& s-c=(125-50) \mathrm{m}=75 \mathrm{~m} .
\end{aligned}
$

$
\begin{aligned}
\text { Therefore, area of the park } & =\sqrt{s(s-a)(s-b)(s-c)} \\
= & \sqrt{125 \times 5 \times 45 \times 75} \mathrm{~m}^2 \\
= & 375 \sqrt{15} \mathrm{~m}^2
\end{aligned}
$

Also, perimeter of the park $=\mathrm{AB}+\mathrm{BC}+\mathrm{CA}=250 \mathrm{~m}$
Therefore, length of the wire needed for fencing $=250 \mathrm{~m}-3 \mathrm{~m}$ (to be left for gate)
$
=247 \mathrm{~m}
$

And so the cost of fencing $=₹ 20 \times 247=₹ 4940$
Example 3 :

The sides of a triangular plot are in the ratio of $3: 5: 7$ and its perimeter is $300 \mathrm{~m}$. Find its area.
Solution :

Suppose that the sides, in metres, are $3 x, 5 x$ and $7 x$ (see Fig. 10.5).
Then, we know that $3 x+5 x+7 x=300$ (perimeter of the triangle)
Therefore, $15 x=300$, which gives $x=20$.
So the sides of the triangle are $3 \times 20 \mathrm{~m}, 5 \times 20 \mathrm{~m}$ and $7 \times 20 \mathrm{~m}$ i.e., $\quad 60 \mathrm{~m}, 100 \mathrm{~m}$ and $140 \mathrm{~m}$.

Can you now find the area [Using Heron's formula]?
We have $s=\frac{60+100+140}{2} \mathrm{~m}=150 \mathrm{~m}$,
Fig. 10.5 and area will be $\sqrt{150(150-60)(150-100)(150-140)} \mathrm{m}^2$

$
\begin{aligned}
& =\sqrt{150 \times 90 \times 50 \times 10} \mathrm{~m}^2 \\
& =1500 \sqrt{3} \mathrm{~m}^2
\end{aligned}
$