Find the area of a triangle whose sides are $9\ cm, 12\ cm$ and $15\ cm$.

Given:

The sides of a triangle are $9\ cm, 12\ cm$ and $15\ cm$.

To do:

We have to find the area of the triangle. 

Solution:

Let $a=9\ cm, b=12\ cm$ and $c=15\ cm$

Therefore,

$s=\frac{a+b+c}{2}$

$=\frac{9+12+15}{2}$

$=\frac{36}{2}$

$=18$

Area of the triangle $=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{18(18-9)(18-12)(18-15)}$

$=\sqrt{18 \times 9 \times 6 \times 3}$

$=\sqrt{9 \times 2 \times 9 \times 3 \times 2 \times 3}$

$=9 \times 2 \times 3 \mathrm{~cm}^{2}$

$=54 \mathrm{~cm}^{2}$

The area of the triangle is $54\ cm^2$.

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Updated on: 10-Oct-2022

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