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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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