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Find the area of a triangle two sides of which are $18\ cm$ and $10\ cm$ and the perimeter is $42\ cm$.
Given:
The two sides of a triangle are $18\ cm$ and $10\ cm$ and the perimeter is $42\ cm$.
To do:
We have to find the area of the triangle.
Solution:
Perimeter of the triangle $= 42\ cm$
Two sides of the triangle are $a=18\ cm$ and $b=10\ cm$
Third side $c= 42 - (18 + 10)\ cm$
$= 42 - 28\ cm$
$= 14\ cm$
Therefore,
$s=\frac{\text { Perimeter }}{2}$
$=\frac{42}{2}$
$=21$
Area $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{21 \times(21-18)(21-10)(21-14)}$
$=\sqrt{21 \times 3 \times 11 \times 7} \mathrm{~cm}^{2}$
$=\sqrt{3 \times 7 \times 7 \times 3 \times 11}$
$=3 \times 7 \times \sqrt{11}$
$=21 \sqrt{11} \mathrm{~cm}^{2}$
$=21 \times 3.316 \mathrm{~cm}^{2}$
$=69.65 \mathrm{~cm}^{2}$
The area of the triangle is $69.65\ cm^2$.