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 $69.65\ cm^2$.