Solve the triangle in which

Given: $a=2 \mathrm{~cm}, b=1 \mathrm{~cm}$ and $c=\sqrt{3} \mathrm{~cm}$

Perimeter $=a+b+c=3+\sqrt{3} \mathrm{~cm}$

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

$=\sqrt{\frac{3+\sqrt{3}}{2}\left(\frac{3+\sqrt{3}}{2}-2\right)\left(\frac{3+\sqrt{3}}{2}-1\right)\left(\frac{3+\sqrt{3}}{2}-\sqrt{3}\right)}$

$=\sqrt{\frac{3+\sqrt{3}}{2} \cdot \frac{\sqrt{3}-1}{2} \cdot \frac{\sqrt{3}+1}{2} \cdot \frac{3-\sqrt{3}}{2}}$

$=\sqrt{\frac{(9-3)(3-1)}{16}}$

$=\sqrt{\frac{12}{16}}=\frac{2 \sqrt{3}}{4}=\frac{\sqrt{3}}{2} \mathrm{~cm}^{2}$ [Proved]