Find perimeter and area of a triangle if it's sides are $6\mathrm{cm}$, $10\mathrm{cm}$ and $12\mathrm{cm}$.

Given: Sides of triangle: $6\mathrm{cm}$, $10\mathrm{cm}$, $12\mathrm{cm}$

To find: Perimeter($P$) and Area($A$) of triangle

Solution: First lets find the perimeter of the triangle. $$\begin{array}{rcll}\because & P& =& a+b+c\\ \therefore & P& =& 6+10+12\\ \therefore & P& =& 28\mathrm{cm}\end{array}$$Therefore, the perimeter of the given triangle is $28\mathrm{cm}$.

Now lets find the semi perimeter of the triangle.
As we know that, semi perimeter is half of the perimeter,

$$\begin{array}{rcll}\because & s& =& \frac{P}{2}\\ \therefore & s& =& \frac{28}{2}\\ \therefore & s& =& 14\mathrm{cm}\end{array}$$

Now, as per the Heron’s formula, we know;

$$\begin{array}{rcl}A& =& \sqrt{s(s-a)(s-b)(s-c)}\\ & =& \sqrt{14(14-6)(14-10)(14-12)}\\ & =& \sqrt{14\times 8\times 4\times 2}\\ & =& \sqrt{2^2\times 224}\\ & =& 2\times \sqrt{224}\\ & =& 2\times \sqrt{16\times 14}\\ & =& 2\times \sqrt{4^2\times 14}\\ & =& 2\times 4\times \sqrt{14}\\ & =& 8\sqrt{14}{\mathrm{cm}}^2\end{array}$$Therefore, the area of the given triangle is $8\sqrt{14}{\mathrm{cm}}^2$.