Find perimeter and area of a triangle if it's sides are $8\mathrm{cm}$, $10\mathrm{cm}$ and $14\mathrm{cm}$.

Given: Sides of triangle: $8\mathrm{cm}$, $10\mathrm{cm}$, $14\mathrm{cm}$

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

Solution: First lets find the perimeter of the triangle. $$\begin{array}{rcll}\because & P& =& a+b+c\\ \therefore & P& =& 8+10+14\\ \therefore & P& =& 32\mathrm{cm}\end{array}$$Therefore, the perimeter of the given triangle is $32\mathrm{cm}$.

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

$$\begin{array}{rcll}\because & s& =& \frac{P}{2}\\ \therefore & s& =& \frac{32}{2}\\ \therefore & s& =& 16\mathrm{cm}\end{array}$$

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

$$\begin{array}{rcl}A& =& \sqrt{s(s-a)(s-b)(s-c)}\\ & =& \sqrt{16(16-8)(16-10)(16-14)}\\ & =& \sqrt{16\times 8\times 6\times 2}\\ & =& \sqrt{4^2\times 96}\\ & =& 4\times \sqrt{96}\\ & =& 4\times \sqrt{16\times 6}\\ & =& 4\times \sqrt{4^2\times 6}\\ & =& 4\times 4\times \sqrt{6}\\ & =& 16\sqrt{6}{\mathrm{cm}}^2\end{array}$$Therefore, the area of the given triangle is $16\sqrt{6}{\mathrm{cm}}^2$.