Find perimeter and area of a triangle if it's sides are $6\mathrm{cm}$, $10\mathrm{cm}$ and $14\mathrm{cm}$.

Given: Sides of triangle: $6\mathrm{cm}$, $10\mathrm{cm}$, $14\mathrm{cm}$

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

Solution: First lets find the perimeter of the triangle. $$\begin{array}{rcll}\because & P& =& a+b+c\\ \therefore & P& =& 6+10+14\\ \therefore & P& =& 30\mathrm{cm}\end{array}$$Therefore, the perimeter of the given triangle is $30\mathrm{cm}$.

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

$$\begin{array}{rcll}\because & s& =& \frac{P}{2}\\ \therefore & s& =& \frac{30}{2}\\ \therefore & s& =& 15\mathrm{cm}\end{array}$$

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

$$\begin{array}{rcl}A& =& \sqrt{s(s-a)(s-b)(s-c)}\\ & =& \sqrt{15(15-6)(15-10)(15-14)}\\ & =& \sqrt{15\times 9\times 5\times 1}\\ & =& \sqrt{3^2\times 75}\\ & =& 3\times \sqrt{75}\\ & =& 3\times \sqrt{25\times 3}\\ & =& 3\times \sqrt{5^2\times 3}\\ & =& 3\times 5\times \sqrt{3}\\ & =& 15\sqrt{3}{\mathrm{cm}}^2\end{array}$$Therefore, the area of the given triangle is $15\sqrt{3}{\mathrm{cm}}^2$.