Find perimeter and area of a triangle if it's sides are $6\mathrm{cm}$, $11\mathrm{cm}$ and $15\mathrm{cm}$.

Given: Sides of triangle: $6\mathrm{cm}$, $11\mathrm{cm}$, $15\mathrm{cm}$

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

Solution: First lets find the perimeter of the triangle. $$\begin{array}{rcll}\because & P& =& a+b+c\\ \therefore & P& =& 6+11+15\\ \therefore & P& =& 32\mathrm{cm}\end{array}$$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;

$$\begin{array}{rcl}A& =& \sqrt{s(s-a)(s-b)(s-c)}\\ & =& \sqrt{16(16-6)(16-11)(16-15)}\\ & =& \sqrt{16\times 10\times 5\times 1}\\ & =& \sqrt{4^2\times 50}\\ & =& 4\times \sqrt{50}\\ & =& 4\times \sqrt{25\times 2}\\ & =& 4\times \sqrt{5^2\times 2}\\ & =& 4\times 5\times \sqrt{2}\\ & =& 20\sqrt{2}{\mathrm{cm}}^2\end{array}$$Therefore, the area of the given triangle is $20\sqrt{2}{\mathrm{cm}}^2$.