Find perimeter and area of a triangle if it's sides are $8\mathrm{cm}$, $11\mathrm{cm}$ and $15\mathrm{cm}$.

Given: Sides of triangle: $8\mathrm{cm}$, $11\mathrm{cm}$, $15\mathrm{cm}$

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

Solution: First lets find the perimeter of the triangle. $$\begin{array}{rcll}\because & P& =& a+b+c\\ \therefore & P& =& 8+11+15\\ \therefore & P& =& 34\mathrm{cm}\end{array}$$Therefore, the perimeter of the given triangle is $34\mathrm{cm}$.

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

$$\begin{array}{rcll}\because & s& =& \frac{P}{2}\\ \therefore & s& =& \frac{34}{2}\\ \therefore & s& =& 17\mathrm{cm}\end{array}$$

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

$$\begin{array}{rcl}A& =& \sqrt{s(s-a)(s-b)(s-c)}\\ & =& \sqrt{17(17-8)(17-11)(17-15)}\\ & =& \sqrt{17\times 9\times 6\times 2}\\ & =& \sqrt{3^2\times 204}\\ & =& 3\times \sqrt{204}\\ & =& 3\times \sqrt{4\times 51}\\ & =& 3\times \sqrt{2^2\times 51}\\ & =& 3\times 2\times \sqrt{51}\\ & =& 6\sqrt{51}{\mathrm{cm}}^2\end{array}$$Therefore, the area of the given triangle is $6\sqrt{51}{\mathrm{cm}}^2$.