Find perimeter and area of a triangle if it's sides are $5\mathrm{cm}$, $6\mathrm{cm}$ and $9\mathrm{cm}$.

Given: Sides of triangle: $5\mathrm{cm}$, $6\mathrm{cm}$, $9\mathrm{cm}$

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

Solution: First lets find the perimeter of the triangle. $$\begin{array}{rcll}\because & P& =& a+b+c\\ \therefore & P& =& 5+6+9\\ \therefore & P& =& 20\mathrm{cm}\end{array}$$Therefore, the perimeter of the given triangle is $20\mathrm{cm}$.

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

$$\begin{array}{rcll}\because & s& =& \frac{P}{2}\\ \therefore & s& =& \frac{20}{2}\\ \therefore & s& =& 10\mathrm{cm}\end{array}$$

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

$$\begin{array}{rcl}A& =& \sqrt{s(s-a)(s-b)(s-c)}\\ & =& \sqrt{10(10-5)(10-6)(10-9)}\\ & =& \sqrt{10\times 5\times 4\times 1}\\ & =& \sqrt{2^2\times 50}\\ & =& 2\times \sqrt{50}\\ & =& 2\times \sqrt{25\times 2}\\ & =& 2\times \sqrt{5^2\times 2}\\ & =& 2\times 5\times \sqrt{2}\\ & =& 10\sqrt{2}{\mathrm{cm}}^2\end{array}$$Therefore, the area of the given triangle is $10\sqrt{2}{\mathrm{cm}}^2$.