Find perimeter and area of a triangle if it's sides are $5\mathrm{cm}$, $9\mathrm{cm}$ and $10\mathrm{cm}$.

Given: Sides of triangle: $5\mathrm{cm}$, $9\mathrm{cm}$, $10\mathrm{cm}$

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

Solution: First lets find the perimeter of the triangle. $$\begin{array}{rcll}\because & P& =& a+b+c\\ \therefore & P& =& 5+9+10\\ \therefore & P& =& 24\mathrm{cm}\end{array}$$Therefore, the perimeter of the given triangle is $24\mathrm{cm}$.

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

$$\begin{array}{rcll}\because & s& =& \frac{P}{2}\\ \therefore & s& =& \frac{24}{2}\\ \therefore & s& =& 12\mathrm{cm}\end{array}$$

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

$$\begin{array}{rcl}A& =& \sqrt{s(s-a)(s-b)(s-c)}\\ & =& \sqrt{12(12-5)(12-9)(12-10)}\\ & =& \sqrt{12\times 7\times 3\times 2}\\ & =& \sqrt{504}\\ & =& \sqrt{36\times 14}\\ & =& \sqrt{6^2\times 14}\\ & =& 6\times \sqrt{14}\\ & =& 6\sqrt{14}{\mathrm{cm}}^2\end{array}$$Therefore, the area of the given triangle is $6\sqrt{14}{\mathrm{cm}}^2$.