Find perimeter and area of a triangle if it's sides are $7\mathrm{cm}$, $9\mathrm{cm}$ and $10\mathrm{cm}$.

Given: Sides of triangle: $7\mathrm{cm}$, $9\mathrm{cm}$, $10\mathrm{cm}$

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

Solution: First lets find the perimeter of the triangle. $$\begin{array}{rcll}\because & P& =& a+b+c\\ \therefore & P& =& 7+9+10\\ \therefore & P& =& 26\mathrm{cm}\end{array}$$Therefore, the perimeter of the given triangle is $26\mathrm{cm}$.

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

$$\begin{array}{rcll}\because & s& =& \frac{P}{2}\\ \therefore & s& =& \frac{26}{2}\\ \therefore & s& =& 13\mathrm{cm}\end{array}$$

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

$$\begin{array}{rcl}A& =& \sqrt{s(s-a)(s-b)(s-c)}\\ & =& \sqrt{13(13-7)(13-9)(13-10)}\\ & =& \sqrt{13\times 6\times 4\times 3}\\ & =& \sqrt{2^2\times 234}\\ & =& 2\times \sqrt{234}\\ & =& 2\times \sqrt{9\times 26}\\ & =& 2\times \sqrt{3^2\times 26}\\ & =& 2\times 3\times \sqrt{26}\\ & =& 6\sqrt{26}{\mathrm{cm}}^2\end{array}$$Therefore, the area of the given triangle is $6\sqrt{26}{\mathrm{cm}}^2$.