Find perimeter and area of a triangle if it's sides are $4\mathrm{cm}$, $5\mathrm{cm}$ and $7\mathrm{cm}$.

Given: Sides of triangle: $4\mathrm{cm}$, $5\mathrm{cm}$, $7\mathrm{cm}$

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

Solution: First lets find the perimeter of the triangle. $$\begin{array}{rcll}\because & P& =& a+b+c\\ \therefore & P& =& 4+5+7\\ \therefore & P& =& 16\mathrm{cm}\end{array}$$Therefore, the perimeter of the given triangle is $16\mathrm{cm}$.

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

$$\begin{array}{rcll}\because & s& =& \frac{P}{2}\\ \therefore & s& =& \frac{16}{2}\\ \therefore & s& =& 8\mathrm{cm}\end{array}$$

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

$$\begin{array}{rcl}A& =& \sqrt{s(s-a)(s-b)(s-c)}\\ & =& \sqrt{8(8-4)(8-5)(8-7)}\\ & =& \sqrt{8\times 4\times 3\times 1}\\ & =& \sqrt{2^2\times 24}\\ & =& 2\times \sqrt{24}\\ & =& 2\times \sqrt{4\times 6}\\ & =& 2\times \sqrt{2^2\times 6}\\ & =& 2\times 2\times \sqrt{6}\\ & =& 4\sqrt{6}{\mathrm{cm}}^2\end{array}$$Therefore, the area of the given triangle is $4\sqrt{6}{\mathrm{cm}}^2$.