Find perimeter, area and altitude of an equilateral triangle whose sides measure $18\mathrm{cm}$.

Given: Sides of equilateral triangle: $18\mathrm{cm}$

To find: Perimeter, Area and Altitude of an equilateral triangle

Solution: First lets find the perimeter of the given equilateral triangle. $$\begin{array}{rcll}\because & P& =& 3a\\ \therefore & P& =& 3\times 18\\ \therefore & P& =& 54\mathrm{cm}\end{array}$$Therefore, the perimeter of the given equilateral triangle is $54\mathrm{cm}$.

Now, lets find area of the given equilateral triangle. $$\begin{array}{lrcl}\because & A& =& \frac{\sqrt{3}}{4}\times a^2\\ \therefore & A& =& \frac{\sqrt{3}}{4}\times {18}^2\\ \therefore & A& =& \frac{\sqrt{3}}{4}\times 324\\ \therefore & A& =& \sqrt{3}\times 81\\ \therefore & A& =& 81\sqrt{3}{\mathrm{cm}}^2\end{array}$$Therefore, the area of the given equilateral triangle is $81\sqrt{3}{\mathrm{cm}}^2$.

Now, lets find altitude of the given equilateral triangle. $$\begin{array}{lrcl}\because & h& =& \frac{\sqrt{3}}{2}a\\ \therefore & h& =& \frac{\sqrt{3}}{2}\times 18\\ \therefore & h& =& \sqrt{3}\times 9\\ \therefore & h& =& 9\sqrt{3}\mathrm{cm}\end{array}$$Therefore, the altitude of the given equilateral triangle is $9\sqrt{3}\mathrm{cm}$.