Find perimeter, area and altitude of an equilateral triangle whose sides measure $54\mathrm{cm}$.

Given: Sides of equilateral triangle: $54\mathrm{cm}$

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

Solution: First lets find the perimeter of the given equilateral triangle. $$\begin{array}{rcll}\because & P& =& 3a\\ \therefore & P& =& 3\times 54\\ \therefore & P& =& 162\mathrm{cm}\end{array}$$Therefore, the perimeter of the given equilateral triangle is $162\mathrm{cm}$.

Now, lets find area of the given equilateral triangle. $$\begin{array}{lrcl}\because & A& =& \frac{\sqrt{3}}{4}\times a^2\\ \therefore & A& =& \frac{\sqrt{3}}{4}\times {54}^2\\ \therefore & A& =& \frac{\sqrt{3}}{4}\times 2916\\ \therefore & A& =& \sqrt{3}\times 729\\ \therefore & A& =& 729\sqrt{3}{\mathrm{cm}}^2\end{array}$$Therefore, the area of the given equilateral triangle is $729\sqrt{3}{\mathrm{cm}}^2$.

Now, lets find altitude of the given equilateral triangle. $$\begin{array}{lrcl}\because & h& =& \frac{\sqrt{3}}{2}a\\ \therefore & h& =& \frac{\sqrt{3}}{2}\times 54\\ \therefore & h& =& \sqrt{3}\times 27\\ \therefore & h& =& 27\sqrt{3}\mathrm{cm}\end{array}$$Therefore, the altitude of the given equilateral triangle is $27\sqrt{3}\mathrm{cm}$.