Find perimeter, area and altitude of an equilateral triangle whose sides measure $24\mathrm{cm}$.

Given: Sides of equilateral triangle: $24\mathrm{cm}$

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

Solution: First lets find the perimeter of the given equilateral triangle. $$\begin{array}{rcll}\because & P& =& 3a\\ \therefore & P& =& 3\times 24\\ \therefore & P& =& 72\mathrm{cm}\end{array}$$Therefore, the perimeter of the given equilateral triangle is $72\mathrm{cm}$.

Now, lets find area of the given equilateral triangle. $$\begin{array}{lrcl}\because & A& =& \frac{\sqrt{3}}{4}\times a^2\\ \therefore & A& =& \frac{\sqrt{3}}{4}\times {24}^2\\ \therefore & A& =& \frac{\sqrt{3}}{4}\times 576\\ \therefore & A& =& \sqrt{3}\times 144\\ \therefore & A& =& 144\sqrt{3}{\mathrm{cm}}^2\end{array}$$Therefore, the area of the given equilateral triangle is $144\sqrt{3}{\mathrm{cm}}^2$.

Now, lets find altitude of the given equilateral triangle. $$\begin{array}{lrcl}\because & h& =& \frac{\sqrt{3}}{2}a\\ \therefore & h& =& \frac{\sqrt{3}}{2}\times 24\\ \therefore & h& =& \sqrt{3}\times 12\\ \therefore & h& =& 12\sqrt{3}\mathrm{cm}\end{array}$$Therefore, the altitude of the given equilateral triangle is $12\sqrt{3}\mathrm{cm}$.