Question

$△\mathrm{ABC}$ is an equilateral triangle of side $2\mathrm{a}$. Find each of its altitudes.

Answer 1

Finding the altitude of the equilateral triangle:

Given:

$△ABC$ is an equilateral triangle of each $2\mathrm{a}$ units

Construction: Draw$\mathrm{AD}\perp \mathrm{BC}$

In right $∆\mathrm{ADB}$

$\begin{array}{rcl}{\left(AB\right)}^2& =& {\left(AD\right)}^2+{\left(BD\right)}^2\\ {\left(2a\right)}^2& =& {\left(AD\right)}^2+{\left(a\right)}^2\\ 4a^2& =& {\left(AD\right)}^2+a^2\\ 4a^2–a^2& =& {\left(AD\right)}^2\\ 3a^2& =& {\left(AD\right)}^2\end{array}$

$\surd 3\mathrm{a}=\mathrm{AD}$

∴ Each of its altitude$=\surd 3\mathrm{a}\mathrm{unit}$

Therefore, the altitude of the equilateral triangle is $\mathrm{AD}=\sqrt{3}\mathrm{a}\mathrm{unit}$