Question

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

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Solution

## Finding the altitude of the equilateral triangle:Given:$△ABC$ is an equilateral triangle of each $2\mathrm{a}$ unitsConstruction: 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}\\ 4{a}^{2}& =& {\left(AD\right)}^{2}+{a}^{2}\\ 4{a}^{2}–{a}^{2}& =& {\left(AD\right)}^{2}\\ 3{a}^{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 $\mathrm{AD}=\sqrt{3}\mathrm{a}\mathrm{unit}$

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