In the equilateral \[ \Delta ABC\], let AB=BC=CA='a' units.
Draw \[AD \perp BC\]. LET AD= 'h' units.
\[\therefore BD=DC\] (\[\because \] RHS theorem)
\[\therefore BD=DC=\frac{a}{2}\]
Now there are two right angled triangles, \[ \Delta ADB and \Delta ADC.\]
In \[ \Delta ADB , \angle ADB= {90}^{\circ}\] (\[\because AD \perp BC\])
\[AB^2=AD^2+BD^2\] (\[\because \]Pythagoras theorem)
\[\therefore a^2=h^2+\left ( \frac{a}{2} \right ) ^2\]
\[a^2=h^2+\frac{a}{4} ^2\]
\[\frac{a^2}{1}-\frac{a^2}{4}=h^2\]
\[\frac{4a^2-a^2}{4}=h^2\]
\[\frac{3a^2}{4}=h^2\]
Take square root on either side.
\[\sqrt{\frac{3a^2}{4}}=\sqrt{h^2} \]
\[\frac{\sqrt{3}\times a}{2}=h \]
\[\therefore h=\frac{a\sqrt{3}}{2}\]
\[\therefore \] Height of an equilateral triangle of side 'a' units, \[h=\frac{a\sqrt{3}}{2} \]
