ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Given, ABC is an equilateral triangle of side 2a.



Draw, AD \(\perp\) BC

In \(\triangle\) ADB and \(\triangle\) ADC,
AB = AC
AD = AD
\(\angle\) ADB = \(\angle\) ADC [Both are 90°]

Therefore, \(\triangle\) ADB \( {\displaystyle \cong }\) \(\triangle\) ADC by RHS congruence.
Hence, BD = DC [by CPCT]

In right angled \(\triangle\) ADB,
\(AB^2 = AD^2 + BD^2\)
\((2a)^2 = AD^2 + a^2\)
\(AD^2 = 4a^2 – a^2\)
\( AD^2 = 3a^2\)
\(AD = \sqrt{3} a\)