As `Delta ABC` is an equilateral triangle, its altitude will also be the median.
If we draw a median `AD` to `BC`. Then,
`BD = DC and /_ADB=90^@`
We are given,
`AB = AC = BC = 2a`
So,`BD = DC = (BC)/2 = a`
As, `Delta ABD` is a right angle triangle,
`AB^2 = AD^2+BD^2`
`AD^2 = (2a)^2-a^2 = 3a^2`
So,`AD = sqrt3a`
As, it is an equilateral triangle, all its altitudes will be equal to `sqrt3a`.
