Find the length of altitude AD of an isosceles `Delta ABC` in which `AB=AC=2a` units and BC=a units.

Answer

`(sqrt(15))/(2)` units

Explanation

In `Delta ABC, AB= AC =2a and BC=a`
Draw `AD bot BC`. Then, D is the midpoint of BC and so `BD=(a)/(2)`
`AD^(2)=AB^(2)-BD^(2)=4a^(2)-(a^(2))/(4)=(15a^(2))/(4) rArr AD=(asqrt(15))/(2)`