Question

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

Answer 1

In isosceles Δ ABC, we have:
AB = AC = 2a units and BC = a units
Let AD be the altitude drawn from A that meets BC at D.
Then, D is the midpoint of BC.  

`BD=BC=a/2` units  

Applying Pythagoras theorem in right-angled ΔABD, we have:  

 

`AB^2=AD^2+BD^2` 

`AD^2=AB^2-BD^2=(2a)^2-(a/2)^2` 

`AD^2=4a^2-a^2/4=(15a^2)/4` 

`AD= sqrt((15a^2)/4
)= (asqrt15)/2` unit