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