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Question

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

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Solution

In isosceles ∆ 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 = a2 units
Applying Pythagoras theorem in right-angled ∆ABD, we have:

AB2 = AD2 + BD2AD2 = AB2 - BD2 = (2a)2 - a22AD2 = 4a2 - a24 = 15a24AD = 15a24 = a152 units

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