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Question
In an isosceles triangle ABC, AB = AC = 25 cm, BC = 14 cm. Calculate the altitude from A on BC.
Solution
We have
AB = AC = 25 cm and BC = 14 cm
In ΔABD and ΔACD
∠ADB = ∠ADC [Each 90°]
AB = AC [Each 25 cm]
AD = AD [Common]
Then, ΔABD ≅ ΔACD [By RHS condition]
∴ BD = CD = 7 cm [By c.p.c.t]
In ΔADB, by Pythagoras theorem
AD2 + BD2 = AB2
⇒ AD2 + 72 = 252
⇒ AD2 = 625 − 49 = 576
⇒ AD = `sqrt576` = 24 cm
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