Q13 of 95 Page 5

ΔABC is an isosceles triangle with AB = AC = 13 cm. The length of altitude from A on BC is 5 cm. Find BC.


Given: ABC is an isosceles triangle with AB = AC = 13 cm


Suppose the altitude from A on Bc meets BC at M.


∴ M is the midpoint of BC. AM = 5 cm


In ∆AMB, using Pythagoras theorem, we have


(Perpendicular)2 + (Base)2 = (Hypotenuse)2


⇒ (AM)2 + (BM)2 = (AB)2


⇒ (5)2 + (BM)2 = (13)2


⇒ (BM)2 = (13)2 – (5)2


⇒ (BM)2 = (13 – 5)(13+5)


[∵ (a2 – b2) = (a + b)(a – b)]


⇒ (BM)2 = (8)(18)


⇒ (BM)2 = 144


⇒ BM = ±12


⇒ BM = 12 [taking positive square root]


∴ BC = 2BM or 2MC = 2×12 = 24cm


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