The area of an isosceles triangle is 60cm2 the length of each one of its equal side is 13cm, find its base.

ΔABC is isosceles

AB = AC =13cm

Lets drop a perpendicular AD from A to BC with height h

Let CD = x


Area of ΔADC = 1/2(xh)

Area of ΔABC = xh = 60

h = 60/x

In ΔADC by pythagoras theorem


Solving we get

u = 144 or 25

⇒ x = 12 or 5


CD = x

BC = 2x

⇒ BC could take two possible values 24 cm and 10 cm

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