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
⇒
Let
⇒
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