Question

the area of an isosceles triangle is 60 CM square and the length of each one of its equal side is 13 cm find its base​

Answer 1

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

13 square = x square + h square

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