Question

If area of an isosceles triangle is 60cm^2 and each its equal side is 13cm Find base of triangle and its height

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

Answer 2

Let us divide base into two halves of isoceles triangle .Do,we get 2 right angle triangle whose area is equal to 30cm^2.
Now,we get area of triangle 1/2×b×h=30cm^2........1
By Pythagoras theorem we get b^2+h^2=13^2.....2
Now we get two equation so by solving equation 1 we get b*h=60cm.
On solving further we get 5,12
We have divided the base in two equal parts so we need to multiply with 2
So we get Base =10cm and height =24cm
OR
Base =24cm and height=10cm