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