if the area of an isosceles triangle is 60 cm² and the length of each of its equal sides is 12 cm , find its base....
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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.
Hope it will help ya :)
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