find the area of isosceles triangle with equal sides A unit and third side B unit using herons formula.
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area of a triangle by heron's formula is
√(s(s-a)(s-b)(s-c))
given a=b=Aunit , c=Bunit
so s=A+A+B/2 = (2A+B)/2
by substituting these values we get
area = √[(2A+B)/2][(2A+B)/2-A]^2[(2A+B)/2-B]
= 1/4B√(4A^2-B^2)
hence the answer is 1/4B√(4A^2-B^2)sq.units
√(s(s-a)(s-b)(s-c))
given a=b=Aunit , c=Bunit
so s=A+A+B/2 = (2A+B)/2
by substituting these values we get
area = √[(2A+B)/2][(2A+B)/2-A]^2[(2A+B)/2-B]
= 1/4B√(4A^2-B^2)
hence the answer is 1/4B√(4A^2-B^2)sq.units
vedha03:
can you mark my answer as brainliest answer
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