Question

Prove that
ar(ABC)= √{s(s-a)(s-b)(s-c)}
where s=(a+b+c)÷2​

Answer 1

Answer:

Well, are you kidding me? This is derived from high level trigonometry and Brahmagupta's formula. I don't know high level trigonometry, but I can derive this formula from Brahmagupta's formula.

HEY MATE HERE IS YOUR DERIVATION!

In a cyclic quadrilateral ABCD with sides AB=a, BC=b, CD=c, AD= d, and semi-perimeter s=(a+b+c+d) / 2

Area of cyclic quadrilateral ABCD= √{(s-a) (s-b) (s-c) (s-d)}

This is known as Brahmagupta's formula.

Now, if you reduce the 4th side I. e. d to 0, you will get a triangle ABC, which is always cyclic as all triangles are cyclic.

•°• The modified formula becomes,

ar(∆ABC) =√{s(s-a) (s-b) (s-c)}

This is the derivation of Heron's formula from the famous Brahmagupta's Formula.

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#BAL

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