can anyone give an appropriate proof to the heron's formula using geometric approach??????

Answer 1

The first poster copy pasted from Wikipedia (without giving the link, tss tss...):

http://en.wikipedia.org/wiki/Heron%27s_formula#Proof

Look it up, it is much more readable over there.

Here is an excellent geometrical proof:

http://mathforum.org/library/drmath/view/54686.html

Here is another one with step by step explanation:

http://mathforum.org/library/drmath/view/54686.html

Good luck!

Answer 2

Heron's Formula Proof

Click on the URL below for additional information concerning Heron's proof

agutie.homestead.com/files/Heron/index.html

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Answer 3

you should use the geometric techniques-set while the subject to be solved is in touch with figures, like proving 2 triangles congruent or comparable, you should use algebraic techniques-set once you could desire to detect lacking quantites in the subject. nicely specially you will continually might desire to apply the two those procedures in at present's international. for e.g coach that the quantity small cone shaped by employing a plane slicing a frustum parallely to the backside is1/thirtieth the quantity of the massive cone shaped

Answer 4

A modern proof, which uses algebra and trigonometry and is quite unlike the one provided by Heron, follows. Let a, b, c be the sides of the triangle and A, B, C the angles opposite those sides. We have

\cos(C) = \frac{a^2+b^2-c^2}{2ab}

by the law of cosines. From this we get with some algebra

\sin(C) = \sqrt{1-\cos^2(C)} = \frac{\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2 }}{2ab}.

The altitude of the triangle on base a has length bsin(C), and it follows

S\, = \frac{1}{2} (\mbox{base}) (\mbox{altitude})
= \frac{1}{2} ab\sin(C)
= \frac{1}{4}\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2}
= \frac{1}{4}\sqrt{(2a b -(a^2 +b^2 -c^2))(2a b +(a^2 +b^2 -c^2))}
= \frac{1}{4}\sqrt{(c^2 -(a -b)^2)((a +b)^2 -c^2)}
= \frac{1}{4}\sqrt{(c -(a -b))((c +(a -b))((a +b) -c))((a +b) +c)}
= \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}.