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Heron formula for area of a triangle

Posted July 11th, 2010 by Structure

Let ABC be a triangle, and a=length(BC), b=length(CA) and c=length(AB).Let $ p=\frac{a+b+c}{2} $
Then area S of the triangle is

$$S=\sqrt{p(p-a)(p-b)(p-c)}$$

We know that

$$\cos A=\frac{b^2+c^2-a^2}{2bc}$$

so

$$2\sin^2\frac{A}{2}=1-\cos A=1-\frac{b^2+c^2-a^2}{2bc}=\frac{a^2-(b-c)^2}{2bc}=\frac{(a-b+c)(a+b-c)}{2bc}=2\frac{(p-b)(p-c)}{bc}$$

Similar

$$2\cos^2\frac{A}{2}=1+\cos A=1+\frac{b^2+c^2-a^2}{2bc}=\frac{(b+c)^2-a^2}{2bc}=\frac{(a+b+c)(-a+b+c)}{2bc}=2\frac{p(p-a)}{bc}$$

so we have

$$\sin\frac{A}{2}=\sqrt{\frac{(p-b)(p-c)}{bc}}$$

and

$$\cos\frac{A}{2}=\sqrt{\frac{p(p-a)}{bc}}$$

Now we know

$$S=\frac{bc\sin A}{2}=bc \sin\frac{A}{2}\cos\frac{A}{2}=bc\sqrt{\frac{(p-b)(p-c)}{bc}}\sqrt{\frac{p(p-a)}{bc}}=\sqrt{p(p-a)(p-b)(p-c)}$$
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