Find Area of Triangle Using No Formula

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How can I go about finding the area of a triangle with sides a, b, and c, assuming we do not already know a formula for it (such as Heron's Formula)? Kind of like a proof, I suppose.
 
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You could split it into 2 right angle triangles. It's failry easy to calculate the area of a right angle triangle.
 
define a line as a side and integrate...
 
PowerIso said:
define a line as a side and integrate...
I take it you mean set the triangle up so that one vertex is at the origin and another is along the positive x-axis. The difficulty with that is that, since you know only the lengths of the sides, you would have to break it into right triangles to determine the limits of integration. Once you have done that, it is simpler to find the area of the right triangles, mgb_phys' recomendation, than to integrate.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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