Check

[Garfield]

I came up with a cheap formula to solve the surfice of any triangle. Though I'm not sure if it is correct. I wonder if you 1337 guys could check it.*points to sig*

[EvoN1020v]

I checked the formula by using c=5, b=3, a=4, but first, I used the [math]c^{2}=a^{2}+b^{2}[/math]. The surface area should be 5. Then I put in the variable with the known numbers, using this formula:[math]S_{triangle}=\frac{c\sqrt{b^2-\left(\frac{b^2-a^2+c^2}{2c}\right)^2}}{2}[/math]. I got a 6 for an answer following my calculation, so the formula would be right, if it have minus 1 like this: [math]S_{triangle}=\frac{c\sqrt{b^2-\left(\frac{b^2-a^2+c^2}{2c}\right)^2}}{2}-1[/math]. Therefore the answer would be 5, thus the surface area for a triangle with c=5, b=3, a=4.

[Garfield]

Ahem...[math]

S_{triangle}=\frac{3*4}{2}=6

[/math]

Don't you just hate it when that happens when doing calculations in your head.

[Rakdos]

[math']c^{2}=a^{2}+b^{2}[/math]

 

That is the pythagorean theorem, not area.

[EvoN1020v]

WHOOPS! I mistakely used the pythagorean theorem to find the surface area of a triangle! Stupid me. The formula to find the surface area for a triangle is [math]A=\frac{1}{2}bh[/math]. Therefore, [math]\frac{1}{2}4*3[/math] is 6. So the big crazy formula is correct.