I would like to calculate the area for a triangle such that $a^2+b^2-c^2=1$ (an almost Pythagorean triple).

I know that the triangle is non-right, so I would like to use $\text{Area}=\frac{1}{2}ab\sin C$... but I do not know how to represent $\sin C$ since I don't have any actual values.

I know about Heron's formula where $S=\frac{a+b+c}{2}$ and $\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$, but I feel like that gets too lengthy with our side lengths?

Edit to add: For $Area = \frac{1}{4}\sqrt{4a^2b^2-1}$ as shown by @zipirovich, can this area ever be an integer if $a,b,c$ are positive integers and $a,b >1$? Or, is this impossible?


Your equation can be rewritten as $c^2=a^2+b^2-1$. Comparing it with the Law of Cosines $c^2=a^2+b^2-2ab\cos C$, we can see that $2ab\cos C=1$ or $\cos C=\frac{1}{2ab}$. Then $$\sin C=\sqrt{1-\cos^2C}=\sqrt{1-\left(\frac{1}{2ab}\right)^2}=\sqrt{1-\frac{1}{4a^2b^2}},$$ and the area is $$\text{Area}=\frac{1}{2}ab\sin C=\frac{1}{2}ab\sqrt{1-\frac{1}{4a^2b^2}}=\frac{1}{2}\sqrt{a^2b^2-\frac{1}{4}}=\frac{1}{4}\sqrt{4a^2b^2-1}.$$ So there's no single answer because there are many such triangles, but here's an answer in the sense of having a formula for it. Going back to where we found that $\cos C=\frac{1}{2ab}$: as long as $2ab>1$, we can always find the angle $C$ and build such a triangle.

  • $\begingroup$ I do have, in another statement that I didn't include, that $a,b>1$. So, we would in fact be able to find angle C and build that triangle. I'm wondering now, though, if I'd ever find a value for the area that is an integer? I'll edit this question in. $\endgroup$ – Math1 Nov 26 '16 at 5:31
  • $\begingroup$ Are there any other constraints on $a,b,c$? Are they, or any of them, required to be integers or not? If not, then you can make the value of the area anything you want: pick an integer value for the area and then solve for the product $ab$. (And you'll still have the flexibility to choose their individual values, as long as their product is what you found.) $\endgroup$ – zipirovich Nov 26 '16 at 5:37
  • $\begingroup$ a,b and c are positive integers, and $a,b >1$. $\endgroup$ – Math1 Nov 26 '16 at 5:39
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    $\begingroup$ Then, no. Reason: $4a^2b^2=(2ab)^2$ is a perfect square of a number greater than $1$. (In fact, $2ab$ is at least $8$, but that doesn't matter right now.) Therefore the preceding integer $4a^2b^2-1$ can't be a perfect square, so the square root in the expression for the area can't be an integer. $\endgroup$ – zipirovich Nov 26 '16 at 5:47
  • $\begingroup$ Thank-you, I was trying to stick in some numbers to make it work - this explains why I couldn't make it happen. It makes sense. I wasn't sure if it had to do with $cosC=\frac{1}{2ab}$, but your answer clearly explains it. $\endgroup$ – Math1 Nov 26 '16 at 5:50

You can use:

\begin{equation} c^2=a^2+b^2-2ab\cos\hat C\\c^2=c^2+1-2ab\cos\hat C\\2ab\cos\hat C=1\\\cos\hat C=\frac{1}{2ab}\\ \sin\hat C=\sqrt{1-\left(\frac{1}{2ab}\right)^2} \end{equation}

So we have: $A=\frac{1}{2}ab\sqrt{1-\left(\frac{1}{2ab}\right)^2}$


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