# Alternative way to calculate the area of a right angled triangle

## Main Question or Discussion Point

Hello all!

New to the forums, and I have a question for you. In my classes, we have been dealing a lot with proofs lately, so when I was working on an assignment, I figured I would try and find my own proof for something, just for the hell of it. I decided to tacle the area of a right angled triangle, because I was using it in one of my assignments, and I wanted to skip some of the legwork I had to do to find the area. On my third try of finding a proof, I finally arrived at something useful.

Starting with Area=½ab, I ended up with:
Area=¼ $c^{2}$sin(2A)=¼ $c^{2}$cos(2B)​

I verified my findings by comparing the results with the ½ab version, and they match. I also wanted to doube check it with some trusted source (and that I wasn't just getting lucky), so I searched around on the web, but can't find it listed anywhere.

So now the question: Has anyone seen this before? And where?

tiny-tim
Homework Helper
welcome to pf!

hello kenneth! welcome to pf!

(actually, it's ¼ $c^{2}$sin(2A)=¼ $c^{2}$sin(2B) )

i don't think it has a name, since it's not particularly useful

but if you double the triangle, by adding a reflection to itself, making a triangle with angles 2A B and B, then it's the well-known formula for the area of a triangle, 1/2 side*side*sin(anglebetween) = 1/2 c*c*sin2A

AlephZero
Homework Helper
Your result shows one (slightly) interesting thing. For a fixed length of c, the maximum area is when sin(2A) = 1, or A = B = 45 degrees.

You can see that a different way, if you know that the right angle always lies on the circle with c as the diameter. If you use the formula "base x height / 2" with c as the base, so you can see where the right angle must be to get the maximum area.

Don't worry about the fact that we don't find this incredibly exciting, or useful. You will learn a lot more by getting the habit of "playing around" with math, rather than just "remembering the right formula" to solve problems.