It is tempting to think that the triangle would need to be Pythagorean. Certainly for any Pythagorean triangle, the values of a and b would be whole numbers, and c would be a rational number. Suitably scaled up, this too can become a whole number. For instance, if we started with a 3-4-5 triangle, a would be 2 and b would be 3. c would work out to be 2.4. Scaling everything by 5 would give a solution of (10,15,12) with a total of 37. This is not however the minimum solution.

It's a nice little fact that the area of such a triangle is simply a*b. It’s simple enough to prove with a bit of algebra; I’ll leave it to the reader to try it. And if you find a nice geometric proof, let me know as thus far one has eluded me!

That being the case, (a+b)*c/2, the standard way of finding the area, must be equal to a*b. Rearranging, we need to find two numbers a and b such that their product divided by their sum is either a whole number or a half number. The first such pair is 2 and 6, yielding the minimum solution of (2,6,3). The area is 12, which is both 2*6 and (2+6)*3/2. So the answer is 11. The resulting triangle is certainly not Pythagorean, having sides (2*sqrt(7)-2, 2*sqrt(7)+2,8).