Area of a triangle given its circumradius and inradius. A triangle has inradius $4$ cm and a circumradius of $\frac{65}{8}$ cm. Find its area.
I let the sides of the triangle be $a$, $b$, and $c$. After using their corresponding formulas and some manipulation, I got $65 = \frac{abc}{a+b+c}$. I'm kind of stuck here and I don't know how to proceed.
 A: You are correct in the formula you found and in the particular example you found for the sides. But @VasilyMitch is correct in his comment that the area is not uniquely determined.
The diagram below may help to show what is going on.
Let's use the usual notation of $R$ for circumradius and $r$ for inradius, and let's call the circumcentre $O$ and the incentre $I$. Euler's formula gives the distance $d=OI$ as satisfying: $$d^2=R(R-2r)$$
In this case $d$ is much smaller than $r$ or $R$, so we get two circles which are nearly concentric. Hence the triangle doesn't vary much from equilateral. In fact, $\angle A$ is smallest, about $52^o$ when $A$ lies on the ray $IO$ and largest, about $68^o$ when it lies diametrically opposite on the ray $OI$.

The only case when all the sides are integral is the one you found with sides $13,14,15$. You might enjoy figuring out $\angle AIO$ in that case.
The area can take any value in a narrow range around 84. It cannot vary much, but to be unique we would need $d=0$, in which case the circles would be concentric and the triangle would be equilateral.
