The points $(0,0)\,$, $(a,11)\,$, and $(b,37)\,$ are the vertices of an equilateral triangle. Find the value of $ab\,$.
Identify the vertices of the triangle with $a + 11i,$ $b + 37i,$ and $0.$

[asy]
unitsize(0.1 cm);

pair A, B, O;

A = (21*sqrt(3),11);
B = (5*sqrt(3),37);
O = (0,0);

draw(A--B--O--cycle);
draw((-5,0)--(40,0));
draw((0,-5)--(0,40));

label("$a + 11i$", A, E);
label("$b + 37i$", B, N);
label("$O$", O, SW);
[/asy]

Then we can obtain $b + 37i$ by rotating $a + 11i$ about the origin by $60^\circ$ counter-clockwise, so
\begin{align*}
b + 37i &= (a + 11i) \cdot \operatorname{cis} 60^\circ \\
&= (a + 11i) \cdot \frac{1 + i \sqrt{3}}{2} \\
&= \left( \frac{a - 11 \sqrt{3}}{2} \right) + i \left( \frac{11 + a \sqrt{3}}{2} \right).
\end{align*}Hence, $2b = a - 11 \sqrt{3}$ and $11 + a \sqrt{3} = 74.$  Solving this system, we find $a = 21 \sqrt{3}$ and $b = 5 \sqrt{3},$ so $ab = \boxed{315}.$