Numerical coincidence? (Nobody's answered this one on stackexchange after several days.)
My brother built a garage whose horizontal cross-section is a rectangle that measures $45$ feet by $30$ feet.  To make sure the right angles were accurate, he measured the two diagonals of the rectangle to see that they were equal.  In  inches,
\begin{align}
& \sqrt{540^2+360^2} \approx 648.999229548\text{ inches} \\[10pt]
= {} & 54\text{ feet}+1\text{ inch} - \text{less than $0.001$ inches}.
\end{align}
It's a bit odd to come within a thousandth of an inch when rounding to the nearest inch, but there's more: The rounding error is about $1/1298$ inches.  But $1298$ is of course not exact; it's
$$
1298+\frac{1}{24073+\text{fraction}}.
$$
Coming within $1/24074$ units when rounding to the nearest unit is even rarer.
$1298\cdot24073>31\text{ million}.$  At that rate you can win lotteries.
(I found that $540^2+360^2 + 1^2 = 649^2$.)
Is this merely a case of an archangel descending from the supernatural realm to have some fun messing with our brains, or is something remarkable happening?  Or more prosaically, can something intelligent be said about this or is that the whole story?
 A: I'm going to put my comments into an answer.  The first thing to note is that $\sqrt{x^2-1}$, written as a negative continued fraction, has a simple form:
$$\sqrt{x^2-1} = x - \cfrac{1}{2x-\cfrac{1}{2x-\cfrac{1}{2x-\cfrac{1}{\ldots}}}}$$
so I'm not sure where the OP got the number $24073$.  The significance of $\sqrt{x^2-1}$ for the problem is, as the OP observed, the fact that $540^2+360^2 = 649^2-1$.  What seems to be going on is that
\begin{align}
& 649^2-1=(649-1)(649+1)=648\cdot650 \\[8pt]
= {} & (2^4\cdot3^4)(2\cdot5^2\cdot13)=12^2\cdot3^2(5^2\cdot13),
\end{align}
where the $5^2\cdot13$, being a product of primes not congruent to $3$ mod $4$, can be written (in various ways) as the sum of two squares.  
What's important is that from two consecutive even numbers ($648$ and $650$) we got a factor of $12^2$ to come out (which converts square feet to square inches), and any remaining primes congruent to $3$ mod $4$ (in this case, just $3$) are raised to an even power.  As I remarked in comments, $808=2^3\cdot101$ and $810=2\cdot3^4\cdot5$ is another such pair.  I found a few others in a table of factorizations up to $1000$, and I would imagine there are more beyond $1000$.  Perhaps a good question is whether there are infinitely many such pairs.
A: This probably falls short of answering the question of why a freaky coincidence happens, but at another level it explains it.  As I said in comments, the stupid calculator I was using gave me that result repeatedly; now the other stupid calculator I'm using is giving me what Barry Cipra got.  Since $540^2+360^2+1^2=649^2$, we have $\sqrt{540^2+360^2} = \sqrt{649^2-1}$, and notice that since $2\cdot649=1298$, we get that
$$-649+\sqrt{649^2-1}=\frac{-1298+\sqrt{1298^2-4}}{2}\tag{1}$$
is a solution to
$$
x^2+1298x+1=0.
$$
Rearrange the quadratic equation:
$$
x = \frac{-1}{1298+x}
$$
and then substituting the expression on the right for $x$ within that very expression gives us
$$
x=\cfrac{-1}{1298-\cfrac{1}{1298-\cfrac{1}{1298-\cfrac{1}{1298-\cdots\cdots\cdots}}}}
$$
and from $(1)$ we have
$$
\sqrt{649^2-1} = 649+x.
$$
That's ONE WAY OF LOOKING AT IT, and at one level it explains it and at another it doesn't.  But it proves that this expansion is right.  I'll look at that dumb calculator again.
