1. For a right-angled triangle with a hypotenuse of 2009, if the two legs are also integers, then its area is $\qquad$ .
$-.1 .432180$.
Let the two legs be $x, y$. Then $x^{2}+y^{2}=2009^{2}$. According to Pythagorean triples and the symmetry of $x, y$, there exist positive integers $m, n, k (m>n)$, such that
$$
\begin{array}{l}
x=2 m n k, y=\left(m^{2}-n^{2}\right) k, \\
2009=\left(m^{2}+n^{2}\right) k .
\end{array}
$$

Since $2009=41 \times 7^{2}$, and $1,7,7^{2}, 7 \times 41$ cannot be expressed as the sum of squares of two positive integers, we have
$$
k=49, m^{2}+n^{2}=41 \text{. }
$$

Solving gives $m=5, n=4$.
Thus, $x=1960, y=441$.
Then $S=\frac{1}{2} x y=432180$.