Let $ \triangle{XOY}$ be a right-angled triangle with $ m\angle{XOY}\equal{}90^\circ$. Let $ M$ and $ N$ be the midpoints of legs $ OX$ and $ OY$, respectively. Given that $ XN\equal{}19$ and $ YM\equal{}22$, find $ XY$.

$ \textbf{(A)}\ 24 \qquad
\textbf{(B)}\ 26 \qquad
\textbf{(C)}\ 28 \qquad
\textbf{(D)}\ 30 \qquad
\textbf{(E)}\ 32$
1. Let \( OX = a \) and \( OY = b \). Since \( M \) and \( N \) are the midpoints of \( OX \) and \( OY \) respectively, we have \( OM = \frac{a}{2} \) and \( ON = \frac{b}{2} \).

2. Given \( XN = 19 \) and \( YM = 22 \), we can use the Pythagorean theorem in the triangles \( \triangle XON \) and \( \triangle YOM \).

3. For \( \triangle XON \):
   \[
   XN^2 = OX^2 + ON^2 \implies 19^2 = a^2 + \left(\frac{b}{2}\right)^2
   \]
   \[
   361 = a^2 + \frac{b^2}{4}
   \]
   \[
   1444 = 4a^2 + b^2 \quad \text{(Multiplying through by 4)}
   \]

4. For \( \triangle YOM \):
   \[
   YM^2 = OY^2 + OM^2 \implies 22^2 = b^2 + \left(\frac{a}{2}\right)^2
   \]
   \[
   484 = b^2 + \frac{a^2}{4}
   \]
   \[
   1936 = 4b^2 + a^2 \quad \text{(Multiplying through by 4)}
   \]

5. We now have the system of equations:
   \[
   4a^2 + b^2 = 1444
   \]
   \[
   a^2 + 4b^2 = 1936
   \]

6. Adding these two equations:
   \[
   4a^2 + b^2 + a^2 + 4b^2 = 1444 + 1936
   \]
   \[
   5a^2 + 5b^2 = 3380
   \]
   \[
   a^2 + b^2 = 676 \quad \text{(Dividing through by 5)}
   \]

7. Since \( XY \) is the hypotenuse of the right triangle \( \triangle XOY \), we use the Pythagorean theorem:
   \[
   XY^2 = a^2 + b^2 = 676
   \]
   \[
   XY = \sqrt{676} = 26
   \]

The final answer is \(\boxed{26}\).