Medians are drawn from point $A$ and point $B$ in this right triangle to divide segments $\overline{BC}$ and $\overline{AC}$ in half, respectively. The lengths of the medians are 6 and $2\sqrt{11}$ units, respectively. How many units are in the length of segment $\overline{AB}$?

[asy]
draw((0,0)--(7,0)--(0,4)--(0,0)--cycle,linewidth(2));
draw((0,1/2)--(1/2,1/2)--(1/2,0),linewidth(1));

label("$A$",(0,4),NW);
label("$B$",(7,0),E);
label("$C$",(0,0),SW);
[/asy]
We draw the medians as shown below.

[asy]
draw((0,0)--(7,0)--(0,4)--(0,0)--cycle,linewidth(2));
draw((0,1/2)--(1/2,1/2)--(1/2,0),linewidth(1));
draw((0,4)--(3.5,0));
draw((0,2)--(7,0));
label("$A$",(0,4),NW);
label("$B$",(7,0),E);
label("$C$",(0,0),SW);
label("$M$",(3.5,0),S);
label("$N$",(0,2),W);
[/asy]

From right triangles $ACM$ and $BCN$, we have \begin{align*}
AC^2 + CM^2 &= 36,\\
BC^2 + CN^2 &= (2\sqrt{11})^2 = 44.\end{align*}

However, we have $CM = BC/2$ and $CN = AC/2$, so the equations above become \begin{align*}
AC^2 + \frac14BC^2 &= 36,\\
BC^2 + \frac14AC^2 &=44.
\end{align*}

Adding these equations gives  \[\frac54(AC^2 + BC^2) = 80,\] so $AC^2 + BC^2 = 64$.  But the Pythagorean Theorem gives us $AB^2 = AC^2 + BC^2$, so $AB^2 = 64$, which means $AB = \boxed{8}$.
8
Level 5
Geometry
Problem: Medians are drawn from point $A$ and point $B$ in this right triangle to divide segments $\overline{BC}$ and $\overline{AC}$ in half, respectively. The lengths of the medians are 6 and $2\sqrt{11}$ units, respectively. How many units are in the length of segment $\overline{AB}$?

[asy]
draw((0,0)--(7,0)--(0,4)--(0,0)--cycle,linewidth(2));
draw((0,1/2)--(1/2,1/2)--(1/2,0),linewidth(1));

label("$A$",(0,4),NW);
label("$B$",(7,0),E);
label("$C$",(0,0),SW);
[/asy]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the length of segment \(\overline{AB}\), we denote the lengths of the sides of the triangle as follows: let \(AC = b\), \(BC = a\), and \(AB = c\). The triangle is a right triangle with the right angle at point \(C\). 

The lengths of the medians from points \(A\) and \(B\) can be calculated using the median formula. The median \(m_a\) from vertex \(A\) to side \(BC\) is given by:

\[
m_a = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2}
\]

Given that \(m_a = 6\), we have:

\[
6 = \frac{1}{2} \sqrt{2b^2 + 2c^2 - a^2}
\]

Squaring both sides results in:

\[
36 = \frac{1}{4} (2b^2 + 2c^2 - a^2)
\]

Multiplying through by 4 gives:

\[
144 = 2b^2 + 2c^2 - a^2
\]

Rearranging this, we find:

\[
a^2 = 2b^2 + 2c^2 - 144 \quad (1)
\]

Next, we calculate the median \(m_b\) from vertex \(B\) to side \(AC\):

\[
m_b = \frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2}
\]

Given that \(m_b = 2\sqrt{11}\), we have:

\[
2\sqrt{11} = \frac{1}{2} \sqrt{2a^2 + 2c^2 - b^2}
\]

Squaring both sides results in:

\[
4 \cdot 11 = \frac{1}{4} (2a^2 + 2c^2 - b^2)
\]

This simplifies to:

\[
44 = \frac{1}{4} (2a^2 + 2c^2 - b^2)
\]

Multiplying through by 4 gives:

\[
176 = 2a^2 + 2c^2 - b^2
\]

Rearranging this, we find:

\[
b^2 = 2a^2 + 2c^2 - 176 \quad (2)
\]

Now we have two equations, (1) and (2). We can substitute \(b^2\) from equation (2) into equation (1):

\[
a^2 = 2\left(2a^2 + 2c^2 - 176\right) + 2c^2 - 144
\]

Expanding this gives:

\[
a^2 = 4a^2 + 4c^2 - 352 + 2c^2 - 144
\]

Combining like terms results in:

\[
a^2 = 4a^2 + 6c^2 - 496
\]

Rearranging yields:

\[
0 = 3a^2 + 6c^2 - 496
\]

Dividing through by 3 gives:

\[
0 = a^2 + 2c^2 - \frac{496}{3}
\]

Thus, we can express \(a^2\) in terms of \(c^2\):

\[
a^2 = \frac{496}{3} - 2c^2 \quad (3)
\]

Next, we substitute \(a^2\) from equation (3) back into equation (2):

\[
b^2 = 2\left(\frac{496}{3} - 2c^2\right) + 2c^2 - 176
\]

This simplifies to:

\[
b^2 = \frac{992}{3} - 4c^2 + 2c^2 - 176
\]

Converting \(176\) to a fraction gives:

\[
b^2 = \frac{992}{3} - 4c^2 + 2c^2 - \frac{528}{3}
\]

Combining the fractions results in:

\[
b^2 = \frac{464}{3} - 2c^2
\]

Now we have expressions for \(a^2\) and \(b^2\) in terms of \(c^2\). To find \(c\), we can use the Pythagorean theorem:

\[
a^2 + b^2 = c^2
\]

Substituting the expressions for \(a^2\) and \(b^2\):

\[
\left(\frac{496}{3} - 2c^2\right) + \left(\frac{464}{3} - 2c^2\right) = c^2
\]

Combining gives:

\[
\frac{960}{3} - 4c^2 = c^2
\]

Rearranging yields:

\[
\frac{960}{3} = 5c^2
\]

Thus:

\[
c^2 = \frac{960}{15} = 64
\]

Taking the square root gives:

\[
c = 8
\]

The final answer is \(8\).<|END_OF_LATENT|> The final answer is $8$.