In ΔABC, ∠C = 90° and Q is the midpoint of BC. If AB = 10 cm and AC = $2\sqrt{10}$ cm, then the length of AQ is:

$\sqrt{55}$ cm

Q is midpoint of BC

Therefore, CQ = BQ = \(\frac{1}{2}\) x BC

\( { 10}^{2 } \) = \( { (2√10)}^{2 } \) + \( { BC}^{2 } \)

= BC = 100 - 40 = √60

Therefore, CQ = \(\frac{2√15}{2}\) = √15

In triangle ACQ, \( { AQ}^{2 } \) = \( { CQ}^{2 } \) + \( { AC}^{2 } \)

= 15 + 4 x 10 = \( { AQ}^{2 } \)

= \( { AQ}^{2 } \) = 55

= AQ = √55

So, the length of AQ is √55cm.