In $$\triangle$$ABC, $$\angle$$C = 90$$^\circ$$ and Q is the midpoint of BC. If AB = 10 cm and AC = $$2\sqrt{10}$$ cm, then the length of AQ is:

From right angled triangle ABC,

AC$$^2$$ + BC$$^2$$ = AB$$^2$$

$$\left(2\sqrt{10}\right)^2$$ + BC$$^2$$ = $$\left(10\right)^2$$

40 + BC$$^2$$ = 100

BC$$^2$$ = 60

BC = $$2\sqrt{15}$$ cm

Q is the midpoint of BC.

CQ = $$\frac{BC}{2}$$ = $$\sqrt{15}$$ cm

From right angled triangle ACQ,

AC$$^2$$ + CQ$$^2$$ = AQ$$^2$$

$$\left(2\sqrt{10}\right)^2$$ + $$\left(\sqrt{15}\right)^2$$ = AQ$$^2$$

40 + 15 = AQ$$^2$$

AQ$$^2$$ = 55

AQ = $$\sqrt{55}$$ cm

Hence, the correct answer is Option A