Let's Play Mathematics

If O is the centre of the circle, we have that AOB is an isosceles triangle with sides AO = BO = 5 (the radius) and AB = 6.

The same for BOC with BO = CO = 5 and BC = 6.

If we consider the right triangle BOH (where H is the midpoint of AB) we find that:

OH = $\sqrt{BO^{2}-(AB/2)^{2}}$ = $\sqrt{5^{2}-3^{2}}$ = 4

Being BOH a right triangle and being x the angle OBH, we have:

$OH = BO * \sin(x)$

and thus:

$\sin(x) = \frac{OH}{BO} = \frac{4}{5}$

Being K the midpoint of CA, we notice that triangle ABC is formed by two identical right triangles ABK and CBK and thus:

$AK = AB \times \sin(ABK) = 6 \times \frac{4}{5} = \frac{24}{5}$

but $AK = 2 \times AK = \frac{48}{5} = 9,6$

i did it using pythagoras theoram IN TRIANGLE AOM AO^2=AM^2+OM^2 IN TRIANGLE ABM AB^2=AM^2+BM^2 AB^2=AM^2+(BO-OM)^2 and solving further OM comes to 1.96 and AM =4.8 doubling it will comes 9.6