Two circles of radii 7 cm and 5 cm intersect each other at A and B, and the distance between their centres is 10 cm. The length (in cm) of the common chord AB is:

AP = 5 cm

AQ = 7 cm

PQ = 10 cm

Let the CQ be x. So,

PC = 10 - x cm

$$\triangle$$ ACQ is right angle triangle so,

$$(AQ)^2 = (AC)^2 + (CQ)^2$$

$$(7)^2 = (AC)^2 + (x)^2$$

$$(AC)^2 = 49 - (x)^2$$ ---(1)

$$\triangle$$ APC is right angle triangle so,

$$(AP)^2 = (AC)^2 + (PC)^2$$

$$(5)^2 = (AC)^2 + (10 - x)^2$$

$$(AC)^2 = 25 - (10 - x)^2$$ ---(2)

From eq(1) and (2),

$$ 49 - (x)^2 = 25 - (10 - x)^2$$

$$ 24 - (x)^2 = -(10 - x)^2$$

$$ (x)^2 = 24 + 100 + x^2 - 20x$$

20x = 124

x = 124/20 = 6.2 cm

From eq(1),

$$(AC)^2 = 49 - (6.2)^2$$

$$(AC)^2 = 49 - 38.44$$

$(AC)^2 = 10.56 = 1056/100 = 264/25$$

$$AC = \frac{2\sqrt66}{5}$$

The common chord AB = 2 $$\times \frac{2\sqrt{66}}{5} =  \frac{4\sqrt{66}}{5}$$