The circles of same radius 13 cm intersect each other at A and B. If AB = 10 cm, then the distance between their centres is:

Given, AB = 10 cm

AB is the common chord for both the circles and OD will be the perpendicular bisector of AB.

$$\Rightarrow$$ AE = EB = 5 cm

In $$\triangle\ $$AOE,

OE$$^2$$ + AE$$^2$$ = OA$$^2$$

$$\Rightarrow$$  OE$$^2$$ + 5$$^2$$ = 13$$^2$$

$$\Rightarrow$$  OE$$^2$$ + 25 = 169

$$\Rightarrow$$  OE$$^2$$ = 144

$$\Rightarrow$$  OE = 12 cm

Similarly, in $$\triangle$$AED

ED = 12 cm

$$\therefore\ $$Distance between the centres of the circles = OD = OE + ED = 12 + 12 = 24 cm

Hence, the correct answer is Option C