Two circles of radii 5 cm and 3 cm intersect each other at A and B,and the distance betweentheir centres is 6 cm. The length (in cm) of the common chord AB is:

The two circles with centre O and O' intersect each other at A and B and OO' = 6 cm.

Also, OC and O'C are the perpendicular bisectors of AB.

Let $$OC=x$$ cm and $$O'C=(6-x)$$ cm

In right $$\triangle$$ ACO,

=> $$(OA)^2=(OC)^2+(AC)^2$$

=> $$(OA)^2=(OC)^2+[(O'A)^2-(O'C)^2]$$

=> $$25=x^2+9-(6-x)^2$$

=> $$16=12x-36$$

=> $$x=\frac{52}{12}=\frac{13}{3}$$

$$\therefore$$ $$(AC)^2=25-\frac{169}{9}$$

=> $$AC=\sqrt{\frac{56}{9}}=\frac{2\sqrt{14}}{3}$$

$$\therefore$$ AB = 2 AC

= $$\frac{4\sqrt{14}}{3}$$

=> Ans - (D)