Given : $$C_1$$ and $$C_2$$ be the two concentric circles having radius $$r_1=13$$ cm and $$r_2=12$$ cm respectively. To find : AB = ? Solution : AB is the the tangent to the circle $$C_1$$, hence $$angle$$ OPB = $$90^circ$$ Also, the perpendicular from the centre of a circle to a chord bisects the chord. Thus, in $$ riangle$$ OPB, => $$(PB)^2=(OB)^2-(OP)^2$$ => $$(PB)^2=(13)^2-(12)^2$$ => $$(PB)^2=169-144=25$$ => $$PB=sqrt{25}=5$$ cm $$ herefore$$ AB = $$2 imes5=10$$ cm => Ans - (C)