Answer :

Two chords AB and CD of a circle intersect each other at P outside the circle.


AP × BP = CP × PD


(AB + BP) × BP = (CD + PD) × PD


(6 + 2) × 2 = (CD + 2.5) × 2.5


16 = 2.5 CD + 6.25


2.5 CD = 9.75


CD = 3.9 cm


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