The two chords AB and CD of a circle intersect at point P, such that BP=4 cm, PD=5 cm, and CP=8 cm. find the length of chord AB.
Given : AB and CD are chords of the circle which intersect at point P. BP = 4 cm, CP = 8 cm and PD = 5 cm
To find : AB = ?
Solution : Let AP = $$x$$ cm
Now, when two chords intersect each other inside a circle, the products of their segments are equal.
=> $$(AP)\times(BP)=(CP)\times(DP)$$
=> $$x\times4=8\times5$$
=> $$x=\frac{40}{4}=10$$
$$\therefore$$ AB = AP + PB = $$10+4=14$$ cm
=> Ans - (D)
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