Two chords AB and CD of a circle are produced to intersect each other at a point P outside the circle. If AB = 7 cm, BP = 4.2 cm and PD = 2.8 cm, then the length of CD is:

Correct option is D

Given:
AB = 7 cm, BP = 4.2 cm, and PD = 2.8 cm
Solution:

AP = AB + BP
AP = 7 + 4.2 = 11.2 cm
Let, CD be x.
So, PC = CD + PD
PC = 2.8 + x
As we know,
AP × PB = PC × PD
11.2 × 4.2 = (2.8 + x) × 2.8
2.8 + x =  $$\frac{47.04}{2.8}$$​​
x = 16.8 – 2.8 = 14 cm
So, CD = 14 cm