Answer :


Given: AB = 5 cm, AP = 8 cm, and CD = 2 cm


We know that two chords AB and CD intersect at P inside the circle with centre at O.


Then PA× PB = PC× PD.


PB =AP – AB = 8 – 5 =3 cm


Let PD = x then PC = PD + CD = 2 + x


8 × 3 = x (2+x)


2x +x2 =24


x2 + (6–4)x – 24 = 0


(x+6)(x –4) =0


x = 4 and –6


x = –6 is not possible because length is always positive


PD = 4 cm

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