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We know that, ⊥ drawn from the centre to the chord bisects the chord . Since, OP ⊥ CD, then CP = PD = \frac{CD}{2} = \frac{4 \sqrt{5}}{2} = 2 \sqrt{5} cm We know that, if two chords of a circle intersect internally, then product of the lengths of their segments are equal . Here AB and CD intersect at P, then AP \times PB = CP \times PD \Rightarrow 2 \times PB = 2 \sqrt{5} \times 2 \sqrt{5} \Rightarrow PB = 10 cm Now, AB = AP + PB = 2 + 10 = 12 cm

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