Two chords AB and CD of a circle intersect each other at a point E outside the circle. If AB = 11 cm, BE = 3 cm and DE = 3.5 cm, then CD = ?
Given: AB = 11cm, BE = 3cm and DE = 3.5cm
Construction: Join AC
Here,
AE: CE = DE: BE
AE×BE = DE×CE
(AB + BE)×BE = DE×(CD + DE)
(11 + 3)×3 = 3.5×(CD + 3.5)
14×3 = 3.5×(CD + 3.5)
3.5×(CD + 3.5) = 42
(CD + 3.5) = = 12
CD = 12—3.5 = 8.5
∴ CD = 8.5