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The chords AB and CD ofa circle intersect at E. IfAE = 12 cm, BE = 20.25 cm and CE = 3 DE,thenthe length (in cm) of CE is:
27
25.5
18
28.5
By the property,
AE × EB = DE × CE 12 × 20.25 = DE × 3DE 243 = 3DE2 DE2= 81 DE = 9 cm CE = 3DE = 3 $$\times 9 = 27 cm
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