The chords AB and CD of a circle intersect at E. If AE = 12 cm, BE = 20.25 cm and CE = 3 DE, then the length (in cm) of CE is :

27 25.5 18 28.5

27

We know that , AE × EB = CE × ED Let us consider that DE = a and CE = 3a 12 × 20.25 = 3a × a a² = 4 × 20.25 a = \(\sqrt { 4 × 20.25 }\) a = 9 And it is given that , CE = 3 DE = 3 a =  3 x 9 = 27 cm