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 :
Answer & explanation
Correct answer: option 1
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