Question 54
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:
Solution
By the property,
$$AE \times EB = DE \times CE$$
$$12 \times 20.25 = DE \times 3DE$$
$$243 = 3DE^2$$
$$DE^2 = 81$$
DE = 9 cm
CE = 3DE = 3 $$\times 9 = 27 cm
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