Question : Chords AB and CD of a circle intersect at E. If AE = 9 cm, BE = 12 cm, and CE = 3DE, then the length of DE(in cm) is:
Option 1: $\frac{9}{4}$
Option 2: $4$
Option 3: $6$
Option 4: $7$
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Correct Answer: $6$
Solution : Given: AE = 9 cm, BE = 12 cm , CE = 3DE We know that when chords intersect inside a circle the product of the segment formed by the circle is equal. So, CE × DE = AE × BE ⇒ 3DE × DE = 9 × 12 ⇒ DE 2 = 36 $\therefore$ DE = 6 cm Hence, the correct answer is $6$.
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