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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
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- 9/4
- 4
- 6
- 7
- 9/4
Correct Option: C
On the basis of question we draw a figure of a circle in which chords AB and CD intersect at E , 
Given , AE = 9 cm, BE = 12 cm and CE = 3DE
Using theorem,
CE × ED = AE × EB
⇒ 3 DE × DE = 9 × 12
| ⇒ DE² = | = 36 | |
| 3 |
⇒ DE = √36 = 6 cm.
