MCQ
Two chords $AB$ and $CD$ of a circle intersect each other at a point $E$ outside the circle. If $AB = 11\ cm, BE = 3\ cm$ and $DE = 3.5\ cm,$ then $CD = ?$
- A$10.5\ cm$
- B$9.5\ cm$
- ✓$8.5\ cm$
- D$7.5\ cm$
✓
Answer
Correct option: C.
$8.5\ cm$
Join $AC.$
$\frac{\text{AE}}{\text{CE}}=\frac{\text{DE}}{\text{BE}}$
$⇒ AE × BE = DE × CE ...(i)$
Then,
$AE = AB + BE = 11 + 3 = 14\ cm, BE = 3\ cm, CE = (x + 3.5)\ cm$ and $DE = 3.5\ cm$
So, from $(i),$ we get
$14 × 3 = 3.5 × (CD + 3.5)$
$\Rightarrow\ \frac{14\times3}{3.5}=\text{CD}+3.5$
$⇒ 12 = CD + 3.5$
$⇒ CD = 8.5\ cm$
$\frac{\text{AE}}{\text{CE}}=\frac{\text{DE}}{\text{BE}}$
$⇒ AE × BE = DE × CE ...(i)$
Then,
$AE = AB + BE = 11 + 3 = 14\ cm, BE = 3\ cm, CE = (x + 3.5)\ cm$ and $DE = 3.5\ cm$
So, from $(i),$ we get
$14 × 3 = 3.5 × (CD + 3.5)$
$\Rightarrow\ \frac{14\times3}{3.5}=\text{CD}+3.5$
$⇒ 12 = CD + 3.5$
$⇒ CD = 8.5\ cm$
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