Two chords AB and CD of a circle intersect each other at a point E outside the circle

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Text Solution

10.5 cm9.5 cm8.5 cm7.5 cm

Answer : C

Solution : Join AC . Then,<br> `AE: CE = DE : BE` <br> `:. AE xx BE = CE xxDE` <br> Let CD `= x cm `. Then, <br> `AE = (AB+ BE) = ( 11+3) cm = 14 cm, BE = 3 cm, CE = ( x + 3.5 ) cm ` and `DE = 3.5 cm` <br> `:. 14 xx 3 = ( x+ .3.5) implies x + 3.5 = (14 xx 3)/( 3.5) = 12` <br> `implies x = (12 - 3.5 ) = 8.5 cm` <br> `:. CD = 8.5 cm`

Two chords AB and CD of a circle intersect each other at a point E outside the circle.

Question:

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
(c) 8.5 cm
(d) 7.5 cm

Solution:

(c) 8.5 cm
Join AC.

Then AE : CE = DE : BE     (Intersecting secant theorem)
∴ AE × BE = DE × CE
Let CD = x cm
Then AE = (AB + BE) = (11 + 3) cm = 14 cm; BE = 3cm; CE = (x + 3.5) cm; DE = 3.5 cm
∴ 14 × 3 = (x + 3.5) × 3.5

$\Rightarrow x+3.5=\frac{14 \times 3}{3.5}=\frac{42}{3.5}=12$

⇒ x = (12 – 3.5) cm = 8.5 cm
Hence, CD = 8.5 cm

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 c 8.5 cm d 7.5 cm

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