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What will be the radius of a circle with center O in which a diameter AB bisects the chord CD at a point E such that EB=4cm?

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10cm

20cm

15cm

25cm

answer is A.

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Detailed Solution

It is given that diameter AB bisects the chord CD at a point E such that EB=4cm.
We have to find the radius.
Using the Pythagoras Theorem:

(H y p o t e n u s e) 2 = (P e r p e n d i c u l a r) 2 + (B a s e) 2

Also,

(a − b) 2 = a 2 − 2 a b + b 2

Now joining the diagonal OC:
Question Image

Suppose radius OC=OB=r cm
And, EB=4 cm
Also, OB=OE+EB
OE=OB-EB
OE=(r-4) cm
We know that AB is the bisector of chord CD, so OEC will be a right angled-triangle.
Using Pythagoras Theorem in ∆OEC:

(O C) 2 = (O E) 2 + (E C) 2

⇒ r 2 = (r − 4) 2 + 82 ⇒ r 2 = r 2 − 8 r + 16 + 64 ⇒ r 2 − r 2 + 8 r = 80 ⇒ 8 r = 80 ⇒ r = 808 ⇒ r = 10 c m

So, the radius of the circle is r = 10cm.
Therefore, option 1 is correct.

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