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In the figure, O and O’ are the centres of the bigger and smaller circles respectively and small circle touches the square ABCD at the mid point of side AD. The radius of the bigger circle is equal to 15 cm and the side of the square ABCD is 18 cm. Find the radius of the smaller circle.
(2015)
  • a)
    4.25 cm
  • b)
    4.5 cm
  • c)
    4.75 cm
  • d)
    5 cm
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
In the figure, O and O’ are the centres of the bigger and smalle...

Let the radius of the bigger circle be R and the smaller circle be r and the side of the square is 2a.
∴ OE = R – EF
= R – [2R – (2r + 2a)]
OE2 + EB2 = OB2
i.e  [2a +2r – R]2 +a2  = R2
a = 9 (∵ 2a = 18); R = 15
∴ (18 + 2r – 15)2 + 92 =152
∴ 2r + 3 = 12

Radius of smaller circle = 4.5 cm.
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Most Upvoted Answer
In the figure, O and O’ are the centres of the bigger and smalle...

Let the radius of the bigger circle be R and the smaller circle be r and the side of the square is 2a.
∴ OE = R – EF
= R – [2R – (2r + 2a)]
OE2 + EB2 = OB2
i.e  [2a +2r – R]2 +a2  = R2
a = 9 (∵ 2a = 18); R = 15
∴ (18 + 2r – 15)2 + 92 =152
∴ 2r + 3 = 12

Radius of smaller circle = 4.5 cm.
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Community Answer
In the figure, O and O’ are the centres of the bigger and smalle...

Let the radius of the bigger circle be R and the smaller circle be r and the side of the square is 2a.
∴ OE = R – EF
= R – [2R – (2r + 2a)]
OE2 + EB2 = OB2
i.e  [2a +2r – R]2 +a2  = R2
a = 9 (∵ 2a = 18); R = 15
∴ (18 + 2r – 15)2 + 92 =152
∴ 2r + 3 = 12

Radius of smaller circle = 4.5 cm.
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