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In the square shown below, the side is 4 units. The three semicircles with their diameters along the sides of the square touch each other as shown. What is the diameter of the smaller semicircle?
  • a)
    √2 − 1
  • b)
    2√2 − 2
  • c)
    √2
  • d)
    4√2 − 4
  • e)
    √3
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
In the square shown below, the side is 4 units. The three semicircles ...

In the above diagram, S and R are the centers of the semicircles.
PS = SQ = QR = Half the side of the square = 2
Thus: ST = Radius of the bigger semicircle = 2
Let the radius of the smaller semicircle = r
⇒ SR = 2 + r
Thus, from Pythagoras' theorem in triangle SQR:
SR2 = SQ2 + QR2
⇒ (2 + r)2 = 22 + 22 = 8
Taking square-root throughout:
⇒ 2 + r = √8 = 2√2
⇒ r = 2√2 − 2
⇒ Diameter = 4√2 − 4
The correct answer is option D.
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Most Upvoted Answer
In the square shown below, the side is 4 units. The three semicircles ...

In the above diagram, S and R are the centers of the semicircles.
PS = SQ = QR = Half the side of the square = 2
Thus: ST = Radius of the bigger semicircle = 2
Let the radius of the smaller semicircle = r
⇒ SR = 2 + r
Thus, from Pythagoras' theorem in triangle SQR:
SR2 = SQ2 + QR2
⇒ (2 + r)2 = 22 + 22 = 8
Taking square-root throughout:
⇒ 2 + r = √8 = 2√2
⇒ r = 2√2 − 2
⇒ Diameter = 4√2 − 4
The correct answer is option D.
Free Test
Community Answer
In the square shown below, the side is 4 units. The three semicircles ...

In the above diagram, S and R are the centers of the semicircles.
PS = SQ = QR = Half the side of the square = 2
Thus: ST = Radius of the bigger semicircle = 2
Let the radius of the smaller semicircle = r
⇒ SR = 2 + r
Thus, from Pythagoras' theorem in triangle SQR:
SR2 = SQ2 + QR2
⇒ (2 + r)2 = 22 + 22 = 8
Taking square-root throughout:
⇒ 2 + r = √8 = 2√2
⇒ r = 2√2 − 2
⇒ Diameter = 4√2 − 4
The correct answer is option D.
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