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AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB is 6 cm, and the length of AP is twice that of AQ. Then the length, in cm, of QB is nearest to
  • a)
    9.3
  • b)
    7.8
  • c)
    9.1
  • d)
    8.5
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
AB is a diameter of a circle of radius 5 cm. Let P and Q be two points...

Since AB is a diameter, AQB and APB will right angles.
In right triangle APB, AP = √102 - √62 = 8
Now, 2AQ=AP => AQ= 8/2=4
In right triangle AQB, AP = 
√102 - √42 = 9.165 = 9.1(Approx)
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AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB is 6 cm, and the length of AP is twice that of AQ. Then the length, in cm, of QB is nearest toa)9.3b)7.8c)9.1d)8.5Correct answer is option 'C'. Can you explain this answer?
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