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  9. Q22

SSC GD Constable 2011 · Question 22 of 97

Let C be a point on a straight line AB. Circles are drawn with diameters AC and AB. Let P be any point on the circumference of the circle with diameter AB. If AP meets the other circle at Q, then

  1. AQC // PBCorrect
  2. BQC is never parallel to PB 1
  3. CQC = 2 PB 1
  4. DQC // PB and QC = 2 PB

Answer: A. QC // PB

Explanation

In △ AQC, ∠ AQC = 90∘ (∵ Angle in a semi circle is 90∘ ) and in △ APB, ∠ APB = 90∘ (∵ Angle in a semi circle is 90∘ ) Comparing two triangles △ APB and △ AQC, ∠ QAC = ∠P AB ∠ AQC = ∠AP B ∴ △AP B = △AQC ∴ QC // PB 1 Since we cannot prove that C is exactly midpoint of AB, QC = 2 PB cannot be proved

Source: Solved SSC GD 2011 Paper with Solutions — Testbook · reliable-secondary

← Q21View full paper (97 questions)Q23 →

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