CDS 2026 (I) Elementary Mathematics, question 36

Topic: Geometry

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Q36

ABC is a triangle right-angled at C. A semicircle is drawn on AB as diameter. Let P be any point on AC produced such that AP=AB=10AP=AB=10 cm. Further, B and P are joined. If BC=8BC=8 cm, then what is BP equal to ?

  1. (a) 454\sqrt{5} cmCorrect answer
  2. (b) 99 cm
  3. (c) 1010 cm
  4. (d) 555\sqrt{5} cm

Solution.

Answer: (a) 454\sqrt{5} cm

  1. From the right triangle, AC=102−82=6cmAC = \sqrt{10^{2} - 8^{2}} = 6cm.
  2. Since P lies beyond C on AC and AP=10cm,CP=4cmAP = 10cm,CP = 4cm.
  3. Triangle BCP is right-angled at C, so BP=82+42=80=45cmBP = \sqrt{8^{2} + 4^{2}} = \sqrt{80} = 4\sqrt{5}cm.
  4. The semicircle is not needed for this calculation.