CDS 2026 (I) Elementary Mathematics, question 50

Topic: Geometry

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Q50

ABC is an equilateral triangle. Let PQRS be a square inscribed in it such that P is on AB and Q is on AC. Which of the following is/are correct ?

I. AP:PB=4:3AP:PB=4:3

II. 3AB=(2+3)PQ\sqrt{3}AB=(2+\sqrt{3})PQ

Select the answer using the code given below :

  1. (a) I only
  2. (b) II onlyCorrect answer
  3. (c) Both I and II
  4. (d) Neither I nor II

Solution.

Answer: (b) II only

  1. Let AB=BC=AC=aAB = BC = AC = a, let the square side be s, and let the triangle's height be h=3a/2h = \sqrt{3}a/2.
  2. With the lower side of the inscribed square on BC, similar triangles give s/a=(h−s)/hs/a = (h - s)/h.
  3. Hence s=ah/(a+h)=3a/(2+3)s = ah/(a + h) = \sqrt{3}a/\left( 2 + \sqrt{3} \right), proving II.
  4. Also AP/AB=s/aAP/AB = s/a, so AP/PB=s/(a−s)=3/2AP/PB = s/(a - s) = \sqrt{3}/2, not 4/34/3.