Largest Rectangle Inside a Semicircle

CAT 2023 Slot 3 · QA · Medium · Geometry

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A rectangle with the largest possible area is drawn inside a semicircle of radius 2 cm. Then, the ratio of the lengths of the largest to the smallest side of this rectangle is

  1. A.

    2:12 : 1

  2. B.

    5:1\sqrt{5} : 1

  3. C.

    1:11 : 1

  4. D.

    2:1\sqrt{2} : 1

Answer

A

Explanation

Let the rectangle have base 2x2x along the diameter and height yy. Since the upper vertices lie on the semicircle of radius R=2R = 2: x2+y2=4x^2 + y^2 = 4

Area A=2xyA = 2xy. Maximizing A2=4x2y2=4x2(4−x2)=16x2−4x4A^2 = 4x^2 y^2 = 4x^2(4 - x^2) = 16x^2 - 4x^4. Setting derivative with respect to x2x^2 to zero: 16−8x2=0  ⟹  x2=2  ⟹  x=216 - 8x^2 = 0 \implies x^2 = 2 \implies x = \sqrt{2}

Then height y=4−2=2y = \sqrt{4 - 2} = \sqrt{2}.

The dimensions of the rectangle are:

  • Base =2x=22= 2x = 2\sqrt{2}
  • Height =y=2= y = \sqrt{2}

Ratio of largest to smallest side =222=2:1= \frac{2\sqrt{2}}{\sqrt{2}} = 2 : 1.

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