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CAT 2023 Slot 3 — QA Question 18

Basics of QuadrilateralsEasy

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?

Answer & solution

  • 2 : 1

  • B

    √5 : 1

  • C

    1 : 1

  • D

    √2 : 1

Solution

Easy

By symmetry the maximal rectangle sits with its long side on the diameter. Set the half-length to aa and the height to bb; the top corner lies on the circle, giving a2+b2=4a^2+b^2=4. Maximise the area 2ab2ab with AM–GM.

a b radius 2
1

Set variables. Centre the rectangle on the diameter: long side =2a=2a, short side =b=b. The top corner is on the circle of radius 22:

a2+b2=22=4a^2+b^2=2^2=4
2

Maximise the area =2ab=2ab with AM–GM on a2a^2 and b2b^2:

a2+b22≥a2b2 ⇒ a2+b2≥2ab⇒ 4≥2ab ⇒ ab≤2\begin{aligned} &\frac{a^2+b^2}{2}\ge\sqrt{a^2b^2}\ \Rightarrow\ a^2+b^2\ge 2ab\\ &\Rightarrow\ 4\ge 2ab\ \Rightarrow\ ab\le 2 \end{aligned}

Equality (max area) holds when a=ba=b.

3

Form the ratio. With a=ba=b, the longer side is 2a2a and the shorter side is b=ab=a:

longestshortest=2aa=21\frac{\text{longest}}{\text{shortest}}=\frac{2a}{a}=\frac{2}{1}
ratio=2:1\text{ratio}=\mathbf{2:1}
CAT 2023 Slot 3 QA Q18: A rectangle with the largest possible area is drawn inside a semicircle of radius 2 cm. Then, the ratio of the — Solution | TheCATExam