Question:medium

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

Updated On: Jan 13, 2026
  • \(1 :1\)
  • \(\sqrt5 :1\)
  • \(\sqrt2 :1\)

  • \(2:1\)
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The Correct Option is D

Solution and Explanation

A rectangle with the largest possible area is drawn inside a semicircle of radius 2 cm

Let the length of the rectangle be \( l \) and breadth be \( b \).
The radius of the semicircle, half the length of the rectangle, and the breadth form a right-angled triangle.
Applying the Pythagorean theorem:
\[ \left(\frac{l}{2}\right)^2 + b^2 = 2^2 \]
The area of the rectangle is \( l \times b \).

To maximize the area, we can use the AM-GM inequality or set the two squared terms equal:
\[ \left(\frac{l}{2}\right)^2 = b^2 \] 
This yields:
\[ \frac{l}{2} = b \Rightarrow l = 2b \] 
The ratio of the length to the breadth is therefore:
\[ \frac{l}{b} = \frac{2}{1} \]

The correct option is (D): \( 2:1 \)

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