Question:
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
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
Let the length of the rectangle be \( l \) and breadth be \( b \).
The radius, \( \frac{l}{2} \), and \( b \) in the above diagram form a right-angled triangle.
Using the Pythagorean theorem:
\[ \left(\frac{l}{2}\right)^2 + b^2 = 2^2 \]
Area of the rectangle = \( l \times b \)
The area can be maximized by using the AM-GM (Arithmetic Mean – Geometric Mean) inequality or by setting the two squares equal to make their geometric mean maximum:
\[ \left(\frac{l}{2}\right)^2 = b^2 \]
Solving this gives:
\[ \frac{l}{2} = b \Rightarrow l = 2b \]
So, the ratio of length to breadth is:
\[ \frac{l}{b} = \frac{2}{1} \]
Therefore, the correct option is (D): \( 2:1 \)
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