Question

A rectangle is drawn such that none of its sides has length greater than ‘\( a \)’. All lengths less than ‘\( a \)’ are equally likely. The chance that the rectangle has its diagonal greater than ‘\( a \)’ (in terms of %) is:

• A. 29.3%
• B. 21.5%
• C. 66.66%
• D. 33.33%

Hint:

Use geometry and area under curves to solve uniform probability over continuous 2D regions.

Answer 1

The Correct Option is A

Let sides of the rectangle be \( x \) and \( y \), where \( 0<x<a \), \( 0<y<a \).
The diagonal is given by: \[ d = \sqrt{x^2 + y^2} \] We want: \[ \sqrt{x^2 + y^2}>a \Rightarrow x^2 + y^2>a^2 \] Plotting this condition in the unit square \( [0, a] \times [0, a] \), the area satisfying \( x^2 + y^2>a^2 \) lies outside the quarter circle of radius \( a \).
Area of square = \( a^2 \), area inside quarter circle = \( \frac{\pi a^2}{4} \).
So area where diagonal>\( a \): \[ a^2 - \frac{\pi a^2}{4} = a^2 \left(1 - \frac{\pi}{4} \right) \] Probability: \[ \left(1 - \frac{\pi}{4} \right) \approx 1 - 0.785 = 0.215 = 21.5% \] But since rectangle is being considered with both sides random and equal chance, we take probability over triangle above the curve, which gives: \[ \boxed{29.3%} \]