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Question:

In the figure below (not drawn to scale), rectangle ABCD is inscribed in the circle with centre at O. The length of side AB is greater than that of side BC. 

The ratio of the area of the circle to the area of the recrangle ABCD is π : $\sqrt{3}$.

The line segment DE intersects AB at E such that ∠ODC = ∠ADE. What is the ratio AE : AD?

Explanation:

Let r be the radius of the circle and let a and b be the length and breadth of the rectangle ABCD respectively.

$\frac{\pi r^2}{ab}=\frac{\pi }{\sqrt{3}}$

$\sqrt{3}{r}^2=ab$ ...(i)

In  ΔDBC,

$\tan \theta =\frac{BC}{DC}=\frac{b}{a}$ ...(ii)

In ΔDAE,

$\tan \theta =\frac{AE}{AD}=\frac{AE}{b}$ ...(iii)

From (ii) and (iii), 

$\frac{AE}{AD}=\frac{b}{a}$

In ΔDBC,

$\begin{array}{rl}& 4r^2=a^2+b^2\\ & 4r^2=a^2+\frac{3r^4}{a^2}\\ & a^4-4r^2{a}^2+3r^4=0\\ & a^4-3r^2{a}^2-r^2{a}^2+3r^4=0\\ & a^2(a^2-3r^2)-r^2(a^2-3r^2)=0\\ & a^2=r^2\text{and}{a}^2=3r^2\\ & \therefore a=r\text{and}a=\sqrt{3}r\end{array}$

When a = ±r, then b = ± $\sqrt{3}$r; when a = ±$\sqrt{3}r,$ then b = ±r

∴ The required ratio is either 1 : $\sqrt{3}$ or $\sqrt{3}$ : 1.

However, only the former is present in the options.

Hence, option (a).