The Ratio of AE:AD

Let the angle ADE = x.Line DO is extended to B, DOB is the diameter of the circle (<A= 90°). The ratio of area of the circle to the triangle is πr^2/(2rsinx . 2rcosx) = π/√3. sin2x = ½ √3, x=30°, AE/AD = tan 30°=1/√3

Join OB. Let the length of the rectangle $(AB = DC)$ be denoted by $l$ and its breadth $(AD = BC)$ be denoted by $b$ . Let $r$ be the radius of the circle.

We notice: $BD^{2}$ = $BC^{2}$ + $DC^{2}$

$\implies$ $(2r)^{2}$ = $b^{2}$ + $l^{2}$

$\implies$ $4r^{2}$ = $b^{2}$ + $l^{2}$ ... (a)

Also, $\frac{Ar Circle}{Ar \square ABCD}$ = $\frac{\pi}{\sqrt3}$

$\implies$ $\frac{\pi r^{2}}{lb}$ = $\frac{\pi}{\sqrt3}$

$\implies$ $\frac{r^{2}}{lb}$ = $\frac{1}{\sqrt3}$ ... (b)

We observe that $\triangle DEA$ $\sim$ $\triangle DBC$ by AA similarity.

$\frac{AD}{DC}$ = $\frac{AE}{BC}$

$\frac{b}{l}$ = $\frac{AE}{b}$

AE = $\frac{b^{2}}{l}$

So, $\frac{AE}{AD}$ = $\frac{b}{l}$ = $x$ ... (c)

Coming back to (b) and using (a), we get:

$\frac{r^{2}}{lb}$ = $\frac{1}{\sqrt3}$

$\frac{b^{2} + l^{2}}{4bl}$ = $\frac{1}{\sqrt3}$

$\frac{b}{4l}$ + $\frac{l}{4b}$ = $\frac{1}{\sqrt3}$

Using (c):

$\frac{x}{4}$ + $\frac{1}{4x}$ = $\frac{1}{\sqrt3}$

$x$ + $\frac{1}{x}$ = $\frac{4}{\sqrt3}$

Solving for $x$ , we get:

$x$ = $\sqrt3$ or $\frac{1}{\sqrt3}$

Among the options given, we realize the answer is $\frac{AE}{AD}$ = $\frac{1}{\sqrt3}$ as length has to be greater than breadth for the rectangle.