In the figure (not drawn to scale), rectangle $ABCD$ is inscribed in the circle with center at $O$.The length of side $AB$ is greater than side $BC$.The ratio of area of the circle to the rectangle $ABCD$ is $\pi:\sqrt3$. The line segment $DE$ intersects $AB$ at $E$ such that $\angle ODC=\angle ADE$. What is the ratio of $AE:AD$.
Options
$a.)\quad \dfrac{1}{\sqrt3}\\ b.)\quad \dfrac{1}{2\sqrt3}\\ c.)\quad \dfrac{1}{\sqrt2}\\ d.)\quad \dfrac{1}{2}\\$
I constructed $OX$ perpendicular to $DC$.
So now $\triangle ADE\sim \triangle ODX$ , So ratio
$AE:AD=OX:DX$
From the ratio's of area i have,
$\dfrac{OD^2}{AD\cdot DC}=\dfrac{1}{\sqrt3}$
and by Pythagorus i have,
$OD^2=OX^2+\dfrac{DC^2}{4}$.
I am stucked trying this .
Let $AB=a$ and $BC=b$. If those green angles are $\theta$ then you'll realize that the question is just asking you what is $\dfrac ba =\tan\theta$. Use the given area ratio: $$\dfrac{\pi r^2}{ab}=\frac {\pi}{\sqrt 3}$$ $$\frac ar \times \frac br = \sqrt 3$$ $$\frac a {2r} \times \frac b {2r} = \frac{\sqrt 3}{4}$$ $$2\sin\theta\cos\theta=\frac{\sqrt 3}{2}$$ $$\sin 2\theta=\frac{\sqrt 3}{2}$$
I suppose the answer is quite obvious now.