# In the figure,What is the ratio of $AE:AD$?

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 .

• Is trigonometry allowed? Mar 27, 2015 at 18:26
• @g-man : every thing is allowed upto graduation level.
– R K
Mar 27, 2015 at 18:26
• That's all right, but just a tip: graduation level means different things in different things places. Mar 27, 2015 at 18:28
• I mean graduation level (my point of view) upto undergraduate level.
– R K
Mar 27, 2015 at 18:30

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: $$\dfrac{\pi r^2}{ab}=\frac {\pi}{\sqrt 3}$$ $$\frac ar \times \frac br = \sqrt 3$$ $$\frac a {2r} \times \frac b {2r} = \frac{\sqrt 3}{4}$$ $$2\sin\theta\cos\theta=\frac{\sqrt 3}{2}$$ $$\sin 2\theta=\frac{\sqrt 3}{2}$$
• is the ratio $\dfrac ab=\dfrac {AD}{AE}$ , how ?
• Sorry that was a typo, its $\frac ba = \frac{AE}{AD}$ since $\triangle DAE\sim \triangle DCB$ Mar 27, 2015 at 19:00
• Ok, i got, $\tan\theta =\dfrac{AE}{AD}=\dfrac{1}{\sqrt3}$ , from that $\sin2\theta$ is that correct ?