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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 $AE:AD$.

enter image description here

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 $DC$.

enter image description here

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 .

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

1 Answer 1

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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.

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  • $\begingroup$ is the ratio $\dfrac ab=\dfrac {AD}{AE}$ , how ? $\endgroup$
    – R K
    Mar 27, 2015 at 18:55
  • $\begingroup$ Sorry that was a typo, its $\frac ba = \frac{AE}{AD}$ since $\triangle DAE\sim \triangle DCB$ $\endgroup$
    – najayaz
    Mar 27, 2015 at 19:00
  • $\begingroup$ Ok, i got, $\tan\theta =\dfrac{AE}{AD}=\dfrac{1}{\sqrt3}$ , from that $\sin2\theta$ is that correct ? $\endgroup$
    – R K
    Mar 27, 2015 at 19:15
  • $\begingroup$ Yeah that's right. And..you're welcome! And...an up vote suffices in showing one's appreciation, no need for thank-you comments. $\endgroup$
    – najayaz
    Mar 27, 2015 at 19:48

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