Question

In the figure below $\left(notdrawntoscale\right),rectangle$ABCD$isinscribedinthecirclewithcentreofO.Thelengthofside$AB$isgreaterthanthatofside$BC$.Theratiooftheareaofthecircletotherectangle$ABCD$is$\pi : \sqrt {3}$.Thelinesegment$DE$intersects$AB$atEsuchthat$\angle ODC = \angle ADE$.Whatistheratio$AE : AD$?

• A. $1:\sqrt{3}$
• B. $1:\sqrt{2}$
• C. $1:2\sqrt{3}$
• D. $1:2$

Answer 1

Answer: (A)

$1:\sqrt{3}$

Let the radius of the circle be ‘R’ and $\angle ODC=\angle ADE=\theta$. If OM is drawn perpendicular to DC,

$DM=R\times \cos \theta$

$OM=R\times \sin \theta$

Length of rec\tan gle ABCD,

$AB=CD=2DM=2R\times \cos \theta$

$AD=BC=2OM=2R\times \sin \theta$

Area of rectangle $=AB\times BC=2R\times \cos \theta \times 2R\times \sin \theta =2R^2\sin 2\theta$

Area of circle $=\pi R^2$

According to the question,

$\pi R^2:2R^2\sin 2\theta =\pi :\sqrt{3}$

$\sin 2\theta =\frac{\sqrt{3}}{2}=\sin {60}^{\circ }.$

$\theta ={30}^{\circ }$

In $\Delta ADE,\tan \theta =\frac{AE}{AD}$

$AE:AD=\tan {30}^o=\frac{1}{\sqrt{3}}=1:\sqrt{3}$