Question

In the figure above (not drawn to scale), rectangle ABCD is inscribed in the circle with centre at O. The length of side AB is greater than that of side BC. The ratio of the area of the circle to the area of the rectangle ABCD is $\pi :\sqrt{3}.$ The line segment DE interests AB at E such that $\angle ODC=\angle ADE$ What is the ratio AE : AD?

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

Answer 1

The correct option is A $1:\sqrt{3}$

Wehave to find the ratio of AE:AD i.e. tan x $=\frac{AE}{AD}$ (let $\angle ODC=\angle ADE=x$)
Let r be the radius of the circle
and, Let l and b be the length and breadth of the rectangle.
Then, by Pythagoras theorem in $△ABD$
$l^2+b^2=(2r{)}^2$
$l^2+b^2=4r^2$ (1)
Now, acc to question: Area of circle:area of rectangle $=\pi :\sqrt{3}$
$\Rightarrow \frac{\pi r^2}{lb}=\frac{\pi }{\sqrt{3}}$
$\Rightarrow lb=\sqrt{3}{r}^2$ (2)
Dividing (1) by (2) we get,
$\frac{l^2}{lb}+\frac{b^2}{lb}=\frac{4r^2}{\sqrt{3}{r}^2}$
$\Rightarrow \frac{l}{b}+\frac{b}{l}=\frac{4}{\sqrt{3}}$
$\frac{l}{b}=\sqrt{3}$
$\Rightarrow \frac{b}{l}=\frac{1}{\sqrt{3}}$
Now, $\frac{AE}{AD}=tanx=\frac{1}{\sqrt{3}}$