CAT 2003 Slot 2 — QA Question 42

Let r be the radius of the circle and let a and b be the length and breadth of the rectangle ABCD respectively.

$\frac{\pi r^2}{ab}=\frac{\pi }{\sqrt{3}}$

$\sqrt{3}{r}^2=ab$ ...(i)

In  ΔDBC,

$\tan \theta =\frac{BC}{DC}=\frac{b}{a}$ ...(ii)

In ΔDAE,

$\tan \theta =\frac{AE}{AD}=\frac{AE}{b}$ ...(iii)

From (ii) and (iii), 

$\frac{AE}{AD}=\frac{b}{a}$

In ΔDBC,

$\begin{array}{rl}& 4r^2=a^2+b^2\\ & 4r^2=a^2+\frac{3r^4}{a^2}\\ & a^4-4r^2{a}^2+3r^4=0\\ & a^4-3r^2{a}^2-r^2{a}^2+3r^4=0\\ & a^2(a^2-3r^2)-r^2(a^2-3r^2)=0\\ & a^2=r^2\text{and}{a}^2=3r^2\\ & \therefore a=r\text{and}a=\sqrt{3}r\end{array}$

When a = ±r, then b = ± $\sqrt{3}$r; when a = ±$\sqrt{3}r,$ then b = ±r

∴ The required ratio is either 1 : $\sqrt{3}$ or $\sqrt{3}$ : 1.

However, only the former is present in the options.

Hence, option (a).