CAT 2003 Slot 2 — QA Question 47

Area and perimeter of square S₁ is a², 4a respectively.

Now, the side of S₂ will be $\frac{a\sqrt{2}}{2}=\frac{a}{\sqrt{2}}$

Thus, the area and perimeter of S₂ = $\frac{a^2}{2},\frac{4a}{\sqrt{2}}$

The side of S₃ will be $\frac{a}{2\sqrt{2}}\times \sqrt{2}=\frac{a}{2}$

Thus, area and perimeter of S₃ = $\frac{a^2}{4},\frac{4a}{{\left(\sqrt{2}\right)}^2}$

Similarly, the area and perimeter of S₄ = $\frac{a^2}{8},\frac{4a}{{\left(\sqrt{2}\right)}^3}$

∴ The required ratio

$\begin{array}{rl}& =\frac{4a+\frac{4a}{\sqrt{2}}+\frac{4a}{(\sqrt{2}{)}^2}+\frac{4a}{(\sqrt{2}{)}^3}+\cdots }{a^2+\frac{a^2}{2}+\frac{a^2}{4}+\frac{a^2}{8}+\cdots }\\ & =\frac{4a\left(1+\frac{1}{\sqrt{2}}+\frac{1}{(\sqrt{2}{)}^2}+\frac{1}{(\sqrt{2}{)}^3}\dots \right)}{a^2\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\dots \right)}\\ & =\frac{4a\left(\frac{1}{1-\frac{1}{\sqrt{2}}}\right)}{a^2\left(\frac{1}{1-\frac{1}{2}}\right)}=\frac{4a\left(\frac{\sqrt{2}}{\sqrt{2}-1}\right)}{a^2\times 2}\\ & =\frac{4a\times \sqrt{2}(\sqrt{2}+1)}{2a^2}=\frac{2(2+\sqrt{2})}{a}\end{array}$

Hence, option (c).