Question

Let $S_1$ be a square of side a. Another square $S_2$ is formed by joining the mid-points of the sides of $S_1$ . The same process is applied to $S_2$ to form yet another square $S_3$, and so on. If $A_1,A_2,A_3,.......$ be the areas and $P_1,P_2,P_3,.......$ be the perimeters of $S_1,S_2,S_3,.......$ , respectively, then the ratio $\frac{P_1+P_2+P_3+P_4+....}{A_1+A_2+A_3+A_4+....}$ equals : (CAT 2003)

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Option (c)

From the given condition in question

Area and perimeter of $S_1=a^2,4a$

Area and perimeter of $S_2=\frac{a^2}{2},\frac{4a}{\sqrt{2}}$

Area and perimeter of $S_3=\frac{a^2}{4},\frac{4a}{(\sqrt{2}{)}^2}$

Area and perimeter of $S_4=\frac{a^2}{8},\frac{4a}{(\sqrt{2}{)}^3}$

These are 2 infinite GPs , which can be solved as follows

Required ratio

$=\frac{[4a+\frac{4a}{\sqrt{2}}+\frac{4a}{(\sqrt{2}{)}^2}+\frac{4a}{(\sqrt{2}{)}^3}+.....]}{a^2+\frac{a^2}{2}+\frac{a^2}{4}+\frac{a^2}{8}+....}$

$=\frac{4a[1+\frac{1}{\sqrt{2}}+\frac{1}{(\sqrt{2}{)}^2}+\frac{1}{(\sqrt{2}{)}^3}+.....]}{a^2(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....)}$

$=\frac{[4a\times \sqrt{2}(\sqrt{2}+1)]}{2a^2}\to \frac{\left[2\sqrt{2}(\sqrt{2}+1)\right]}{a}\to \frac{2(2+\sqrt{2})}{a}$