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 \(A_1, A_2, A_3, \ldots\) be the areas and \(P_1, P_2, P_3, \ldots\) be the perimeters of \(S_1, S_2, S_3, \ldots\), respectively, then the ratio \(\dfrac{P_1 + P_2 + P_3 + \cdots}{A_1 + A_2 + A_3 + \cdots}\) equals

• A. \(\dfrac{2(1+\sqrt{2})}{a}\)
• B. \(\dfrac{2(2-\sqrt{2})}{a}\)
• C. \(\dfrac{2(2+\sqrt{2})}{a}\)
• D. \(\dfrac{2(1+2\sqrt{2})}{a}\)

Hint:

Each new square has side \(1/\sqrt{2}\) times the previous side. Both sums are infinite geometric series.

Answer 1

The Correct Option is C

Step 1: Relation between consecutive squares:
Joining the mid-points of the sides of a square of side \(s\) gives a square of side \(s/\sqrt{2}\) (the hypotenuse of a right triangle with both legs \(s/2\)). So the sides are \(a, \dfrac{a}{\sqrt{2}}, \dfrac{a}{2}, \ldots\).

Step 2: Sum of the areas:
\(A_1 = a^2\), and each area is half the previous one (common ratio \(\dfrac{1}{2}\)). So \(A_1 + A_2 + \cdots = \dfrac{a^2}{1 - \frac{1}{2}} = 2a^2\).

Step 3: Sum of the perimeters:
\(P_1 = 4a\), and each perimeter is \(\dfrac{1}{\sqrt{2}}\) times the previous one. So \(P_1 + P_2 + \cdots = \dfrac{4a}{1 - \frac{1}{\sqrt{2}}} = \dfrac{4a\sqrt{2}}{\sqrt{2} - 1} = 4a\sqrt{2}(\sqrt{2} + 1) = 4a(2 + \sqrt{2})\).

Step 4: Take the ratio:
\(\dfrac{4a(2+\sqrt{2})}{2a^2} = \dfrac{2(2+\sqrt{2})}{a}\).

Final Answer:
The correct option is (3): \(\dfrac{2(2+\sqrt{2})}{a}\).