Question

Let S₁ be a square of side a. Another square S₂ is formed by joining the mid-points of the sides of S₁. The same process is applied to S₂ to form yet another square S₃ and so on . If A₁, A₂, A₃, ......... be the areas and P₁, P₂, P₃, .......... be the perimeters of S₁, S₂, S₃, ............ respectively, then what does the ratio 

\(\frac{P_1+P_2+P_3+...}{A_1+A_2+A_3+...}\) equal to ?

Answer 1

Let each side of S₁ = a units

∴ Each side of S₂ = \(\sqrt{\big(\frac{a}{2}\big)^2+\big(\frac{a}{2}\big)^2}\) = \(\frac{a}{\sqrt{2}}\) units

⇒ Each side of S₃ = \(\sqrt{\big(\frac{a}{2\sqrt{2}}\big)^2+\big(\frac{a}{2\sqrt{2}}\big)^2}\) = \(\sqrt{\frac{2a^2}{8}}\) = \(\frac{a}{2}\) units

⇒ Each side of S₄ = \(\sqrt{\big(\frac{a}{4}\big)^2+\big(\frac{a}{4}\big)^2}\)  = \(\sqrt{\frac{2a^2}{16}}\) = \(\frac{a}{2\sqrt{2}}\) unit and so on.

∴ P₁ + P₂ + P₃ + P₄ + ......... = 4a +\(\frac{4a}{\sqrt{2}}\) + \(\frac{4a}{2}\) + \(\frac{4a}{2\sqrt{2}}\) +...

This is an infinite GP with 1st term 4a and common ratio \(\frac{1}{\sqrt{2}}\)

∴ P₁ + P₂ + P₃ + P₄ + ......... = \(\frac{4a}{1-\frac{1}{\sqrt{2}}}\) = \(\frac{4a\times \sqrt{2}}{(\sqrt{2}-1)}\) = \(\frac{4 \sqrt{2}a}{(\sqrt{2}-1)}\)

(Sum of infinite GP = \(\frac{First\,term}{1-Common\,Ratio}\) )

Similarly, A₁ + A₂ + A₃ + A₄ + ......... = a² + \(\frac{a^2}{2}\) +\(\frac{a^2}{4}\)+ \(\frac{a^2}{8}\)+...

= \(\frac{a^2}{1-\frac{1}{2}}\) = 2a² 

= \(\frac{4+2\sqrt{2}}{a}\) 

= \(\frac{2(2+\sqrt{2})}{a}\).