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:
 $ \dfrac{{{P_1} + {P_2} + {P_3} + ............}}{{{A_1} + {A_2} + {A_3} + ...........}} $ equals
A. $ \dfrac{{2\left( {1 + \sqrt 2 } \right)}}{a} $
B. $ \dfrac{{2\left( {2 - \sqrt 2 } \right)}}{a} $
C. $ \dfrac{{2\left( {2 + \sqrt 2 } \right)}}{a} $
D. $ \dfrac{{2\left( {1 + 2\sqrt 2 } \right)}}{a} $

Answer 1

Hint: For solving this particular question, we have to consider the sides of the squares then we have to find the sum of the perimeters of the squares and the sum of the areas of the squares, then we have to calculate the required ratio.

Complete step-by-step answer:
We can understand the given conditions better with the following figure

Let us consider the side of square $ {S_1} = a $ units ,
And sides of the square \[{S_2}\] can be calculated by assuming the isosceles right triangle in the figure whose height and base are $ \dfrac{a}{2} $ and by finding its hypotenuse with help of Pythagoras theorem, the required side of \[{S_2}\] will be calculated,
So, each side of the square \[{S_2}\] will be given as
\[
   = \sqrt {{{\left( {\dfrac{a}{2}} \right)}^2} + {{\left( {\dfrac{a}{2}} \right)}^2}} \\
   = \sqrt {\dfrac{{{a^2}}}{4} + \dfrac{{{a^2}}}{4}} \\
   = \sqrt {\dfrac{{2{a^2}}}{4}} \\
   = \sqrt {\dfrac{{{a^2}}}{2}} \\
   = \dfrac{a}{{\sqrt 2 }} \;
 \]
Similarly for the square \[{S_3}\] taking base and height equals $ \dfrac{a}{{2\sqrt 2 }} $
so its sides will be given as
\[
   = \sqrt {{{\left( {\dfrac{a}{{2\sqrt 2 }}} \right)}^2} + {{\left( {\dfrac{a}{{2\sqrt 2 }}} \right)}^2}} \\
   = \sqrt {\dfrac{{{a^2}}}{8} + \dfrac{{{a^2}}}{8}} \\
   = \sqrt {\dfrac{{2{a^2}}}{8}} \\
   = \sqrt {\dfrac{{{a^2}}}{4}} \\
   = \dfrac{a}{2} \;
 \]
Now we will substitute these values into the sum of perimeters,
Perimeter of a square of side $ a $ is given as
 $ P = 4 \times a $
And sum of perimeters will be given as,
 $ {P_1} + {P_2} + {P_3} + ............ $
Substitute the values,
We will get,
 $
   = {P_1} + {P_2} + {P_3} + ............ \\
   = 4a + 4\dfrac{a}{{\sqrt 2 }} + 4\dfrac{a}{2} + ....... \\
   = 4a + 2\sqrt 2 a + 2a + ....... \;
  $
Here we are getting a geometric progression whose common ratio is $ \dfrac{1}{{\sqrt 2 }} $ and first term is $ 4a $ so its sum will be given as
 $
   = \dfrac{{4a}}{{1 - \dfrac{1}{{\sqrt 2 }}}} \\
   = \dfrac{{4\sqrt {2a} }}{{\sqrt 2 - 1}} \;
  $
Rationalising it,
 $
   = \dfrac{{4\sqrt 2 \left( {\sqrt 2 + 1} \right)a}}{{\left( {\sqrt 2 - 1} \right)\left( {\sqrt 2 + 1} \right)}} \\
   = \dfrac{{4\sqrt 2 \left( {\sqrt 2 + 1} \right)a}}{{\left( {{{\left( {\sqrt 2 } \right)}^2} - {1^2}} \right)}} \\
   = \dfrac{{4\sqrt 2 \left( {\sqrt 2 + 1} \right)a}}{{2 - 1}} \\
   = 4\sqrt 2 \left( {\sqrt 2 + 1} \right)a \;
  $
Similarly, we will substitute values of sides to get the areas of the squares and their sum,
Sum of areas is given as ,
 $ {A_1} + {A_2} + {A_3} + ............ $
Area of a square of side $ a $ is given as
 $ A = {a^2} $
Substitute the values,
We will get ,
 $
   \Rightarrow {A_1} + {A_2} + {A_3} + ............ \\
   = {a^2} + {\left( {\dfrac{a}{{\sqrt 2 }}} \right)^2} + {\left( {\dfrac{a}{2}} \right)^2} + ....... \\
   = {a^2} + \dfrac{{{a^2}}}{2} + \dfrac{{{a^2}}}{4} + ....... \;
  $
Again we got a G.P., whose common ratio of the given G.P is given by $ r = \dfrac{1}{2} $ ,
Now, sum of infinite G.P is given as
 $
   = \dfrac{{{a^2}}}{{1 - \dfrac{1}{2}}} \\
   = 2{a^2} \;
  $
Now, according to the question,
We have to find the ratio given below ,
 $ \dfrac{{{P_1} + {P_2} + {P_3} + ............}}{{{A_1} + {A_2} + {A_3} + ...........}} $
Now, substitute the above results , we will get ,
 $
   = \dfrac{{{P_1} + {P_2} + {P_3} + ............}}{{{A_1} + {A_2} + {A_3} + ...........}} \\
   = \dfrac{{4\sqrt 2 \left( {\sqrt 2 + 1} \right)a}}{{2{a^2}}} \\
   = \dfrac{{2\sqrt 2 \left( {\sqrt 2 + 1} \right)}}{a} \\
   = \dfrac{{2\left( {\sqrt 2 + 2} \right)}}{a} \;
  $
Here, we get the required result.
So, the correct answer is “Option C”.

Note: If we have questions similar in nature as that of above can be approached in a similar manner and we can solve it easily and can find the corresponding result. We have to calculate the midpoint, sum of infinite G.P.