Infinite Geometric Progression

The question is quite clear in itself! It is an infinite G.P with Initial term = 100 and common ratio=0.5 THis we can see quite clearly from the facts that the areas are 100 (10*10), 50 ( midpoints are joined to form a square, thus area is halved), and thus area is halved indefinitely.

WE know the formula for infinite G.P to be- $\frac {a}{1-r}$ . This is where A=100 and R=0.5 . Applying this we get 100/0.5= 200. Done!

@Ameya Salankar - you should not have titled the question as "Infinte G.p"

Let $S$ be the required sum.

$B_1B_2=\sqrt{(A_2B_1)^2+(A_2B_2)^2}=\sqrt{5^2+5^2}=5\sqrt{2} cm$ .

Similarly, $C_1C_2=5cm$ &
the side of the fourth such square is $\frac{5}{\sqrt{2}} cm$ .

Therefore, the sum of the areas of all the squares so formed is the sum of an infinite Geometric Progression with first term as $100$ & common ratio as $(\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$ .

Sum of an infinite G.P. = $\frac{a}{1-r}$ where $a$ is the first term & $r$ is the common ratio.
(This formula works only when $|r|<1$ . For $|r|>1$ , the sum tends to infinity.)

$\Rightarrow S = \frac{100}{1-\frac{1}{2}} = \boxed{200}cm^2$ .