Second Square has sides of length of 2*sqrt2

Third Square has sides of length sqrt2 * sqrt2 = 2

Therefore:

Area of Square 1 = 16

Area of Square 2 = 8

Area of Square 3 = 4

Squares continue reduce by half in this way.

IMO 32

Any other clue ?

IMO C

Same reason as yeahdisk. I just drew a little diagram, used Pythag Thm. to find side lengths of 2nd square and extrapolated that to:

A=16
B=8
C=4
D=2
E=1
F=0.5
...
..
.

Total = 32cm^2

first sq area is 16
2nd =8
3rd=4
4th=2
5th=1
6th=0.5
7th=0.25
.....

ans is C

E

What's the OA ?

This can be solved using the formulae for sum of GP of an infinite series.
Sum of the infinite series = First Term/ (1- Common Ratio)
The area is in the order 16 , 8 , 4 = First term is 16 and Common Ratio is 8/16 = 0.5
So sum of the area is 16/(1-0.5) = 32
But who needs to remember one more formula
The series is 16 8 4 after this we can realize it will continue in the same vein i.e . 2 1 then to decimal 0.5 0.25
16+8+4+2+1 = 31 and the decimal part of the series will never exceed 1 so IMO C

Guess what 3 dimensional shape the squares will take if we arrange them in order. (P.S. Out of scope vis-a-vis GMAT)[spoiler][/spoiler]