Question

Starting with a unit square, a sequence of squares is generated. Each square in the sequence has half the side length of its predecessor, and two of sides bisected by its predecessor. This process is repeated indefinitely. The total area enclosed by all the squares in limiting situation is?
Please do not reply if you genuinely don't know. Also, if you know how to do, but are lazy to do it, no problem...just tell me how to do it. It will be appreciated.

Answer 1

Answer:

1.33 sq units

Step-by-step explanation:

Starting with a unit square, a sequence of squares is generated. Each square in the sequence has half the side length of its predecessor, and two of sides bisected by its predecessor. This process is repeated indefinitely.

Area of 1st Square = 1 * 1 = 1 sq unit

Length is halved

Area of next Square = (1/2)² = 1/4

Area of further next square = (1/4)² = 1/16

and so o 1/64 , 1/256..........

Total Area = 1 + 1/4  + 1/16  + 1/64  + 1/256 +...............................

Sum of infinite GP = a/(1 - r)

Here a = 1  r = 1/4

Sum  =  1/(1 - 1/4)  =  4/3  = 1.33 sq units

The total area enclosed by all the squares in limiting situation is 1.33 sq units