Question 3590
 Let the side of the 1st square be s1, consider the diagram below
          s1/2
      A --------- B
       |     
       |    
       |  1/2s1
       |
      C
  the side of the 2nd square s2 = BC = sqrt(2(1/2s1)^2) = s1 sqrt(1/2)
  Let the side of the nth square of this process be sn.
  We have sn = sn-1 sqrt(1/2) for n>=2
  and so sn = s1 [sqrt(1/2)]^(n-1) for all n>=1
                                          oo     oo
  Hence, the summation of the total sides E sn = E s1 [sqrt(1/2)]^(n-1)
                                          n=1    n=1
 = s1/ (1- sqrt(1/2))  = s1/(1 - sqrt(2)/2) = 2s1/(1-sqrt(1/2))
 = 2s1(1+sqrt(1/2))/(1-1/2) = 4s1 (1+sqrt(1/2)) =  4s1 (1+sqrt(2)/2)
 = 2s1 (2+sqrt(2)).
 
 The perimeter of a square = 4 * side,
 Since the sum of all perimeters = 8 s1 (2+sqrt(2))
 = 96 (2+sqrt(2)) cm, when s1 = 12 cm.

 The 2nd one about triangle is similar but easier.

 Let s(1)=10 be the side of the first equilateral triangle. 
 and s(n) be the side of the first equilateral triangle for n>=2.
 Since s(2) = 1/2 s(n)= 1/2 s(n-1) for n>=2 (why?)
 
                                         oo       oo
 Hence, the summation of the total sides E s(n) = E s(1) [1/2)]^(n-1)
                                         n=1      n=1

 = s(1)/(1- 1/2) = 2 s(1) = 20 cm.

 The perimeter of a square = 3 * side,
 Since the sum of all perimeters of the equilateral triangles = 60 cm 

 Kenny