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Suppose that you draw a 2" by 2" square (this is the first square), then you join the midpoints of its sides to draw another square, then you join the midpoints of that square's sides to draw another square, and so on. determine the sum of the perimeters of the first five squares, and then determine the perimeter of the 20th square.

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wuzzleluver
Whenever you have a 45-90-45 triangle. You will have 2 equal legs with a length "x" units long.  The length of the hypotenuse is x[tex] \sqrt{2} [/tex] units long. I hope you under stand that concept because it is how I came up with my answers. 

Perimeter of original square: 2in × 2 in
2+2+2+2=8 inches

Perimeter of second square: 
√2+√2+√2+√2=4√2 inches ≈ 5.66 in

Perimeter of the 3rd square:
1+1+1+1= 4 inches
(This is exactly half the perimeter of the first square)

Perimeter of the 4th square:
((√2)/2)+((√2)/2)+((√2)/2)+((√2)/2) = 4((√2)/2) inches ≈ 2. 83 inches
(This is exactly half the perimeter of the second square)

Perimeter of the 5th square:
0.5+ 0.5+0.5+0.5= 2 inches

First question: What is the sum of the first 5 squares' perimeters?
8+5.66+4+2.83+2≈ 22.49 inches

Answer to the first question: 22.49 inches


Second Question: 20th square's perimeter?
Every even numbered square had a side length with √2 in it 

the 2nd square's perimeter: 5.66in
the 4th square's perimeter: 5.66/2= 2.83in
Therefore, theoretically if the pattern continues saquare 6 swill be half that value and so on.
6th: 1.145
8th: 0.07075
10th: 0.35375
12th:  0.176875
14th: 0.0884375
16th: 0.04421875
18th: 0.022109375
20th: 0.011054688 ≈ 0.011 inches

Any further Questions? I hope to attach a picture of my ork below for you if you need it.
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