SOLUTION: The sides of the outside square in the figure are 6 inches long. The corners of the next smaller square are at the midpoints of the outside square, and so on. Imagine that the pro

Algebra ->  Length-and-distance -> SOLUTION: The sides of the outside square in the figure are 6 inches long. The corners of the next smaller square are at the midpoints of the outside square, and so on. Imagine that the pro      Log On


   

Question 568322: The sides of the outside square in the figure are 6 inches long. The corners of the next smaller square are at the midpoints of the outside square, and so on. Imagine that the process of nesting squares continues forever. The sum of the perimeters approaches a finite number. What is that number?
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
this forms a geometric progression.
a reference for the sum of a geometric progression formula that i used is shown here:
http://www.jimloy.com/algebra/gseries.htm
since the geometric progression is infinite, the formula to be used is:
S = a / (1-r)
the trick in this problem is to first determine that you have a geometric series and then to find the common ratio.
the common ratio turns out to be(1/sqrt(2))
a is the perimeter of the square we are starting with, which is 6 * 4 = 24.
the formula becomes:
S = 24 / (1 - (1/sqrt(2))
this becomes:
S = 81.9411255
the logic used is as follows:
the side of the original square is 6.
the subtending square is found by forming another square whose corners are at the midpoint of the sides of the original square.
the corners of the original square form an isosceles right triangle with the side of the subtending square (attached diagram shows this).
the side of the subtending square is the hypotenuse of this isosceles right triangle.
the formula to find the hypotenuse of this isosceles right triangle uses the formula of c^2 = a^2 + b^2 where c is the hypotenuse of the right triangle and a and b are legs of the right triangle.
since the legs are equal to each other, the formula becomes:
c^2 = a^2 + a^2 which is then equal to 2 * a^2
the formula becomes:
c^2 = 2a^2
the length of the side of the subtending square is therefore equal to sqrt(2a^2)
half the length of the side of the subtending square is then used to find the length of the side of it's subtending square.
this process goes on indefinitely.
a table was created to capture this relationship.
this table is shown below:
columns are identified as follows:
roe = the particular row that you are on in the table.
A is the side of the original square.
B is the perimeter of the original square.
C is half the side of the original square.
D is the side of the subtending square.
E is the perimeter of the subtending square.
F is the ratio of the perimeter of the original square to the perimeter of the subtending square.
row     A         B          C         D         E          F
1	6.00000   24.00000   3.00000   4.24264   16.97056   1.41421
2	4.24264   16.97056   2.12132   3.00000   12.00000   1.41421
3	3.00000   12.00000   1.50000   2.12132   8.48528    1.41421
4	2.12132   8.48528    1.06066   1.50000   6.00000    1.41421
5	1.50000   6.00000    0.75000   1.06066   4.24264    1.41421
6	1.06066   4.24264    0.53033   0.75000   3.00000    1.41421
7	0.75000   3.00000    0.37500   0.53033   2.12132    1.41421

the side of subtending square in row 1 becomes the side of the original square in row 2 and so on down the line.
this means that column A row 2 is the same value as column D row 1, and that column A row 3 is the same value as column D row 2, etc. down the line.
the calculation for the side of the subtending square is based on the formula of:
value in column D = sqrt(2 * value in column C squared).
For example:
in row 3, value in column C is 1.5 and value in column D is 2.12132.
column C is half the length of the side of the original square.
column D is the length of the side of the subtending square.
in row 3, column D is calculated as sqrt(2*1.5^2) which equals 2.121320344 which was shown in the table to 5 decimal places of 2.12132

one the common ratio of the perimeters was found, it was just a matter of plugging the values into the formula to get the sum of the infinite series which resulted in the answer of 81.9411255.

the diagram is shown below:
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