SOLUTION: A side of a square is 16in. The midpoints of its sides are joined to form an inscribed square. Another square is drawn in such a way that its vertices would also lie at the midpoin

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Question 234891: A side of a square is 16in. The midpoints of its sides are joined to form an inscribed square. Another square is drawn in such a way that its vertices would also lie at the midpoints of the sides of the second square. This process is continued infinitely. Find the sum of the areas of these infinite squares.
>>I don't have any idea to solve this one. Please show how you solve it. Thanks
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
It appears that this forms a geometric series with a ratio of .5

The formula for an infinite geometric series is:

S[infinity] = a[1]/(1-r)

where:

a[1] = the first area = 16*16 = 256
r = the ratio = .5

This formula becomes:

S[infinity] = 256/(1-.5) = 256/.5 = 512

Sum should equal 512.

Here's how it was derived.

The first square is 16 * 16 = 256 square inches.

The first inscribed square is 11.3137085 * 11.3137085 = 128 square inches.

The second inscribed square is 8 * 8 = 64 square inches.

The third inscribed square is 5.656854249 * 5.656854249 = 32 square inches.

The fourth inscribed square is 4 * 4 = 16 square inches.

Each succeeding inscribed square is 1/2 * the area of the one above it.

Adding them up, you get:

256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 511 + ever decreasing fractions which will converge on 512 as the number of iterations increases towards infinity.

The formula for finding the side of the inscribed square is the side of the original square times cos(45).

original is 16
16 * cos(45) = 11.3137085
11.3137085 * cos(45) = 8
8 * cos(45) = .........

Check this Picture of Inscribed Square to see how the length of each side of the inscribed square was derived.

Each side of the inscribed square formed a triangle with the corner of the original square.

The side of the inscribed square was the hypotenuse of the triangle.

The length of the hypotenuse of each triangle was equal to half the length of the original side * cos(45).

This became .707106781 * the length of the original side.

Original Side was 16
Inscribed square side became 11.3137....
Next Inscribed square side became .707 * 11.313 = 8
etc.