Question 981630
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I already answered this one.


Consider a square with sides that measure *[tex \Large x].  If you create a new square by joining the midpoints of the adjacent sides, the side of the new square will be the hypotenuse of an isosceles right triangle with legs that measure *[tex \Large \frac{x}{2}], from which it is a simple application of the Pythagorean Theorem to determine that the measure of the side of the new square is *[tex \Large \frac{x\sqrt{2}}{2}].  From that you can see that the area of the new square is *[tex \Large \frac{x^2}{2}], exactly half of the area of the original square.


Your original square is 8 by 8 or 64 square inches.  The next one is half of that or 32 square inches, and so on...


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \