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December 20, 2014

December 20, 2014

Posted by **Me & u** on Saturday, February 25, 2012 at 4:10pm.

Can anyone help with this using quadratics? Please!

- Math 12A -
**Steve**, Saturday, February 25, 2012 at 4:31pmIf the points are located so that each side is divided into sections of length x and (20-x), then the side (s) of the inner square can be calculated

s^2 = x^2 + (20-x)^2

the are of the inner square is thus

a(x) = s^2 = x^2 + (20-x)^2 = 2x^2 - 40x + 400

That is just a parabola with vertex at x = 40/4 = 10

So, the minimum size of the inner square is just 10*10 = 100, or 1/4 the area of the outer square, when its sides are half as long as the outer sides.

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