If 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.
Each side of a square is 20cm. Four points on the sides are joined as shown to form an inner square. What is the minimum area of the inner square?
Can anyone help with this using quadratics? Please!
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