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Sloping Square

Problem

In unit square, ABCD, A is joined to the midpoint of BC, B is joined to the midpoint of CD, C is joined to the midpoint of DA, and D is joined to the midpoint of AB.


Find the area of the shaded square formed by this construction.


Solution

We shall solve this in two different ways. First the hard way... (c;


Using the Pythagorean Theorem, BE2 = 12 + (1/2)2 implies BE = radical5/2.

As ΔBEC is similar to ΔFEC,

BE/BC = CE/FC, radical5/2 = (1/2)/FC implies FC = 1/radical5.

FE/CE = CE/BE, FE/(1/2) = (1/2)/(radical5/2) implies FE = 1/(2radical5).

HC = HG + GF + FC, and as HG = FE, radical5/2 = 1/(2radical5) + GF + 1/radical5.

therefore 5/(2radical5) = 3/(2radical5) + GF implies GF = 1/radical5.

Hence the area of the shaded square = 1/5.

Now the easy way...


It can be seen that ΔEFC is congruent with ΔEID.

therefore Area five squares = Area of unit square, ABCD = 1

therefore Area of one square = Area of the shaded square = 1/5

What about the area of the shaded octagon in the diagram below?


Is it a regular octagon?

Problem ID: 179 (Oct 2004)     Difficulty: 2 Star

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