 ## 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 BE = 5/2.

As ΔBEC is similar to ΔFEC,

BE/BC = CE/FC, 5/2 = (1/2)/FC FC = 1/ 5.

FE/CE = CE/BE, FE/(1/2) = (1/2)/( 5/2) FE = 1/(2 5).

HC = HG + GF + FC, and as HG = FE, 5/2 = 1/(2 5) + GF + 1/ 5. 5/(2 5) = 3/(2 5) + GF GF = 1/ 5.

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

Now the easy way... It can be seen that ΔEFC is congruent with ΔEID. Area five squares = Area of unit square, ABCD = 1 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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