## 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, BE^{2} = 1^{2} + (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/(25).

HC = HG + GF + FC, and as HG = FE, 5/2 = 1/(25) + GF + 1/5.

5/(25) = 3/(25) + 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?