SOLUTION: Two squares, each of side length 1-sqrt( 2 ) unit, overlap. The overlapping region is a regular octagon. What is the area (in square units) of the octagon?

The side of a square cannot be 1-√2 because that's a negative number, 1-1.414 = -0.414,
so I will assume you meant √2-1 instead, which is positive.



All 8 little triangles around the points of the "star" are congruent,
so if we add the areas of the two squares we will have twice the
area of the octagon + 8 times the area of one of those triangles. So
we will seek to find the area of one of those little triangles.

The area of one of those squares is (√2-1)² = 2-2√2+1 = 3-2√2

So if we add both of them we get 2(3-2√2)

Let the area of the octagon be x, and let the area of each triangle be t.

Then 

2(3-2√2) = 2x-8t

Divide through by 2:

3-2√2 = x-4t

Solve for x:

(1)    x = 3-2√2+4t

So we will now seek to find t, the area of each of those little triangles.

Draw a diagonal AB of the red square cutting the green square
at D and E.  We label C,F, and G 



The area of the octagon is the area of a square minus 4 times
the area of the ΔAFG

We will calculate diagonal AB by using the Pythagorean theorem on
isosceles right triangle ΔABC:

AB² = AC² + BC²

AB² = (√2-1)² + (√2-1)²

AB² = 2(√2-1)² 

AB  = √2(√2-1)

AB = 2-√2

DE = √2-1, the same
as a side of a square.

AD+DE+BE = AB and since AD=BE

2AD+DE = AB

2AD = AB-DE

2AD = (2-√2)-(√2-1)

2AD = 2-√2-√2+1

2AD = 3-2√2

AD = 

ΔADF is an isosceles right triangle so DF=AD

Area of ΔADF =  =  =  =  =  =  =  

The area of ΔAFG = t = twice the area of ΔADF = 

Now we go back to equation (1) back near the top:

(1)    x = 3-2√2+4t

       x = 3-2√2+4

Cancel the 4's

       x = 3-2√2-17+12√2

       x = -14+10√2
 
       x = 10√2-14

Answer: Area = 10√2-14  

Edwin