Question 805623
Question 805623
<pre>
The side of a square cannot be 1-&#8730;<span style="text-decoration: overline">2</span> because that's a negative number, 1-1.414 = -0.414,
so I will assume you meant &#8730;<span style="text-decoration: overline">2</span>-1 instead, which is positive.

{{{drawing(300,300,-5,5,-5,5,

green(line(-3,-3,3,-3),line(3,-3,3,3),line(3,3,-3,3),line(-3,3,-3,-3)),

red(line(-sqrt(18),0,0,sqrt(18)), line(sqrt(18),0,0,sqrt(18)),
line(sqrt(18),0,0,-sqrt(18)),line(-sqrt(18),0,0,-sqrt(18)))


)}}}

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 (&#8730;<span style="text-decoration: overline">2</span>-1)² = 2-2&#8730;<span style="text-decoration: overline">2</span>+1 = 3-2&#8730;<span style="text-decoration: overline">2</span>

So if we add both of them we get 2(3-2&#8730;<span style="text-decoration: overline">2</span>)

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

Then 

2(3-2&#8730;<span style="text-decoration: overline">2</span>) = 2x-8t

Divide through by 2:

3-2&#8730;<span style="text-decoration: overline">2</span> = x-4t

Solve for x:

(1)    x = 3-2&#8730;<span style="text-decoration: overline">2</span>+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 

{{{drawing(300,300,-5,5,-5,5,locate(.1,4.8,C),
line(-sqrt(18),0,sqrt(18),0),locate(-4.6,.3,A),locate(4.4,.3,B),
green(line(-3,-3,3,-3),line(3,-3,3,3),line(3,3,-3,3),line(-3,3,-3,-3)),
locate(-2.85,0,D),locate(2.7,0,E),
red(line(-sqrt(18),0,0,sqrt(18)), line(sqrt(18),0,0,sqrt(18)),
line(sqrt(18),0,0,-sqrt(18)),line(-sqrt(18),0,0,-sqrt(18))),
locate(-2.85,1.4,F), locate(-2.9,-.8,G)


)}}}

The area of the octagon is the area of a square minus 4 times
the area of the &#916;AFG

We will calculate diagonal AB by using the Pythagorean theorem on
isosceles right triangle &#916;ABC:

AB² = AC² + BC²

AB² = (&#8730;<span style="text-decoration: overline">2</span>-1)² + (&#8730;<span style="text-decoration: overline">2</span>-1)²

AB² = 2(&#8730;<span style="text-decoration: overline">2</span>-1)² 

AB  = &#8730;<span style="text-decoration: overline">2</span>(&#8730;<span style="text-decoration: overline">2</span>-1)

AB = 2-&#8730;<span style="text-decoration: overline">2</span>

DE = &#8730;<span style="text-decoration: overline">2</span>-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-&#8730;<span style="text-decoration: overline">2</span>)-(&#8730;<span style="text-decoration: overline">2</span>-1)

2AD = 2-&#8730;<span style="text-decoration: overline">2</span>-&#8730;<span style="text-decoration: overline">2</span>+1

2AD = 3-2&#8730;<span style="text-decoration: overline">2</span>

AD = {{{(3-2sqrt(2))/2}}}

&#916;ADF is an isosceles right triangle so DF=AD

Area of &#916;ADF = {{{expr(1/2)AD*DF)}}} = {{{expr(1/2)AD*AD)}}} = {{{expr(1/2)AD^2}}} = {{{expr(1/2)((3-2sqrt(2))/2)^2}}} = {{{expr(1/2)((9-12sqrt(2)+4*2)/4)}}} = {{{expr(1/2)((9-12sqrt(2)+8)/4)}}} = {{{expr(1/2)((17-12sqrt(2))/4)}}} 

The area of &#916;AFG = t = twice the area of &#916;ADF = {{{((17-12sqrt(2))/4)}}}

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

(1)    x = 3-2&#8730;<span style="text-decoration: overline">2</span>+4t

       x = 3-2&#8730;<span style="text-decoration: overline">2</span>+4{{{((17-12sqrt(2))/4)}}}

Cancel the 4's

       x = 3-2&#8730;<span style="text-decoration: overline">2</span>-17+12&#8730;<span style="text-decoration: overline">2</span>

       x = -14+10&#8730;<span style="text-decoration: overline">2</span>
 
       x = 10&#8730;<span style="text-decoration: overline">2</span>-14

Answer: Area = 10&#8730;<span style="text-decoration: overline">2</span>-14  

Edwin</pre>