Calculate the area of center square in the following figure 
Calculate the area of center square in the following figure:(the big square has a side length of 1 and each vertex of big square has been connected to the midpoint of opposite side)
       answer choices  : $\color{blue}{\frac14 ,  \frac15 , \frac16 , \frac17 , \frac18}$


Please review and correct my solution:
$$S_{BigSquare}=2S_{BigTriangle}+S_{Paralleogram} \Rightarrow$$
$$S_{Paralleogram}=\frac12 $$
Now let the side lentgh of center square be $x$ , then we have:
$$S_{Paralleogram}=\frac12 =2\times Base\times x =2\times\frac{\sqrt5}{2}\times x \Rightarrow $$
$$x=\frac{\sqrt5}{10} \Rightarrow $$
$$S_{CenterSquare}=\frac{1}{20}$$ ???!!!
Which is not among the choice!!
What's wrong with my solution??!!!
 A: The area of a parallelogram is base $\times$ height, not $2\times$ base $\times$ height.
A: My First Approach
The small triangles are $\left(\frac{\sqrt5}{10},\frac{\sqrt5}5,\frac12\right)$ and the big triangles are $\left(\frac12,1,\frac{\sqrt5}2\right)$. Thus, the side of the small square is
$$
\frac{\sqrt5}2-\frac{\sqrt5}{10}-\frac{\sqrt5}5=\frac{\sqrt5}5
$$
Thus the area of the small square is $\frac15$.

The Approach in the Question
The area of the parallelogram is $\frac{\sqrt5}2$ (the hypotenuse of the big triangle) times the side of the small square (perpendicular to the hypotenuse). The area of the parallelogram is also $\frac12\cdot1=\frac12$. Thus, we get
$$
\frac{\sqrt5}2\cdot\text{side of small square}=\frac12
$$
A: I'm going to solve this a little different, with a proof.
I recreated your pic so I could edit it to show my proof.  Also, I'm a little rusty on my proofs and math lingo, so take it easy on me, please.
a' = green triangle
b' = red trapezoid
d' = blue square
Line a'F equals b'F because it is the same line.
Line J equals G because line F bisects the edge of the larger square, as given.
The angle between lines E & J is complimentary to the angle between lines G & H.
Because of the last line, we know that line H equals 2F.
Because of the last line, we can draw another line from the intersection between lines F & G to the intersection between lines B & H and know that it is the same length as line G.
From this, we can see that line B equals E.
Because of this, line E equals H, so line H equals B.
This means b' + a' equals a square of the same volume as d'.
As there are 3 more unmarked pairs of a' and b', that means there are a total of 5 equal squares within the original square.
This means d' is 1/5 the area of the original square.
Going further:
Using parts of the above proof, we can see that a' is 1/4 the area of d'.
Which means that b' is 3/4 the area of d'.
I may have skipped parts of the proof, but I considered those parts as "given."  I also didn't want to clutter things by trying to prove that all the triangles are the same and that all the trapezoids are the same.
I haven't done a proof in around 20 years, so this proof might not be 100% correct for a proof, but it is correct for the idea.
I don't have enough reputation to post an image, so it's at https://i.stack.imgur.com/ktQLP.png.
A: By the way, you’re doing an extra step. 
The area of a parallelogram is the width × the height:

so the area of the parallelogram in the middle of the square
is $\frac12\times1 = \frac12$;
you don’t need to subtract the area of the big triangles
from the area of the big square.
A: How is the area of the parellelogram $2bx$? The formula is $bh$, in this case $bx$, giving $$\text{Area}=x^2={1\over 5}$$
