Required to calculate the area of the square problem 
I am trying to calculate the area of the square $ABCD$. I have noticed that there are many similar triangles found inside of the square with the ratio of $BE:AB = 2:3$. I am struggling to get the ratio for $HB$. Please can someone assist me with finishing this question?
 A: I think that the easiest way is analytic geometry. Put $A(0,0)$ and $B(k,0)$. Then, $H(0,2k/3)$.
The equation of the line $BH$ is
$$2x+3y=2k$$
Now, you only have to compute the distance from $D(0,k)$ to this line:
$$\frac{|2\cdot 0+3k-2k|}{\sqrt{2^2+3^2}}=\frac k{\sqrt{13}}$$
Since this is the length of the side of the black square, and its area is $1$, we have the equation $k^2/13=1$. Hence, the area of $ABCD$ is $k^2=13$.
A: Complete the grid:

By Pick's theorem, the area of the large square is $12 + \frac42 - 1 = 13$.
A: The drawing shows the big square consisting of $4$ triangles, $4$ trapezoids, and $1$ small square.  Let's call the areas of interest $S$ (for big square), $t$ (for triangle), $T$ (for trapezoid, which we'll eventually see is larger than $t$), and $s$ (for the little square).  We're given that $s=1$.
From the specification of the points $E$, $F$, $G$, and $H$, we know that the areas of $\triangle ABH$ and $\triangle DCF$ are each ${1\over3}S$.  These means that the area of the parallelogram $BFDH$ is also ${1\over3}S$.  We can write these as
$$2t+T={1\over3}S$$
and
$$2T+s={1\over3}S$$
But it's also easy to see that
$$t=\left(2\over3\right)^2(T+t)$$
since the lengths of the sides of the right triangle of size $t$ (e.g., with hypotenuse $EB$) are two-thirds that of the right triangle of size $T+t$ (e.g., with hypotenuse $AB$).  When you solve these equations with $s=1$, you find $t=4/3$, $T=5/3$, and $S=13$.
A: Let us now consider the square has the area of 1.
Then the triangle $HAB$ has the area $$A_\text{HAB}=\frac{1}{2}\frac{2}{3}1=\frac{1}{3}$$
The length $BH$ is given by
$$d_\text{BH}=\sqrt{1+(2/3)^2}$$ 
The area $A_\text{HAP}$ of triangle $HAP$ (where $P$ is corner of the black square) then fullfills the condition
$$A_\text{HAP}=A_\text{HAB} \frac{(2/3)^2}{d_\text{BH}^2}=\frac{4}{39}$$
Now yo can see, that your big square contains 4 times the area $A_\text{HAB}$ minus 4 times the area $A_\text{HAP}$ plus one black small square $A_\text{small}$.
So you get the condition 
$$4A_\text{HAB}-4A_\text{HAP}+A_\text{small}=1$$
this leads to
$$\frac{12}{13}+A_\text{small}=1$$
thus $$A_\text{small}=\frac{1}{13}$$
When $A_\text{small}=1$ then your big square has the area of $13$
