Calculating area of a shaded region inside a square I'm trying to solve the following problem :  
$ABCD$ is a square of side $4$ units. Find the area of the shaded region as shown in the figure.

The area of square is obviously $16$ , but what after that?
 A: We can compute the area of the white triangle as follows:


*

*There is one black/white boundary from $(0,3)$ to $(4,0)$.  Call the equation of this line $f(x)$.  We can solve for $$f(x)=3-\frac{3}{4}x.$$

*There is one black/white boundary from $(1,0)$ to $(4,4)$.  Call the equation of this line $g(x)$.  We can solve for $$g(x)=\frac{4}{3}x-\frac{4}{3}.$$

*The $y$-coordinate of the intersection of $f(x)$ and $g(x)$ is the height $h$ of the white triangle.  We set $g(x)=f(x)$ and solve for $x$ to obtain the $x$-coordinate of the intersection as $$\frac{52}{25}.$$  We substitute this $x$-coordinate into $f(x)$ to obtain the $y$-coordinate $$h=\frac{36}{25}.$$
The area of the white triangle is $\frac{1}{2} 3h$.
The total area in black is $$4^2-4 \times \frac{3}{2}h=\frac{184}{25}$$ since they have equal area.
A: The area of the outer square is indeed $16$.  The area of the two right triangles on either side of a shaded bar are each $(1/2) (3)(4)=6$, so that the area of one shaded bar is $16-12=4$.  The area of the shaded region is thus $4+4-$ the area of the square center. The side of that square is the width of a shaded bar, which is the sine of the right triangle $4/5$, so that the shaded area is 
$$4+4-\left ( \frac{4}{5}\right)^2 = 8 - \frac{16}{25} = \frac{184}{25}$$
A: White triangles are 3:4:5 triangle scaled down so that the longest side is $3$. Therefore, each one has area $\left(\frac{3}{5}\right)^2 \times \frac 12 \times 3 \times 4 = \frac{54}{25}$. The area of the whole square is $16$. So the area of the shaded region is $\frac{400}{25} - \frac{216}{25} = \frac{184}{25}$.
