How big is the smallest triangle inscribed in a square A square is divided in 7 areas as show on the figure.
The dots show the corners of the square and the middle points on the edges.

How large a fraction is area $D$ and how do I work it out?
I have tried using trigonometry to calculate the area A. If we say each side of the square is $1$, and look at the triangle $ABC$ using Pythagoras, it's hypotenuse must be $ \sqrt {1 \cdot 1 + 0.5 \cdot 0.5} = 1.118$.
Then using the sine relation we see that the $ \hat A$ is $\sin A = 1/1.118 = 63.43°$. It then follows the other angles must be $90°$ and $71.57°$. If I use Heron's formula I can calculate the area of $A = \sqrt {p(p−a)(p−b)(p−c)} = \frac 1 {12}$.
I know $C$ is $ \frac1 {16}$ just by looking at the figure. 
The area of $ABC$ is $ \frac 1 4 $, so $B$ must be $\frac 1 4 - C - A = \frac 5 {48}$.
Now the area of $BD$ must be $ \frac 1 8$. It therefore follows that $D = \frac 1 8 - \frac 5 {48} = \frac 1 {48}$.
The trigonometry part just seems too elaborate, and I was wondering if there is a much more simple solution I am missing?
 A: The base of $D$ is $\dfrac14$.
The height of $A$ is $\dfrac13$.
The height of $D$ is $\dfrac12-\dfrac13=\dfrac16$.
The area of $D$ is $\dfrac12\times\dfrac14\times\dfrac16=\dfrac1{48}$.
A: Alternatively, we can find that the triangle containing $D$ and $E$ has a base of $1$ and a height of $\frac{2}{3}$ (obtained by solving $1-x = \frac{1}{2} + \frac{1}{2}x$), so it has an area of $\frac{1}{2} \times 1 \times \frac{2}{3} = \frac{1}{3}$.
The height of $D$ is $\frac{2}{3} - \frac{1}{2} = \frac{1}{6}$, so the sides are $\frac{1}{4}$ of the big triangle. Since triangles $D$ and $D + E$ are similar by AA, the area of $D$ is $\frac{1}{16}$ times smaller. This gives the area of triangle $D$ as $\frac{1}{3} \times \frac{1}{16} = \frac{1}{48}$.
A: Coordinate geometry can also work out area $D$. If the bottom left corner is at the origin, and the square has a side length of $1$, the equations of the three lines are:
$$y = 1-x \tag{1}$$
$$y = \frac{1}{2} + \frac{1}{2}x \tag{2}$$
$$x = \frac{1}{2} \tag{3}$$
Solving $(1) = (2)$ gives the point $\left(\frac{1}{3}, \frac{2}{3} \right)$, solving $(2) = (3)$ gives $\left(\frac{1}{2}, \frac{3}{4} \right)$, and solving $(3) = (1)$ gives $\left(\frac{1}{2}, \frac{1}{2} \right)$.
Then the area of the triangle is $\frac{1}{2} \times \text{base} \times \text{height}$:
$$\frac{1}{2} \times \left(\frac{3}{4} - \frac{1}{2} \right) \times \left(\frac{1}{2} - \frac{1}{3} \right) $$
$$=\frac{1}{48}$$
A: Here is a solution from geometric ratios. Similar triangles lead to $\frac{OZ}{XT}=\frac{XY}{YW}=\frac12$. Then,
$$D = \frac{OZ^2}{XT^2} A = \frac14\cdot \frac{XY}{XW} \cdot (A+B+C) = \frac14\cdot\frac13\cdot \frac14 = \frac1{48}$$
