Triangles inside Pentagon

Let the point $F$ be the centre of the pentagon, and the point G be the foot of $DF$ on $AB$ , as shown.

It is not difficult to find out that $\angle GFB =36 ^{\circ}$ and that $\angle GDB =18 ^{\circ}$ .

Now, the area of the triangle $ABD$ is $\frac{(AB)(GB)}{2\tan 18 ^{\circ}}$ while the area of the pentagon $ABCD$ is $5\times \frac{(AB)(GB)}{2\tan 36 ^{\circ}}$ .

Hence $\displaystyle \left \lfloor 1000R \right \rfloor =\left \lfloor 1000\times\frac{\tan 36 ^{\circ}}{5\tan 18 ^{\circ}} \right \rfloor= \left \lfloor \frac{1000}{\sqrt{5}} \right \rfloor=447$ .

I used another method. I first let each side of the pentagon to be 2. I combined the two non shaded triangles and make a rhombus, and find out the area using trigonometry ratios ( forming right angles in the non-shaded triangles by dividing them into half ) to find the length of the diagonals and the rhombus area formula (1/2 pq) and found the area of the shaded triangle using 1/2absinC (using the value found from the rhombus, since the longer diagonal of the rhombus is the same as the sides of the shaded isosceles triangle). Then I found the ratio between the triangle and the whole pentagon and multiplied by 1000, and found the answer.