Answer: 21

Small Hint:

The hexagon’s $120^\circ$ angles make the corner triangle at $Q$ equilateral, so $QB = QC = 1;$ set up coordinates with $Q$ at the origin and $QR$ along the $x$-axis

Big Hint:

Vertex $F$ sits directly above $D$ at height $\sqrt{3},$ and the $45^\circ$ line $RP$ through $F$ gives $QR = 2 + \sqrt{3}$

Solution:

Note $\angle R = 45^\circ.$ Because the hexagon’s interior angles are $120^\circ,$ segment $\overline{BC}$ cuts off a corner triangle at $Q$ with two $60^\circ$ base angles, so triangle $BQC$ is equilateral and $QB = QC = 1.$ Put $Q$ at the origin with $QR$ along the positive $x$-axis. Then $C = (1, 0),$ $D = (2, 0),$ and the hexagon’s vertices are $B = \left(\tfrac{1}{2}, \tfrac{\sqrt{3}}{2}\right),$ $A = (1, \sqrt{3}),$ $F = (2, \sqrt{3}),$ $E = \left(\tfrac{5}{2}, \tfrac{\sqrt{3}}{2}\right).$

Since $\angle R = 45^\circ,$ line $RP$ has slope $-1.$ If it passed through $E,$ it would be $x + y = \tfrac{5 + \sqrt{3}}{2},$ which puts $F$ (with $x + y = 2 + \sqrt{3}$) outside the triangle; so the vertex on $\overline{RP}$ is $F,$ and $RP$ is the line $x + y = 2 + \sqrt{3}.$ It meets the $x$-axis at $R = (2 + \sqrt{3},\ 0)$ and the line $y = \sqrt{3}\,x$ (line $QP$) where $x(1 + \sqrt{3}) = 2 + \sqrt{3},$ giving $P$ height $$y = \frac{\sqrt{3}(2 + \sqrt{3})}{1 + \sqrt{3}} = \frac{3 + \sqrt{3}}{2}.$$

The area is $\frac{1}{2} \cdot QR \cdot y$ $= \frac{1}{2}(2 + \sqrt{3}) \cdot \frac{3 + \sqrt{3}}{2}$ $= \frac{9 + 5\sqrt{3}}{4},$ so $a + b + c + d = 9 + 5 + 3 + 4$ $= 21.$