Trapezoid Volume and Surface Area 
Given $AB = 6$, $DC = 12$, $CG = 8$. Find the surface area and volume of the object.
I tried solving for h first. By inspection, you can see that the height is equal to the side of the 45°.
Using $h=(6-h)(\tan 60°)$. I got $h=9-3 \sqrt{3}$ then I plug it in to solve the volume. 
$$V = \left[\frac12(12+6)(9-3 \sqrt{3)})\right]\times 8$$
$$V = 273.877$$
Then for the surface area, I just added the areas of each faces.
$$SA = 2(9*(9-\sqrt{3}) + 2(8(9-\sqrt{3}) + 18 + 96 $$
$$SA = 298.756$$
I am unsure whether this is correct 

The volume is completely correct. Since this is a prism, its volume is simply the area of trapezium $ABCD$ (or $EFGH$) multiplied by $CG$. So the volume is $\frac 12(12+6)(9-3\sqrt 3) \times 8 \approx 273.88$
For the area, just sum the areas of the $4$ rectangles and $2$ (identical) trapeziums. 
$ABEF$ has area $8 \times 6 =48$
$DCGH$ area = $8\times 12 = 96$
Trapeziums $ABCD$ and $EFGH$ are congruent, each with area $\frac 12(12+6)(9-3\sqrt 3)$, so total area of these two is $18(9-3\sqrt 3)$
Finally, "slanted" rectangles $BCGF$ and $ADHE$
$BC = \frac h{\sin 60^{\circ}} = 6(\sqrt 3-1)$
Area $BCGF = 8 \times BC = 48(\sqrt 3-1)$
Similarly, $AD = h\sqrt 2 = 9\sqrt 2 - 3\sqrt 6$
Area $ADHE = 8 \times (9\sqrt 2 - 3\sqrt 6) =  72\sqrt 2 - 24\sqrt 6$
Adding it all up, total surface area $\approx 290.64$
