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 
 A: 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 $\approx 290.64$
A: Hint. This is a prism with a trapezoidal cross-section. Thus the volume is the area of the cross-section multiplied by the length of the prism. Thus, you only need find the area of the section, and then you'd be done. That's easy since you already know the length of the parallel sides, and you can find the distance between them trigonometrically.
