A concrete dam of height 128 feet was built in gorge. One side $AB$ of the gorge slopes at an angle of $60^o$, the other side $CD$ at $45^o$. I was wandering across the internet when a wild question appears. It goes like this:


  A concrete dam of height 128 feet was built in gorge. One side $AB$ of the gorge slopes at an angle of $60^o$, the other side $CD$ at $45^o$. 
  The bases of the dam are horizontal and rectangular in shape. The lower base is 1215 feet by 152 feet, and the upper base is 32 feet wide. 
  How many cubic yards of concrete were required?


My work

I imagine that dam to look like this:



Delving deeper into the picture, it becomes like this:



Positioning the dam so it is easier to get the volume of this dam, it looks like this:



Notice that the it is easier to get the volume of the dam if we separate the shapes that makes up the dam, the red and the green part...



The green part's volume is pretty straightforward. We'll just use the formula $$V = BH$$ where $V$ is the volume of the figure, $B$ is the 
area of the base and $H$ is the height of the figure. Looking at the green part of the figure, the area of it's base is a trapezoid. The area would 
be $$A = \frac{1}{2}h(b_1 + b_2)$$ where $h$ is the height of plane, $b_1$ is the length of the upper base, and $b_2$ is the length of the lower 
base. With that in mind, the volume of the green part would be...

$$V=BH$$
$$V = \left( \frac{1}{2}h(b_1 + b_2)\right)(H)$$
$$V = \left( \frac{1}{2}(128 \space feet)(1013.1 \space feet + 1215 \space feet)\right)(32 \space feet)$$
$$V = 4563148 \space ft^3$$

The red part is pretty tricky, but solvable. Imagining the red part as a one part cut in halves, it becomes easier to get its volume. Getting the 
volume of the whole red part below:



$$V=BH$$
$$V = \left( \frac{1}{2}h(b_1 + b_2)\right)(H)$$
$$V = \left( \frac{1}{2}(128 \space feet)(1013.1 \space feet + 1215 \space feet)\right)(120 \space feet)$$
$$V = 17111808 \space ft^3$$ 

Now getting the volume of the red part we see, it is $\frac{V}{2}$ or $8555904\space ft^3$

The total volume of the dam would be $4563148 \space ft^3 + 8555904 \space ft^3$ or $13119052.8 \space ft^3$

It's not yet the final answer because it must be converted to cubic yards. Now converting from cubic feet to yards:

$$13119052.8 \space ft^3\left(\frac{0.3048 \space meters}{1 \space feet}\right)^3 = 371490.18 \space m^3 $$
$$371490 \space m^3 \left( \frac{1.0936 \space yards}{1 \space meter}\right)^3 = 485873.9 \space cubic \space yards$$

Is my solution, ladies and gentlemen, correct?
It is from  SOLID MENSURATION  book (page 137 problem no 48) by James Bland and you may see that in this link.

https://archive.org/stream/in.ernet.dli.2015.205959/2015.205959.Solid-Mensuration#page/n17/mode/1up

FIRST SOLUTION:    Volume Of Prismatoid
$$\tan 300  =  x/128$$
$$x  =  73.9  \text{ ft}.$$
$$\tan 450  =  y/128$$
$$y  =  128 \text{ ft}.$$



$$L  =   x  +  y  + 1215$$
$$L  =  73.9 +  128  + 1215$$
$$L   =  1416.9  \text{ ft}.$$



$$\text{Mean Width }    =    (32   +  152)/2          =   92 \text{ ft}.$$
$$\text{Mean Length }    =  (1215  + 1416.9)/2    =   1315.95 \text{ ft}.$$



$$V   =  (h/6) [A_{\text{up}}  +  A_\text{down}  +  4A_m ]$$
$$V   =  (128/6) [  (1416.9)(32)+  (1215)(152)+  4(92)(1315.95)  ]$$
$$V   =  (128/6)   [ 45,340.8 +  184,680+  484,269.6 ]$$
$$V   =   15,238,195.2 \text{ ft}.3$$
$$V   =  15,238,195.2 \text{ ft}.3 (〖1 yd〗^3/〖27 ft〗^3 )$$
$$V  =  564,377.6 \text{ yd}.3$$
$$V  =  564,378 \text{ yd}.3\tag{$\neq$ Book answer: $561,230\text{ yd}.3$}$$

Comments:   The length of dam at $1416.9 \text{ ft}$.   It should be $1398.457 \text{ ft}$, 
                        or be reduced by $18.443\text{ ft}$ in order to satisfy book answer. $( V = 561,230 \text{ yd}.3 )$.



SECOND SOLUTION:    Volume of truncated prism:
V  =  Km
V  =  Aright section   x Average Lateral edge
V  =  [1/2  (a+b)h][(sum of lateral edges)/4]
V  =  [1/2  (32+152)(128)][(1215   +    1215   +   1416.9   +  1416.9)/4]
V  =  (11,776)(1,315.95)
V  =  15,496,627.2  ft.3
V  =  15,496,627.2 ft.3(〖1 yd〗^3/〖27 ft〗^3 )
V  =  573,949.1556 yd.3
V  =  573,949 yd.3      ( ≠ Book answer: 561,230yd.3  )



THIRD SOLUTION:    Volume of frustum of Regular Pyramid
V  =  1/3 [b +  B +  √bB]h
V  =  1/3 (32)(1416.9)+  (1215)(152)+  √((45,340.8)(184,680) )
V  =  1/3 [45,340.8 +  184,680 +  91,507.04314]  (128)
V  =  13,718,521.31 ft.3
V  =  13,718,521.31 ft.3(〖1 yd〗^3/〖27 ft〗^3 )
V  =  508,093.3818 yd.3
V  =  508,093 yd.3   ( ≠ Book answer: 561,230yd.3  )
It is from  SOLID MENSURATION  book (page 137 problem no 48) by James Bland and you may see that in this link.

https://archive.org/stream/in.ernet.dli.2015.205959/2015.205959.Solid-Mensuration#page/n17/mode/1up

FIRST SOLUTION:    Volume Of Prismatoid
$$\tan 300  =  x/128$$
$$x  =  73.9  \text{ ft}.$$
$$\tan 450  =  y/128$$
$$y  =  128 \text{ ft}.$$



$$L  =   x  +  y  + 1215$$
$$L  =  73.9 +  128  + 1215$$
$$L   =  1416.9  \text{ ft}.$$



$$\text{Mean Width }    =    (32   +  152)/2          =   92 \text{ ft}.$$
$$\text{Mean Length }    =  (1215  + 1416.9)/2    =   1315.95 \text{ ft}.$$



$$V   =  (h/6) [A_{\text{up}}  +  A_\text{down}  +  4A_m ]$$
$$V   =  (128/6) [  (1416.9)(32)+  (1215)(152)+  4(92)(1315.95)  ]$$
$$V   =  (128/6)   [ 45,340.8 +  184,680+  484,269.6 ]$$
$$V   =   15,238,195.2 \text{ ft}.3$$
$$V   =  15,238,195.2 \text{ ft}.3 (〖1 yd〗^3/〖27 ft〗^3 )$$
$$V  =  564,377.6 \text{ yd}.3$$
$$V  =  564,378 \text{ yd}.3\tag{$\neq$ Book answer: $561,230\text{ yd}.3$}$$

Comments:   The length of dam at $1416.9 \text{ ft}$.   It should be $1398.457 \text{ ft}$, 
                        or be reduced by $18.443\text{ ft}$ in order to satisfy book answer. $( V = 561,230 \text{ yd}.3 )$.



SECOND SOLUTION:    Volume of truncated prism:
V  =  Km
V  =  Aright section   x Average Lateral edge
V  =  [1/2  (a+b)h][(sum of lateral edges)/4]
V  =  [1/2  (32+152)(128)][(1215   +    1215   +   1416.9   +  1416.9)/4]
V  =  (11,776)(1,315.95)
V  =  15,496,627.2  ft.3
V  =  15,496,627.2 ft.3(〖1 yd〗^3/〖27 ft〗^3 )
V  =  573,949.1556 yd.3
V  =  573,949 yd.3      ( ≠ Book answer: 561,230yd.3  )



THIRD SOLUTION:    Volume of frustum of Regular Pyramid
V  =  1/3 [b +  B +  √bB]h
V  =  1/3 (32)(1416.9)+  (1215)(152)+  √((45,340.8)(184,680) )
V  =  1/3 [45,340.8 +  184,680 +  91,507.04314]  (128)
V  =  13,718,521.31 ft.3
V  =  13,718,521.31 ft.3(〖1 yd〗^3/〖27 ft〗^3 )
V  =  508,093.3818 yd.3
V  =  508,093 yd.3   ( ≠ Book answer: 561,230yd.3  )