Answer:
64 cubic feet
Explanation:
You do this type of thing by dividing the figure into parts whose volumes are easy to figure.
This sort of figure offers several choices. We really only need to consider how to divide the front face into appropriate rectangles, since the depth of the figure is 2 feet everywhere. The volume is the area of the front face times that depth.
1) A 2 ft × 4 ft upright atop a 12 ft × 2 ft horizontal base. The frontal area is ...
(2 ft)(4 ft) +(12 ft)(2 ft) = 8 ft² +24 ft² = 32 ft²
so, the total volume is ...
V = frontal area × depth = (32 ft²)(2 ft) = 64 ft³
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2) A vertical post of 2 ft × 6 ft at the left side with a horizontal protrusion of 10 ft × 2 ft from its bottom. The frontal area calculated this way is ...
(2 ft)(6 ft) +(10 ft)(2 ft) = 12 ft² +20 ft² = 32 ft² . . . . . as before
Again, the total volume is (32 ft²)(2 ft) = 64 ft³.
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3) A 6 ft × 12 ft rectangle with a 4 ft × 10 ft cutout from the upper right corner. The area calculated this way is ...
(6 ft)(12 ft) -(4 ft)(10 ft) = 72 ft² -40 ft² = 32 ft² . . . . . same as above
Again, the total volume is (32 ft²)(2 ft) = 64 ft³.