determining box volume of an angled box

53, 8, 18... 8x18= 144... 144x52= 7632/ 2 (b/c it's a tri.) = 3816 in cubed.

Then you do 4x18x53= 3816+3816 + 7632... then divide that by 1728 (1 cubic foot) and you get

4.42 cubes

 
53, 8, 18... 8x18= 144... 144x52= 7632/ 2 (b/c it's a tri.) = 3816 in cubed.
Then you do 4x18x53= 3816+3816 + 7632... then divide that by 1728 (1 cubic foot) and you get

4.42 cubes
Yea i dont think that is right at all.

Im pretty sure his box would be 3.2cubes

 
53, 8, 18... 8x18= 144... 144x52= 7632/ 2 (b/c it's a tri.) = 3816 in cubed.
Then you do 4x18x53= 3816+3816 + 7632... then divide that by 1728 (1 cubic foot) and you get

4.42 cubes
You have to take into account wood thickness.

Yea i dont think that is right at all.
Im pretty sure his box would be 3.2cubes
yes, 3.2 cubes after subtractiong wood volume.

 
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