William W. answered • 02/23/19
Top Prealgebra Tutor
The easiest approach is to draw a 2D view of the critical side of each shape and pack it with squares then expand that to 3 dimensions.
Example, for the cylinder, draw the top (a circle) and see how many squares fit in it. I got 7 (one row of 3 in the middle and then a row of 2 above and below that). Then, since each "layer" of the cylinder can have 7, and there are 6 layers in the height, the total is 7 x 6 = 42 cubes in the cylinder.
For the cone, the bottom is a circle that is the same as the cylinder above. Drawing the side view, you get an isosceles triangle with height of 7. I drew a row of 3 on the bottom layer, the second layer and the third layer each have 2 squares, the 4th and 5th layers have 1 square and the 6th and 7th layers have no squares. Since layer 1 actually has 7 squares in it (looking from the bottom view - which is the same as the top view of the cylinder - then layer 1 has 7, layer 2 has 2 x 2 in it or 4, layer 3 also has 4, and layer 4 and 5 have 1 each. That adds up to 17 cubes in the cone.
For the sphere, just draw a circle and see how many squares that holds, then expand that to 3 dimensions. The circle holds 5 squares in the middle row with another row of 5 above and below that (total of 3 rows of 5 squares). Above and below that is another row of 3 squares. Starting at the bottom, the 1st row of 3 equates to 9 cubes on that layer, the 2nd row of 5 equates to 25 cubes on that layer, the 3rd and 4th layers also have 25, and the top (5th) layer of 3 squares equates to 9 cubes on that layer. The total is then 9+25+25+25+9 = 93
The total of all 3 shapes: 42 + 17 + 93 = 152
But I wouldn't swear on that answer.
