Mr. Akers Counts His Bales KEY Name______

Mr. Akers’ Counts His Bales KEY Name______

Sometimes, a grid represents orthographic projections, or the sides, of an object. In 3 dimensions, this grid becomes stacks of boxes that look like this:

1.) If Mr. Akers stacked his hay bales side-by-side and end-to-end on top of one another, the stack would look like the picture above. A) What would the stack of hay bales look like from the front? B) What would it look like from the side? C) What would it look like from the top? D) How could Mr. Akers figure out how many bales he has?

A.) Front: B.) Side:

C.) Top:

D.) Find the number of bales that make up the side of the stack. 4 x 6 = 24

Multiply that answer by the width of the stack. 24 x 13 = 312 bales.

2.) After some time, Mr. Akers has used some of the hay bales and the stack now looks like the picture above. A) What would the stack of hay bales look like from the front? B) What would it look like from the side? C) What would it look like from the top? D) How could Mr. Akers figure out how many bales he has now?

A.) Front: B.) Side:

C.) Top:

D.) The side of the stack remained the same as the previous problem but the width has changed to 10. The number of bales in a completed stack would be 24 x 10 = 240. However, some are missing so they must be subtracted. 240 – 24 = 216 bales.

3.) Somewhat later, more of the bales have been used resulting in a stack like that above. A) What would the stack of hay bales look like from the front? B) What would it look like from the side? C) What would it look like from the top? D) How could Mr. Akers figure out how many bales he has now?

A.) Front: B.) Side:

C.) Top:

D.) This one is a little harder to calculate because Mr. Akers did not remove the bales evenly. I would count the bales in columns that reached the same height. 22 columns of the bales still reach a height of 6, so those columns total 22 x 6 = 132. 6 of the columns only reach a height of 2, so those columns total 6 x 2 = 12. The total bales in this stack are 132 + 12 = 144 bales.

4.) Finally, Mr. Akers’ stack of hay bales looks like the stack above. A) What would the stack of hay bales look like from the front? B) What would it look like from the side? C) What would it look like from the top? D) How could Mr. Akers figure out how many bales he has left?

A.) Front: B.) Side:

C.) Top:

D.) It seems that Mr. Akers has been moving those bales around and he now has several different heights of bales. 8 columns are 6 high so 8 x 6 = 48. 8 columns are 5 high so 8 x 5 = 40. 4 columns are 4 high so 4 x 4 = 16. 6 columns are 3 high so 6 x 3 = 18. Finally, 2 columns are 2 high so 2 x 2 = 4. The total bales that Mr. Akers has left are 48 + 40 + 16 + 18 + 4 = 126 bales.