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Learning Math Home
Session 8: Solutions
 
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Solutions for Session 8, Part A

See solutions for Problems: A1 | A2 | A3 | A4 | A5 | A6 | A7


Problem A1

a. 

The first net takes 16 cubes. The second net takes 48 cubes. The third net takes 36 cubes.

b. 

Answers will vary. One strategy is to fold up the nets into boxes and fill them with multilink cubes.

<< back to Problem A1


 

Problem A2

Count how many cubes it takes to cover the base in a single layer (that is, the area of the base), and then multiply by how many layers it would take to fill up the box (the height of the box). This formula gives V = l • w • h.

<< back to Problem A2


 

Problem A3

a. 

There are many possible answers; two possibilities are 2 by 2 by 8 and 2 by 4 by 4. These are found by doubling any one of the three dimensions of the original.

b. 

Of the many possible answers, two possibilities are 2 by 4 by 8 and 4 by 4 by 4. These are found by doubling any two of the three dimensions (or multiplying one dimension by a factor of 4).

c. 

Of the many possible answers, a 4-by-4-by-8 box is similar to the original. This is formed by doubling all three of the original's dimensions.

<< back to Problem A3


 

Problem A4

It would be 27 times greater in volume:

Volume of Box B:
(4 • 4 • 3) = 48 cubic units

Volume of Enlarged Box B:
(4 • 3) • (4 • 3) • (3 • 3) = (4 • 4 • 3) • (3 • 3 • 3) = 48 • 27 = 1,296 cubic units

There would be three times as many cubic units in each direction, or 3 • 3 • 3 = 27 times as many in the overall volume.

<< back to Problem A4


 

Problem A5

The ratio is k3, since there are k times as many cubes in all three dimensions. Problem A3 (c) used k = 2, and Problem A4 used k = 3.

<< back to Problem A5


 

Problem A6

Note that Box B is 4 by 4 by 3.

a. 

Only four of Package 1 will fit in Box B.
Sixteen of Package 2 will fit in Box B (vertically), filling the box.
Twelve of Package 3 will fit in Box B, filling the box.
Four of Package 4 will fit in Box B, filling the box.
Package 5 will not fit in Box B at all. Its largest dimension is larger than any of the dimensions of the box!

b. 

Answers will vary, but one strategy is to align the packages along matching dimensions. Recognizing that a package is 1 by 1 by 3 helps to fit it into a 4-by-4-by-3 box.

<< back to Problem A6


 

Problem A7

a. 

One approach is to think about the dimensions of the new box with respect to the dimensions of Packages 1-5. For example, one dimension of the new box needs to be divisible by 5 (because of Package 5). We need all three dimensions to be divisible by 2 (because of Package 1). We need one dimension divisible by 3 (because of Package 2 and Package 4). Based on these observations, the dimensions of the new box are 2 by (2 • 5) by (2 • 3), which is 2 by 10 by 6. Using this same reasoning, you could also decide on a 2-by-2-by-30 box (which also works for all five packages). These are the smallest (in total volume) that will work.

b. 

We know that a 2-by-10-by-6 box works. If we double any of the sides, for example, then the dimensions are all still divisible by the necessary lengths, so it will work to ship all of the packages. By this reasoning, if the sides of a box are "divisible" by 2, 10, and 6, or "divisible" by 2, 2, and 30 from our second example, then it will work to ship all of the packages.

c. 

The dimensions of the larger box must be divisible by the dimensions of the smaller packages. Given a small package with dimensions l, w, and h, we must have one dimension of the box divisible by l, a different dimension divisible by w, and the third dimension divisible by h. For example, for Package 1 to completely fill the box, we see that each dimension must be divisible by 2 (i.e., it must be even).

d. 

Each dimension of the box has to be a common multiple of unique dimension of each of the packages. So for any package with dimensions l, w, and h, we must have one dimension of the box divisible by l, a different dimension divisible by w, and the third dimension divisible by h.

<< back to Problem A7


 

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