Grade 5 · 5.MD.C.5 · 5.MD.C.5A

Volume of a box

You can measure the three edges of a box with a ruler. Is that enough to know how much it holds, without filling it?

You need

Unlocks

DensityMath 6 Surface area, and how it compares with volumeMath 6 Volume of a box with fraction edge lengthsMath 6 Volume of a solid made of boxesGrade 5 +1 more

Where you are 9 in Volume and capacity

Measurement12

Volume and capacity9

▴ 3 above Volume as filling Volume with unit cubes

Volume of a box

Surface area, and how it compares with volume Volume of a box with fraction edge lengths ▾ Show all 9 (1 more)

When you reach for it

Masumi is renting a storage unit. Sanjay is buying an ice chest. Zacarias loads a shipping container at a dock. Each of them wants to know the same thing: how much space is inside?

Nobody is going to fill a storage unit with foot cubes to find out. In Volume with unit cubes the volume of a box was found by counting cubes in layers: the cubes in one layer, times the number of layers. This lesson notices that those two numbers can be read straight off the box’s edges. Then a ruler is enough. You reach for this move whenever you know the edges of a box and want its volume, or you know the volume and want an edge.

The lesson runs over four days. Day 1 turns layers into the rule V = l \times w \times h. Day 2 writes the same rule as base area times height, and shows that any face of the box can be the base. Day 3 runs the rule backward to find a missing edge. Day 4 uses it in real problems, and compares boxes that hold the same amount but have different shapes.

The procedure

Day 1: from layers to length times width times height

A box here means a solid with six flat rectangular faces that meet at square corners, like a shoebox or a brick. Mathematicians call it a rectangular prism. Every box has three kinds of edge: its length l, its width w and its height h.

Figure 1The floor of the box is its base. Length times width gives the base, and the height says how many layers stand on it.

Say the edges are whole numbers of some unit, such as feet. Pack the box with foot cubes.

The floor is a rectangle l feet by w feet. It holds l \times w unit squares, and each square has one cube on it. So one layer has l \times w cubes.
Each layer is 1 foot thick, and the box is h feet tall. So exactly h layers fit.
Every layer is the same, because the walls go straight up. So the cubes in the box are l \times w cubes, taken h times.

That gives the rule for the volume V of a box:

V = l \times w \times h

The volume comes out in cubic units, because it counts cubes. If the edges are in feet, the volume is in cubic feet, \text{ft}^3. The three edges must all be in the same unit, since the same cube has to fit along all three.

Why does the rule work? Not because someone said so. It works because it is a short way of counting the cubes, row by row and layer by layer. A box 3 by 2 by 4 could be counted one cube at a time, and you would reach 24. The rule gets there as 3 \times 2 = 6 cubes per layer and 6 \times 4 = 24. Same cubes, faster count.

Day 2: base area times height

The first two factors, l \times w, are the area of the floor. Call the floor the base of the box, and write B for its area. Then the rule reads

V = B \times h

In words: the area of the base, times the height. This is the same rule, with the first multiplication already done. It is useful when you are told the floor area and not its two sides. Masumi’s storage company might list a unit as “50 square feet, 9 feet tall” without saying how long or wide the floor is. You do not need to know.

Any face can be the base. A box can rest on any of its faces. Whichever face it rests on is the base, and the edge standing up from it is the height. Take a box with edges 7, 4 and 2.

Figure 2One box, three choices of base. 28 \times 2, 14 \times 4 and 8 \times 7 all give 56.

Base	Base area B	Height h	B \times h
7 by 4	28	2	56
7 by 2	14	4	56
4 by 2	8	7	56

All three give 56, as they must. Turning a box over does not change how much space is inside. The numbers agree because of two properties of multiplication.

The associative property says the grouping does not matter: (7 \times 4) \times 2 = 7 \times (4 \times 2).
The commutative property says the order does not matter: 7 \times 2 = 2 \times 7.

Together they say that three numbers multiplied together give the same product whichever two you multiply first. So you may pick the pair that is easiest. That is a real help: 25 \times 4 is 100 in your head, so a box with edges 25, 9 and 4 is 100 \times 9 = 900 cubic units.

Day 3: finding a missing edge

Sometimes you know how much a box holds and need one of its edges. Sanjay’s small ice chest holds 360 cubic inches. Its inside floor is 9 inches by 8 inches. How tall is it inside?

Figure 3The box holds 360 cubic inches. Its base is 9 \times 8 = 72 square inches, so it is 360 \div 72 = 5 inches tall.

The rule says 360 = 9 \times 8 \times h. The base is 9 \times 8 = 72 square inches, so

72 \times h = 360

Each inch of height is one layer of 72 inch cubes. The question is how many layers of 72 make 360. That is a division:

h = 360 \div 72 = 5 \text{ inches}

Check by multiplying back: 9 \times 8 \times 5 = 72 \times 5 = 360. It works.

Division undoes multiplication, so the same thought finds any missing edge.

Know the volume and the base area: the height is V \div B.
Know the volume and the height: the base area is V \div h.
Know the volume and two edges: multiply the two edges, then divide the volume by that product.

Day 4: using the rule, and same volume, different shape

A volume answers “how much fits inside”. Real problems ask more than that: how many boxes to send, how much space is left, which box fits a long object. The rule gives the volume, and then the question decides what to do with it.

One thing the rule shows is that the volume does not fix the shape. Masumi is choosing between three storage units.

Figure 4All three units hold 320 cubic feet, but they are different shapes.

Unit	Edges	Volume
A	10 ft by 4 ft by 8 ft	320\ \text{ft}^3
B	8 ft by 5 ft by 8 ft	320\ \text{ft}^3
C	16 ft by 4 ft by 5 ft	320\ \text{ft}^3

They hold exactly the same amount of space, and yet they are not the same to use. A long object fits in one and not another. A tall stack fits in one and not another. Two boxes with equal volumes can have very different edges.

Worked examples

Examples 1 and 2 go with Day 1, Examples 3 and 4 with Day 2, Examples 5 and 6 with Day 3, and Examples 7 and 8 with Day 4.

Example 1

A storage unit

Masumi’s storage unit is 10 feet long, 6 feet wide and 8 feet tall inside. What is its volume?

Solution

All three edges are in feet, so the volume will be in cubic feet. Multiply length, width and height:

V = 10 \times 6 \times 8

First 10 \times 6 = 60. That is the floor, 60 square feet, and also the number of foot cubes in one layer. The unit is 8 feet tall, so 8 layers fit:

60 \times 8 = 480

The volume is 480\ \text{ft}^3.

Example 2

An ice chest, checked by layers

Sanjay’s ice chest is 18 inches long, 10 inches wide and 12 inches tall inside. How many cubic inches does it hold? Say what each step counts.

Solution

The rule is V = l \times w \times h, with every edge in inches.

V = 18 \times 10 \times 12

The first product, 18 \times 10 = 180, counts the inch cubes in the bottom layer: 10 rows of 18. The second step counts the layers. The chest is 12 inches tall, so 12 layers of 180:

\begin{aligned}180 \times 12 &= 1{,}800 + 360 \\ &= 2{,}160\end{aligned}

The ice chest holds 2{,}160\ \text{in}^3.

Check. Multiply in a different order: 10 \times 12 = 120, and 120 \times 18 = 1{,}200 + 960 = 2{,}160. The same, as the associative and commutative properties promise.

Example 3

Floor area and height only

The listing for a storage unit says: floor 50 square feet, ceiling 9 feet high. Masumi does not know how long or wide the floor is. Can she still find the volume?

Solution

Yes. The volume of a box is the base area times the height, V = B \times h. The floor is the base, so B = 50 square feet, and the height is h = 9 feet:

V = 50 \times 9 = 450

The unit holds 450\ \text{ft}^3.

The floor could be 10 by 5, or 25 by 2, or something else with area 50. Each choice puts 50 foot cubes in a layer, and 9 layers stand on it. The length and width only matter through their product, so the area alone is enough.

Example 4

Choosing the easy base

Zacarias packs a crate 25 inches long, 13 inches wide and 4 inches tall. Find its volume without a long multiplication.

Solution

The volume is the product of the three edges, 25 \times 13 \times 4. By the associative and commutative properties, any two may be multiplied first. Pick the pair that makes a round number:

25 \times 4 = 100

In the picture, that is standing the crate on its 25 by 4 face. That face has area 100 square inches, and the edge standing up from it is 13 inches. So

V = 100 \times 13 = 1{,}300

The crate holds 1{,}300\ \text{in}^3.

Check. The plain order gives 25 \times 13 = 325, then 325 \times 4 = 1{,}300. The same, but with harder steps.

Example 5

How deep is the water?

Sanjay pours 3{,}600 cubic inches of water into an empty fish tank. The tank’s inside floor is 20 inches by 12 inches. How deep is the water?

Solution

The water fills the bottom of the tank as a box, so its volume is base area times height. The base is 20 \times 12 = 240 square inches, so

240 \times h = 3{,}600

Each inch of depth is a layer of 240 inch cubes of water, and there are 3{,}600 of them. The depth is how many 240s make 3{,}600:

h = 3{,}600 \div 240 = 15

The water is 15 inches deep.

Check. 20 \times 12 \times 15 = 240 \times 15 = 3{,}600, the volume poured in.

Example 6

A missing length in two steps

A storage unit holds 960 cubic feet. It is 8 feet tall and 10 feet wide. How long is it?

Solution

Work back one factor at a time. The volume is base area times height, so the base area is the volume divided by the height:

B = 960 \div 8 = 120 \text{ square feet}

That is the floor. The floor is a rectangle with area 120 and width 10, and area is length times width, so the length is the area divided by the width:

l = 120 \div 10 = 12

The unit is 12 feet long.

Check. 12 \times 10 \times 8 = 120 \times 8 = 960.

Example 7

How many containers?

Zacarias has 100 cubic meters of goods to ship. Each shipping container is 12 meters long, 2 meters wide and 2 meters tall inside. How many containers does he need? How much space is left over?

Solution

First the volume of one container. All edges are in meters, so the volume is in cubic meters:

V = 12 \times 2 \times 2 = 48\ \text{m}^3

Now ask how many 48s are in 100:

100 \div 48 = 2 \text{ R } 4

Two containers hold 2 \times 48 = 96\ \text{m}^3, which leaves 4\ \text{m}^3 of goods on the dock. Those goods still have to go, so he needs a third container. The answer is 3 containers.

The third container carries only 4\ \text{m}^3, so it has 48 - 4 = 44\ \text{m}^3 of empty space. Here the remainder decided the answer: leftover goods still need a whole container of their own, so the count rounds up.

Example 8

Three units with the same volume

Masumi is choosing between three storage units: A is 10 ft by 4 ft by 8 ft, B is 8 ft by 5 ft by 8 ft, and C is 16 ft by 4 ft by 5 ft, each given as length by width by height.

(a) Which holds the most?

(b) She has a 12-foot ladder that has to lie flat along the floor. Which units can take it?

Solution

(a) Find each volume with V = l \times w \times h.

\begin{aligned}\text{A:}\ 10 \times 4 \times 8 &= 40 \times 8 = 320 \\ \text{B:}\ 8 \times 5 \times 8 &= 40 \times 8 = 320 \\ \text{C:}\ 16 \times 4 \times 5 &= 64 \times 5 = 320\end{aligned}

They all hold 320\ \text{ft}^3. None holds more.

(b) The ladder lies along the length, so the length must be at least 12 feet. A is 10 feet and B is 8 feet, both too short. Only C, at 16 feet, takes the ladder.

(c) The stack needs a height of at least 7 feet. A and B are 8 feet tall, so they work. C is only 5 feet tall. Only A and B take the stack.

The same volume did not give the same answer to either practical question. Volume tells how much space there is. The edges tell what shape that space is.

Boundaries

Boxes only. V = l \times w \times h needs six rectangular faces with square corners, so that every layer is the same rectangle. A pyramid or a bowl does not have equal layers, and a layer shaped like a triangle or a circle is not l \times w cubes. Other straight-sided solids with the same shape all the way up use base area times height in Volume of prisms and cylinders.
One box at a time. A solid shaped like an L or a set of steps is not one box. Split it into boxes first, as in Volume of a solid made of boxes.
Whole-number edges in this lesson. The argument above packs whole cubes, so it needs whole numbers. The rule still holds when an edge is a fraction, and Volume of a box with fraction edge lengths shows why.
One unit for all three edges. A box 2 feet by 18 inches by 3 feet must be put into a single unit before multiplying. Mixing feet and inches counts no real cube.
Inside, for what it holds. The space a box holds comes from its inside edges. Its outside edges give the space it takes up, which is more when the walls are thick.
Volume alone does not give an edge. To find one missing edge you need the other two, or the area of the face they make. A volume of 320\ \text{ft}^3 by itself fits many different boxes, as Day 4 showed.

Where this goes next

The rule V = l \times w \times h opens several doors. Volume of a solid made of boxes splits a stepped solid into boxes and adds their volumes. Volume of a box with fraction edge lengths packs boxes with tiny fraction cubes and finds that the same rule still holds when an edge is 2\tfrac{1}{2}. Base area times height grows into Volume of prisms and cylinders, where the base can be a triangle or any other shape. Measuring the outside of a box instead of the inside leads to Surface area, and how it compares with volume, and comparing how much a box weighs with how much space it fills leads to Density.

Exercises

Each set goes with one day of the lesson. The last set asks you to explain.

Day 1: length times width times height

Find the volume of a box 8 inches long, 5 inches wide and 3 inches tall.

Masumi’s storage closet is 7 feet long, 4 feet wide and 9 feet tall. What is its volume?

A cube is 8 centimeters on each edge. What is its volume?

An ice chest is 20 inches long, 11 inches wide and 10 inches tall inside. How many cubic inches does it hold?

A box is 11 cm long, 5 cm wide and 6 cm tall. How many centimeter cubes are in one layer? How many layers are there? What is the volume?

Day 2: base area times height

A storage bin has a base with area 18 square feet and is 4 feet tall. What is its volume?

A shipping crate has a floor of 45 square feet and is 6 feet tall inside. What is its volume?

A box has edges 50 cm, 17 cm and 2 cm. Find its volume by choosing which two edges to multiply first. Name the properties that let you choose.

A box is 4 inches by 15 inches by 5 inches. Find its volume three times, using a different face as the base each time.

Sanjay’s ice chest has an inside floor 16 inches by 10 inches. He pours in ice to a depth of 4 inches. What volume of ice is that?

Day 3: finding a missing edge

A box holds 168 cubic inches. Its base is 7 inches by 6 inches. How tall is it?

A storage unit holds 750 cubic feet and is 10 feet tall. (a) What is the area of its floor? (b) The floor is 5 feet wide. How long is it?

A cube holds 343 cubic centimeters. How long is each edge?

An ice chest is 18 inches long and 12 inches wide inside, and it holds 2{,}592 cubic inches. How tall is it inside?

Zacarias’s shipping container is 40 feet long and 8 feet wide inside, and it holds 2{,}560 cubic feet. How tall is it inside?

Day 4: problems and comparisons

Masumi has 200 cubic feet of things to store. Unit A is 5 ft by 5 ft by 8 ft. Unit B is 6 ft by 4 ft by 8 ft. Which unit has enough space?

A container 6 m long, 2 m wide and 3 m tall inside is filled with sand to a depth of 2 m. What volume of sand is in it? How much empty space is above the sand?

List every box with whole-number edges, in feet, whose volume is 12\ \text{ft}^3. Count a box only once, whichever way it is turned.

Ice chest A is 20 in by 12 in by 12 in inside. Ice chest B is 24 in by 15 in by 8 in inside, each given as length by width by height. (a) Which holds more? (b) Which has the larger floor?

Masumi moves from a unit 4 ft by 6 ft by 9 ft to one 12 ft by 6 ft by 9 ft. How many times as much space does she have?

Reasoning

Explain why l \times w counts the cubes in one layer of a box, why h counts the layers, and so why V = l \times w \times h.

Sanjay says a box 2 ft by 3 ft by 4 ft and a box 4 ft by 2 ft by 3 ft hold different amounts, because the numbers come in a different order. Is he right? Explain.

Two boxes have the same volume. Must they have the same base area? Explain with an example.

Explain why dividing the volume by the base area gives the height.

A box is 6 ft by 2 ft by 3 ft. What happens to its volume if you double only its height? What if you double both its length and its height? Explain why.

Answers

120\ \text{in}^3. 8 \times 5 = 40, and 40 \times 3 = 120.

252\ \text{ft}^3. 7 \times 4 = 28 square feet of floor, and 28 \times 9 = 252.

512\ \text{cm}^3. 8 \times 8 = 64 cubes per layer, and 64 \times 8 = 512.

2{,}200\ \text{in}^3. 20 \times 11 = 220, and 220 \times 10 = 2{,}200.

55 cubes in a layer, 6 layers, 330\ \text{cm}^3. The floor is 11 \times 5 = 55 squares, the box is 6 cm tall, and 55 \times 6 = 330.

72\ \text{ft}^3. V = B \times h = 18 \times 4 = 72.

270\ \text{ft}^3. V = 45 \times 6 = 270.

1{,}700\ \text{cm}^3. Multiply 50 \times 2 = 100 first, then 100 \times 17 = 1{,}700. The associative property lets you group the factors as you like, and the commutative property lets you reorder them.

300\ \text{in}^3 each way. On the 4 by 15 face: 60 \times 5 = 300. On the 4 by 5 face: 20 \times 15 = 300. On the 15 by 5 face: 75 \times 4 = 300.

640\ \text{in}^3 of ice. The ice is a box 16 by 10 by 4: 160 \times 4 = 640.

4 inches tall. The base is 7 \times 6 = 42 square inches, and 168 \div 42 = 4.

(a) 75 square feet, since 750 \div 10 = 75. (b) 15 feet long, since 75 \div 5 = 15. Check: 15 \times 5 \times 10 = 750.

7 cm. The three edges of a cube are equal, so try numbers: 6 \times 6 \times 6 = 216 is too small, and 7 \times 7 \times 7 = 49 \times 7 = 343.

12 inches tall. The base is 18 \times 12 = 216 square inches, and 2{,}592 \div 216 = 12. In two easier steps: 2{,}592 \div 18 = 144, then 144 \div 12 = 12.

8 feet tall. The floor is 40 \times 8 = 320 square feet, and 2{,}560 \div 320 = 8.

Unit A. A holds 5 \times 5 \times 8 = 200\ \text{ft}^3, just enough. B holds 6 \times 4 \times 8 = 192\ \text{ft}^3, which is 8\ \text{ft}^3 short.

24\ \text{m}^3 of sand and 12\ \text{m}^3 of empty space. The sand is a box 6 \times 2 \times 2 = 24. The whole container is 6 \times 2 \times 3 = 36, and 36 - 24 = 12. Or: the empty part is 1 m deep, 6 \times 2 \times 1 = 12.

Four boxes: 1 by 1 by 12, 1 by 2 by 6, 1 by 3 by 4, and 2 by 2 by 3. Each product is 12. To find them all, choose the smallest edge first: it is 1 or 2, since 3 \times 3 \times 3 = 27 is already more than 12.

(a) They hold the same, 2{,}880\ \text{in}^3 each: 240 \times 12 = 2{,}880 and 360 \times 8 = 2{,}880. (b) B, with 24 \times 15 = 360 square inches against A’s 20 \times 12 = 240.

3 times as much. The first unit is 4 \times 6 \times 9 = 216\ \text{ft}^3 and the second is 12 \times 6 \times 9 = 648\ \text{ft}^3, and 648 \div 216 = 3. Only the length changed, and it is 3 times as long, so each layer is 3 times as big.

The floor is l units by w units, so it holds l \times w unit squares, one cube on each, and those cubes are one layer. Each layer is 1 unit thick and the box is h units tall, so h layers fit, all the same. Adding l \times w cubes h times is multiplying, so the box holds l \times w \times h cubes.

No. Both hold 24\ \text{ft}^3: 2 \times 3 \times 4 = 24 and 4 \times 2 \times 3 = 24. They are the same box turned a different way. By the commutative and associative properties, the order and grouping of the factors do not change the product.

No. A box 4 by 3 by 1 and a box 2 by 2 by 3 (length by width by height) both hold 12 cubic units. The first has a base of 4 \times 3 = 12 square units and a height of 1. The second has a base of 2 \times 2 = 4 square units and a height of 3. A bigger base is made up for by a smaller height, so equal volumes can sit on very different bases.

The volume is base area times height, V = B \times h. That says the height is the number of layers of B cubes that make V cubes. Finding how many groups of B make V is division, so h = V \div B.

Doubling the height doubles the volume: 36 becomes 72\ \text{ft}^3, because the box gets twice as many layers of the same size. Doubling the length and the height makes it 4 times as big, 144\ \text{ft}^3: each layer is twice as big, and there are twice as many layers, and 2 \times 2 = 4.