Free math worksheets

- Use the rectangular-prism volume formula for Package A. - A cube has three equal dimensions. - Compare the two volumes.

1. Package A's volume is \(60\times30\times20=36{,}000\,\text{cm}^3\). 2. Package B's volume is \(35\times35\times35=42{,}875\,\text{cm}^3\). 3. Since \(42{,}875>36{,}000\), Package B takes up more space.

Package A has volume \(36{,}000\,\text{cm}^3\). Package B has volume \(42{,}875\,\text{cm}^3\). Package B takes up more space.

- Read the length, width, and height of each box from the diagram. - First find how many cubes fit in one bottom layer. - Then multiply by the number of layers.

1. Box A measures \(3 \times 3 \times 4\), so its volume is \(3\times3\times4=36\) cubic units. 2. Box B measures \(5 \times 2 \times 2\), so its volume is \(5\times2\times2=20\) cubic units. 3. Box C measures \(2 \times 4 \times 3\), so its volume is \(2\times4\times3=24\) cubic units.

A: \(36\) unit cubes; \(36\) cubic units B: \(20\) unit cubes; \(20\) cubic units C: \(24\) unit cubes; \(24\) cubic units

- Count the cubes along the width, depth, and height of each box. - A multiplication equation is faster than counting every cube one at a time.

1. Box 1 is \(3\) cubes wide, \(3\) cubes deep, and \(2\) cubes high. A matching equation is \(3 \times 3 \times 2 = 18\). 2. Box 2 is \(4\) cubes wide, \(2\) cubes deep, and \(2\) cubes high. A matching equation is \(4 \times 2 \times 2 = 16\).

1) \(18\) unit cubes; \(3 \times 3 \times 2 = 18\) 2) \(16\) unit cubes; \(4 \times 2 \times 2 = 16\)

- Both solids are rectangular prisms. - Multiply the three dimensions for each box. - Compare the two volumes.

1. Box a) has volume \(8\times3\times3=72\,\text{cm}^3\). 2. Box b) has volume \(4\times6\times3=72\,\text{cm}^3\). 3. The two boxes have the same volume.

Yes. Both boxes have a volume of \(72\,\text{cm}^3\).

- Find each block's volume separately. - A cube has three equal dimensions. - Order the numerical results.

1. Block A's volume is \(4\times6\times5=120\,\text{cm}^3\). 2. Block B's volume is \(5\times5\times5=125\,\text{cm}^3\). 3. Block C's volume is \(8\times2\times8=128\,\text{cm}^3\). 4. Therefore, the order is A, B, C.

The order is \(A<B<C\).

- \(B\) represents the area of the chosen base face, not a side length or a perimeter. - Find the rectangular base area before using the volume formula.

1. The base is a rectangle, so its area is \(B=4\times3=12\,\text{cm}^2\). 2. The height is \(4\,\text{cm}\). Use \(V=B\times h\): \(V=12\times4=48\,\text{cm}^3\).

a) \(B=12\,\text{cm}^2\) b) \(48\,\text{cm}^3\)

- Identify the two dimensions that lie in the bottom face. - First find an area, not a volume. - Then multiply that base area by the perpendicular height.

1. The bottom base measures \(5\,\text{cm}\times3\,\text{cm}\), so \(B=5\times3=15\,\text{cm}^2\). 2. The height is \(4\,\text{cm}\), so \(V=B\times h=15\times4=60\,\text{cm}^3\).

a) \(B=15\,\text{cm}^2\) b) \(V=60\,\text{cm}^3\)

- Treat the base area as the amount in one layer of unit cubes. - Ask what happens to the unit when \(\text{ft}^2\) is multiplied by \(\text{ft}\).

1. Use \(V=B\times h\): \(V=18\times7=126\). 2. The base area measures square feet and the height contributes another factor of feet, so \(\text{ft}^2\times\text{ft}=\text{ft}^3\).

\(126\,\text{ft}^3\). The unit is cubic feet because an area in \(\text{ft}^2\) is multiplied by a length in feet.

- Use the volume formula for a rectangular prism. - Relate cubic centimeters to liters. - Subtract the container's capacity from the amount of juice.

1. The container's volume is \(20\times10\times5=1000\,\text{cm}^3\). 2. Since \(1000\,\text{cm}^3=1\,\text{L}\), the container holds \(1\,\text{L}\). 3. The amount that does not fit is \(1.5\,\text{L}-1\,\text{L}=0.5\,\text{L}\). 4. Since \(0.5\,\text{L}=500\,\text{mL}\), \(500\,\text{mL}\) would overflow.

The container holds \(1\,\text{L}\), so \(500\,\text{mL}\) of the \(1.5\,\text{L}\) would overflow.

- Use the volume formula for a rectangular prism. - Convert cubic centimeters to liters. - Add the mass of the empty aquarium to the mass of the water.

1. The aquarium's volume is \(50\times30\times40=60{,}000\,\text{cm}^3\). 2. Since \(1000\,\text{cm}^3=1\,\text{L}\), its volume is \(60\,\text{L}\). 3. The water has a mass of \(60\,\text{kg}\). 4. The total mass is \(60+12=72\,\text{kg}\).

a) The aquarium's volume is \(60\,\text{L}\). b) The filled aquarium has a total mass of \(72\,\text{kg}\).

- Convert centimeters to decimeters. - Use the volume formula for a cube. - Relate cubic decimeters to liters.

1. Convert the edge length: \(60\,\text{cm}=6\,\text{dm}\). 2. The aquarium's volume is \(6\times6\times6=216\,\text{dm}^3\). 3. Since \(1\,\text{dm}^3=1\,\text{L}\), the capacity is \(216\,\text{L}\).

The aquarium's capacity is \(216\,\text{L}\).

- Use the volume formula for a rectangular prism. - For part b, first find the actual water height. - Convert cubic centimeters to liters.

1. The full aquarium's volume is \(60\times30\times40=72{,}000\,\text{cm}^3\). 2. Since \(1000\,\text{cm}^3=1\,\text{L}\), the full capacity is \(72\,\text{L}\). 3. The current water height is \(40-5=35\,\text{cm}\). 4. The current water volume is \(60\times30\times35=63{,}000\,\text{cm}^3=63\,\text{L}\).

a) The aquarium can hold \(72\,\text{L}\). b) It currently contains \(63\,\text{L}\) of water.

- Choose one length unit for all three dimensions. - Using decimeters makes the final conversion to liters direct. - Multiply only after all dimensions use the same unit.

1. Convert all three lengths to decimeters: \(40\,\text{cm}=4\,\text{dm}\), \(5\,\text{dm}=5\,\text{dm}\), and \(250\,\text{mm}=2.5\,\text{dm}\). 2. Find the volume: \(V=4\times5\times2.5=50\,\text{dm}^3\). 3. Since \(1\,\text{dm}^3=1\,\text{L}\), the volume is \(50\,\text{L}\).

The volume is \(50\,\text{L}\).

- Picture filling the rectangular prism one layer at a time with unit cubes. - What does multiplying two edge lengths find first? - How many unit cubes fit on the chosen base? - How many copies of that layer fill the prism?

1. Using a \(12\,\text{cm}\) by \(5\,\text{cm}\) base, \(V=(12\times5)\times4=60\times4=240\,\text{cm}^3\). 2. Using a \(5\,\text{cm}\) by \(4\,\text{cm}\) base, \(V=12\times(5\times4)=12\times20=240\,\text{cm}^3\). 3. In the first calculation, \(12\times5=60\) is the number of unit cubes in one horizontal layer, and there are \(4\) layers. 4. In the second calculation, \(5\times4=20\) is the number of unit cubes in one vertical layer, and there are \(12\) such layers.

a) \(V=(12\times5)\times4=240\,\text{cm}^3\) \(V=12\times(5\times4)=240\,\text{cm}^3\) b) The product inside the parentheses gives the number of unit cubes in one layer based on the chosen face.

- How many cubes are in one layer? - How many times is that layer repeated? - When the prism is turned, which edge becomes its height? - Are any cubes added or removed when the prism is turned?

1. A row is \(7\,\text{cm}\) long, each layer has \(4\) rows, and there are \(5\) layers. The dimensions are \(7\,\text{cm}\), \(4\,\text{cm}\), and \(5\,\text{cm}\). 2. The volume is \(7\times4\times5=140\,\text{cm}^3\). 3. After the prism is turned, each layer contains \(7\times5=35\) cubes, and the prism has \(4\) layers. 4. The new layer calculation gives \(35\times4=140\,\text{cm}^3\). Turning the prism rearranges the same unit cubes, so its volume does not change.

a) The dimensions are \(7\,\text{cm}\), \(4\,\text{cm}\), and \(5\,\text{cm}\). The volume is \(140\,\text{cm}^3\). b) The statement is correct. In the new position, each layer has \(35\) cubes and there are \(4\) layers, so \(35\times4=140\,\text{cm}^3\).

- Find the number of cubes in one layer by multiplying the length and width. - Determine how many layers are stacked. - Think about what happens when the entire top layer is removed.

1. The original prism contains \(5\times2\times3=30\) cubes. 2. The shorter prism contains \(5\times2\times2=20\) cubes. 3. Jordan must remove \(30-20=10\) cubes.

a) The prism contains \(30\) cubes. b) Jordan must remove \(10\) cubes.

- Find the volume of Luke’s prism. - Find the volume of Sophie’s prism. - Compare the two results.

1. Luke’s prism has volume \(6\times5\times4=120\,\text{cm}^3\), so it contains \(120\) unit cubes. 2. Sophie’s prism has volume \(10\times3\times4=120\,\text{cm}^3\), so it also contains \(120\) unit cubes. 3. The volumes are equal, so neither person used more cubes.

They used the same number of cubes. Each prism has volume \(120\,\text{cm}^3\) and contains \(120\) unit cubes.

- First find the number of cubes in one layer. - The second model uses the same total number of cubes. - Use the relationship among length, width, height, and volume. - Find the missing factor in a multiplication equation with the total number of cubes.

1. Each layer contains \(4\times5=20\) cubes. 2. The first model contains \(20\times3=60\) cubes. 3. For the second model, the length and height account for \(6\times2=12\) cubes in each \(1\,\text{cm}\) of width. 4. The width is \(60\div12=5\,\text{cm}\).

a) \(60\) cubes b) \(5\,\text{cm}\)

- Write the volume formula with the known volume and base area. - Which factor is missing in \(20\times h=120\)? - The missing quantity is a length, so check the answer's unit.

1. Use \(V=B\times h\): \(120=20\times h\). 2. Divide the volume by the base area: \(h=120\div20=6\,\text{cm}\).

\(6\,\text{cm}\)

- For each orientation, identify the two dimensions in the chosen base face. - The height is the remaining dimension, perpendicular to that base. - Compare the two products \(B\times h\) after changing orientation.

1. With the \(6\times4\) face as base, \(B=24\,\text{cm}^2\) and \(h=5\,\text{cm}\). Thus \(V=24\times5=120\,\text{cm}^3\). 2. With the \(4\times5\) face as base, \(B=20\,\text{cm}^2\) and \(h=6\,\text{cm}\). Thus \(V=20\times6=120\,\text{cm}^3\). 3. Changing which face is called the base changes \(B\) and \(h\), but it does not add or remove any part of the prism, so the volume stays the same.

a) \(B=24\,\text{cm}^2\), \(h=5\,\text{cm}\), \(V=120\,\text{cm}^3\) b) \(B=20\,\text{cm}^2\), \(h=6\,\text{cm}\), \(V=120\,\text{cm}^3\)

- What kind of quantity must \(B\) be: a length or an area? - How is the area of a rectangle found from its two side lengths? - Check that the final unit is cubic centimeters.

1. Base area is found by multiplying the base dimensions, not adding them: \(B=8\times3=24\,\text{cm}^2\). 2. Use \(V=B\times h\): \(V=24\times5=120\,\text{cm}^3\). 3. Malik used a sum of side lengths instead of the area of the rectangular base.

Malik's error is using \(8+3\) for the base area. The correct base area is \(24\,\text{cm}^2\), and the volume is \(120\,\text{cm}^3\).

- First use the volume and height to work backward to the base area. - Then treat the base as a rectangle with one known side. - Keep area units and length units distinct in the two parts.

1. From \(V=B\times h\), \(B=324\div4=81\,\text{cm}^2\). 2. The base is a rectangle with area \(81\,\text{cm}^2\) and length \(9\,\text{cm}\), so its width is \(81\div9=9\,\text{cm}\).

a) \(B=81\,\text{cm}^2\) b) \(9\,\text{cm}\)

- Relate liters to cubic decimeters. - Look for three whole-number factors whose product is \(60\). - Convert each dimension separately from decimeters to centimeters.

1. Since \(1\,\text{L}=1\,\text{dm}^3\), the aquarium's volume is \(60\,\text{dm}^3\). 2. One possible set of factors is \(2\times3\times10=60\), so one set of dimensions is \(2\,\text{dm}\times3\,\text{dm}\times10\,\text{dm}\). 3. Another possible set is \(3\times4\times5=60\), so a second set of dimensions is \(3\,\text{dm}\times4\,\text{dm}\times5\,\text{dm}\). 4. Converting the first set using \(1\,\text{dm}=10\,\text{cm}\) gives \(20\,\text{cm}\times30\,\text{cm}\times100\,\text{cm}\).

a) Sample answers are \(2\,\text{dm}\times3\,\text{dm}\times10\,\text{dm}\) and \(3\,\text{dm}\times4\,\text{dm}\times5\,\text{dm}\). b) The first set is \(20\,\text{cm}\times30\,\text{cm}\times100\,\text{cm}\). Answers will vary.

- Find three whole numbers whose product is \(36\). - Think about how many cubes could be in one layer and how many layers would be needed. - Consider a single long row, a flat layer, or a prism with several layers.

1. For a rectangular prism, \(V=l\times w\times h\), so the three dimensions must have a product of \(36\). 2. Three possible factor triples are \(6\times3\times2=36\), \(9\times4\times1=36\), and \(4\times3\times3=36\). 3. Other whole-number factor triples are also possible.

Three possible sets of dimensions are: 1. \(6\,\text{cm}\times3\,\text{cm}\times2\,\text{cm}\) 2. \(9\,\text{cm}\times4\,\text{cm}\times1\,\text{cm}\) 3. \(4\,\text{cm}\times3\,\text{cm}\times3\,\text{cm}\)

- First find the number of cubes in one layer and the number of layers. - If the total number of cubes stays the same but the height decreases, what happens to the number in each layer? - For part c, find two whole numbers whose product is the answer to part b.

1. The original prism contains \(6\times4\times3=72\) cubes. 2. The new prism has \(2\) layers, so each layer contains \(72\div2=36\) cubes. 3. The new length and width must have a product of \(36\). One possible pair is \(6\,\text{cm}\) and \(6\,\text{cm}\), since \(6\times6=36\).

a) \(72\) unit cubes b) \(36\) cubes per layer c) One possible answer is \(6\,\text{cm}\) long and \(6\,\text{cm}\) wide.

- Picture how the cubes are arranged in each prism. - For part b, find how many cubes are in each \(1\,\text{cm}\)-long section. - A cube has equal length, width, and height. - Test the numbers of unit cubes needed for cubes with edge lengths \(2\,\text{cm}\) and \(3\,\text{cm}\).

1. A single row gives dimensions \(12\,\text{cm}\times1\,\text{cm}\times1\,\text{cm}\). 2. A cross section that is \(2\,\text{cm}\) wide and \(2\,\text{cm}\) high contains \(2\times2=4\) cubes for each centimeter of length. Therefore, the length is \(12\div4=3\,\text{cm}\). 3. A cube with edge length \(2\,\text{cm}\) needs \(2\times2\times2=8\) unit cubes, while a cube with edge length \(3\,\text{cm}\) needs \(3\times3\times3=27\). Because \(12\) is between \(8\) and \(27\), no whole-number edge length makes a cube from exactly \(12\) unit cubes.

a) Length \(12\,\text{cm}\), width \(1\,\text{cm}\), height \(1\,\text{cm}\) b) \(3\,\text{cm}\) c) No. A cube with edge length \(2\,\text{cm}\) uses \(8\) unit cubes, and a cube with edge length \(3\,\text{cm}\) uses \(27\) unit cubes.

- Find the bottom-face area of each pictured prism before comparing their heights. - Compare the factor change in \(B\) with the factor change in \(h\). - For the third prism, work backward from the common volume and the new base area. - Any whole-number rectangle with area \(30\,\text{cm}^2\) can serve as the requested base.

1. Prism A has base area \(B=3\times4=12\,\text{cm}^2\) and height \(10\,\text{cm}\), so \(V=12\times10=120\,\text{cm}^3\). 2. Prism B has base area \(B=6\times4=24\,\text{cm}^2\) and height \(5\,\text{cm}\), so \(V=24\times5=120\,\text{cm}^3\). 3. From A to B, the base area doubles from \(12\) to \(24\), while the height is halved from \(10\) to \(5\). These opposite factor changes leave \(B\times h\) unchanged. 4. For the third prism, \(120=30\times h\), so \(h=4\,\text{cm}\). One whole-number base is \(5\,\text{cm}\times6\,\text{cm}\).

a) Prism A: \(B=12\,\text{cm}^2\), \(h=10\,\text{cm}\), \(V=120\,\text{cm}^3\) Prism B: \(B=24\,\text{cm}^2\), \(h=5\,\text{cm}\), \(V=120\,\text{cm}^3\) b) The base area is doubled and the height is halved, so the product \(B\times h\) stays \(120\,\text{cm}^3\). c) \(h=4\,\text{cm}\); one possible base is \(5\,\text{cm}\times6\,\text{cm}\).