- Which three dimensions of a rectangular prism determine its volume? - Multiply the length, width, and height shown in the diagram. - Check that the final unit is cubic centimeters.

1. Multiply the three edge lengths: \(V=4\times3\times0.5=6\,\text{cm}^3\).

\(6\,\text{cm}^3\)

- Count every small cube in the model once. - The prism's volume is the sum of the equal small-cube volumes.

1. The model contains \(6\) small cubes. 2. Each cube has volume \(\frac{1}{8}\,\text{ft}^3\), so the prism's volume is \(6\times\frac{1}{8}=\frac{6}{8}=\frac{3}{4}\,\text{ft}^3\).

\(\frac{3}{4}\,\text{ft}^3\)

- Compare the two edge lengths, not just the picture sizes. - Count how many small cubes fit in one layer. - How many such layers fit through the full height?

1. Each edge of the large cube is twice as long as an edge of the small cube. 2. Two small cubes fit along the length, two along the width, and two along the height. 3. In all, \(2\times2\times2=8\) small cubes fill the large one.

\(2\) fit along each edge, so \(2\times2\times2=8\) small cubes fill cube a).

- Which operation combines length, width, and height into volume? - Multiply the two half-length factors together first.

1. Multiply the three dimensions: \(V=4\times\frac{1}{2}\times\frac{1}{2}=1\,\text{cm}^3\). 2. The half-centimeter width and height together make one-quarter of the corresponding \(4\)-by-\(1\)-by-\(1\) prism.

\(1\,\text{cm}^3\)

- Which measurements do you need to find the volume of a rectangular prism? - After the cut, which two dimensions remain unchanged? - Think about how the two smaller prisms fit together to make the original.

1. The original prism has volume \(V=2.5\times2\times3=15\,\text{cm}^3\). 2. The first smaller prism has dimensions \(2\,\text{cm}\), \(2\,\text{cm}\), and \(3\,\text{cm}\). Its volume is \(2\times2\times3=12\,\text{cm}^3\). 3. The second smaller prism has dimensions \(0.5\,\text{cm}\), \(2\,\text{cm}\), and \(3\,\text{cm}\). Its volume is \(0.5\times2\times3=3\,\text{cm}^3\). 4. The volumes add to \(12+3=15\,\text{cm}^3\). The two pieces fill the original prism without gaps or overlaps.

a) \(15\,\text{cm}^3\) b) \(12\,\text{cm}^3\) and \(3\,\text{cm}^3\). Their sum is \(15\,\text{cm}^3\). The volumes add up because the two pieces together make the original prism without gaps or overlaps.

- Use the rectangular-prism volume formula with all three fractional edge lengths. - Look for factors that can simplify before multiplying everything. - The result should be measured in cubic meters.

1. Multiply the edge lengths: \(V=\frac{3}{4}\times\frac{2}{3}\times\frac{1}{2}\). 2. Simplify: \(\frac{3}{4}\times\frac{2}{3}=\frac{1}{2}\), so \(V=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}\,\text{m}^3\).

\(\frac{1}{4}\,\text{m}^3\)

- Read all three edge lengths from the diagram before calculating. - For part b), replace the middle decimal edge length with the equivalent fraction supplied in the text. - Equivalent forms of the same measurement should not change the prism's volume.

1. For a), \(V=1.5\times0.75\times2=2.25\,\text{ft}^3\). 2. For b), \(1.5=\frac{3}{2}\) and \(0.75=\frac{3}{4}\). Then \(V=\frac{3}{2}\times\frac{3}{4}\times2=\frac{9}{4}=2.25\,\text{ft}^3\). 3. The decimal and fraction forms describe the same edge lengths, so they give the same volume.

a) \(2.25\,\text{ft}^3\) b) \(\frac{9}{4}\,\text{ft}^3=2.25\,\text{ft}^3\)

- Count the rows and cubes per row in just one horizontal layer first. - Then determine how many identical layers are stacked. - A cube's volume uses its edge length three times.

1. One horizontal layer has \(3\times2=6\) cubes. 2. There are \(2\) layers, so the prism contains \(6\times2=12\) cubes. 3. One small cube has volume \(\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4}=\frac{1}{64}\,\text{m}^3\). 4. The prism's volume is \(12\times\frac{1}{64}=\frac{12}{64}=\frac{3}{16}\,\text{m}^3\).

a) \(6\) cubes b) \(2\) layers and \(12\) cubes total c) \(\frac{3}{16}\,\text{m}^3\)

- Compare each prism dimension with the small cube's edge length. - The number of small cubes is a product of the counts in the three perpendicular directions. - Relate the volume of one small cube to the product of the three full prism dimensions.

1. Along the three dimensions, \(3\), \(2\), and \(1\) half-foot cubes fit, so there are \(3\times2\times1=6\) small cubes. 2. One half-foot cube has volume \(\left(\frac{1}{2}\right)^3=\frac{1}{8}\,\text{ft}^3\). Therefore the packed volume is \(6\times\frac{1}{8}=\frac{3}{4}\,\text{ft}^3\). 3. Multiplying the prism dimensions gives \(\frac{3}{2}\times1\times\frac{1}{2}=\frac{3}{4}\,\text{ft}^3\). 4. The products agree because the factors \(3\), \(2\), and \(1\) count cubes along the three directions, while each cube contributes \(\left(\frac{1}{2}\right)^3\) cubic foot.

a) \(3\), \(2\), and \(1\) cubes along the dimensions; \(6\) cubes total b) \(\frac{3}{4}\,\text{ft}^3\) c) \(\frac{3}{2}\times1\times\frac{1}{2}=\frac{3}{4}\,\text{ft}^3\); both methods count the same cubic measure.

- Use the two dimensions that lie on the bottom face to find \(B\). - After finding a square-unit base area, multiply it by the perpendicular prism height. - Replace \(B\) with the product that created the base area to compare the two volume forms.

1. The base dimensions are \(1.25\,\text{in.}\) and \(0.5\,\text{in.}\), so \(B=1.25\times0.5=0.625=\frac{5}{8}\,\text{in.}^2\). 2. The prism height is \(1.5\,\text{in.}=\frac{3}{2}\,\text{in.}\). Thus \(V=Bh=\frac{5}{8}\times\frac{3}{2}=\frac{15}{16}\,\text{in.}^3\). 3. Since \(B=lw\), substituting into \(V=Bh\) gives \(V=(lw)h=lwh\).

a) \(B=\frac{5}{8}\,\text{in.}^2\) b) \(V=\frac{15}{16}\,\text{in.}^3\) c) Because \(B=lw\), \(Bh=(lw)h=lwh\).

- Which two dimensions belong to the bottom face? - Once you know that face's area, what dimension is still missing for volume? - Keep square centimeters and cubic centimeters distinct.

1. The bottom face has area \(2\times1.5=3\,\text{cm}^2\). 2. The prism has height \(0.5\,\text{cm}\), so the volume is \(3\times0.5=1.5\,\text{cm}^3\). 3. A height of half a centimeter multiplies the base-area number by one-half; area and volume still have different units.

Bottom-face area: \(3\,\text{cm}^2\); volume: \(1.5\,\text{cm}^3\). The prism is \(0.5\,\text{cm}\) high, so the area is multiplied by \(0.5\,\text{cm}\). Area uses square centimeters; volume uses cubic centimeters.

- How can you find the volume of one of the small cubes? - How does a half-centimeter edge affect all three factors in the cube's volume? - How can you combine the volumes of equal, nonoverlapping cubes?

1. Each little cube has volume \(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}\,\text{cm}^3\). 2. The twelve cubes fill the box without gaps, giving \(12\times\frac{1}{8}=\frac{3}{2}=1.5\,\text{cm}^3\).

\(1.5\,\text{cm}^3\)

- Make the three measurements use the same length unit before multiplying. - Think about how the tank's base area relates to the amount it holds. - How can you find the part added by the extra half-decimeter without starting over?

1. Convert the length: \(40\,\text{cm}=4\,\text{dm}\). The tank is \(4\,\text{dm}\) by \(5\,\text{dm}\) by \(2.5\,\text{dm}\). 2. The full volume is \(4\times5\times2.5=50\,\text{dm}^3\), or \(50\,\text{L}\). 3. With height \(2\,\text{dm}\), the volume is \(4\times5\times2=40\,\text{dm}^3\), or \(40\,\text{L}\). 4. The added half-decimeter contributes \(4\times5\times0.5=10\,\text{dm}^3\), or \(10\,\text{L}\). This also equals \(50-40=10\,\text{L}\).

a) \(50\,\text{L}\) b) \(40\,\text{L}\); the extra half-decimeter adds \(10\,\text{L}\).

- First combine the two known edge lengths into a base area. - Work backward from volume by dividing by the known base area. - Check the missing edge by multiplying all three dimensions.

1. The area of the rectangular base is \(\frac{5}{4}\times\frac{3}{2}=\frac{15}{8}\,\text{ft}^2\). 2. The height is the volume divided by the base area: \(\frac{15}{16}\div\frac{15}{8}=\frac{15}{16}\times\frac{8}{15}=\frac{1}{2}\,\text{ft}\). 3. Check: \(\frac{5}{4}\times\frac{3}{2}\times\frac{1}{2}=\frac{15}{16}\,\text{ft}^3\).

\(\frac{1}{2}\,\text{ft}\)

- Read all three dimensions of each prism from the diagram. - Compare the products of the two dimensions that differ between the prisms. - What happens when equal base-area products are multiplied by the same third dimension?

1. For prism P, \(V=2.5\times1.2\times3=9\,\text{cm}^3\). 2. For prism Q, \(V=1.5\times2\times3=9\,\text{cm}^3\). 3. The prisms have equal volumes. The common \(3\,\text{cm}\) edge is unchanged, and the products of the other two edge lengths are equal: \(2.5\times1.2=1.5\times2=3\,\text{cm}^2\).

a) Prism P: \(9\,\text{cm}^3\); prism Q: \(9\,\text{cm}^3\) b) Yes. Their different edge-length pairs have the same product, so the full three-factor products are equal.

- What kind of measurement results from multiplying only two edge lengths? - How many dimensions are needed for the volume of a rectangular prism? - Check the unit of the final result: area units and volume units are different.

1. Jordan multiplied only two edge lengths, which gives the area of one rectangular face, not the volume of the prism. 2. Volume requires all three dimensions: \(V=1.25\times2\times3.5\). 3. Since \(1.25\times2=2.5\), the volume is \(2.5\times3.5=8.75\,\text{m}^3\).

Jordan found a face area instead of the volume. The correct volume is \(8.75\,\text{m}^3\).

- Convert the mixed-number dimensions to fractions before multiplying. - Compare the garden-bed volume with the amount of soil in one bag. - If the bag is sufficient, subtract the required volume from the bag's volume to find the remainder.

1. Convert the mixed numbers: \(2\frac{1}{2}=\frac{5}{2}\) and \(1\frac{1}{4}=\frac{5}{4}\). 2. The bed's volume is \(\frac{5}{2}\times\frac{5}{4}\times\frac{3}{4}=\frac{75}{32}=2\frac{11}{32}\,\text{ft}^3\). 3. The bag contains \(2\frac{1}{2}=\frac{80}{32}\,\text{ft}^3\), which is more than the bed requires. 4. The leftover soil is \(\frac{80}{32}-\frac{75}{32}=\frac{5}{32}\,\text{ft}^3\).

Yes. One bag is enough, with \(\frac{5}{32}\,\text{ft}^3\) of soil left over.

- What is the volume of one cube whose edge is only half a centimeter? - The number of cubes and the number of cubic centimeters are not automatically the same. - Multiply the cube count by the volume represented by one small cube.

1. Maya counted the cubes but treated each small cube as though it had volume \(1\,\text{cm}^3\). 2. One small cube actually has volume \(\frac{1}{2}\times\frac{1}{2}\times\frac{1}{2}=\frac{1}{8}\,\text{cm}^3\). 3. The prism's volume is \(10\times\frac{1}{8}=\frac{10}{8}=\frac{5}{4}\,\text{cm}^3\).

Maya's error is treating each fractional-edge cube as a \(1\,\text{cm}^3\) unit cube. The correct volume is \(\frac{5}{4}\,\text{cm}^3\).

- Compare each block dimension with the small cube's edge length. - For the volume method, find the volume of the whole block and one small cube separately. - Dividing total volume by one cube's volume should represent a count of cubes.

1. Along the three dimensions, \(4\), \(3\), and \(2\) quarter-meter cubes fit, so the count is \(4\times3\times2=24\) cubes. 2. The block's volume is \(1\times\frac{3}{4}\times\frac{1}{2}=\frac{3}{8}\,\text{m}^3\). 3. One small cube has volume \(\left(\frac{1}{4}\right)^3=\frac{1}{64}\,\text{m}^3\). 4. Dividing gives \(\frac{3}{8}\div\frac{1}{64}=\frac{3}{8}\times64=24\) cubes. 5. The methods agree because both determine how many nonoverlapping quarter-meter cubes exactly fill the same block.

\(24\) cubes. Counting gives \(4\times3\times2=24\), and the volume quotient also gives \(24\).

- Rewrite the mixed number as both a fraction and a decimal. - Can you express the decimal length as a simple fraction as well? - Check both products separately before comparing the answers.

1. Write \(1\frac{1}{4}=\frac{5}{4}=1.25\) and \(0.8=\frac{4}{5}\). 2. Fraction calculation: \(\frac{5}{4}\times\frac{4}{5}\times2=2\,\text{cm}^3\). 3. Decimal calculation: \(1.25\times0.8\times2=2\,\text{cm}^3\). Both forms describe the same dimensions.

Fractions: \(\frac{5}{4}\times\frac{4}{5}\times2=2\,\text{cm}^3\). Decimals: \(1.25\times0.8\times2=2\,\text{cm}^3\).

- Start with the area of the rectangular base. - What number multiplied by that base area would give the full volume? - Check the units of the missing measurement.

1. The base area is \(\frac{3}{4}\times\frac{2}{3}=\frac{1}{2}\,\text{ft}^2\). 2. Divide the volume by the base area: \(h=\frac{3}{8}\div\frac{1}{2}=\frac{3}{4}\,\text{ft}\). 3. Check: \(\frac{1}{2}\times\frac{3}{4}=\frac{3}{8}\,\text{ft}^3\).

\(\frac{3}{4}\,\text{ft}\)

- Can you decide whether a product is greater or less than a whole without knowing its exact value? - Look at the width and height together before including the length. - After your estimate, check whether the exact result agrees.

1. The width times height is \(\frac{3}{4}\times\frac{1}{2}=\frac{3}{8}<\frac{1}{2}\). Even after multiplying by the \(1\frac{1}{2}\,\text{ft}\) length, the volume is less than \(\frac{3}{4}\,\text{ft}^3\), so it is below \(1\,\text{ft}^3\). 2. Exactly, \(V=\frac{3}{2}\times\frac{3}{4}\times\frac{1}{2}=\frac{9}{16}\,\text{ft}^3\). 3. Since \(\frac{9}{16}<1\), the volume is less than one cubic foot.

Yes. The width times height is less than \(\frac{1}{2}\), so multiplying by \(1\frac{1}{2}\) gives less than \(\frac{3}{4}\,\text{ft}^3\), which is below \(1\,\text{ft}^3\). The exact volume is \(\frac{9}{16}\,\text{ft}^3\).

- Use the known volume and length to determine what the product of width and height must be. - List the positive quarter-meter pairs that have the required sum. - Test which candidate pair also has the required product, then use the width-height ordering condition.

1. The width-height product must be \(\frac{3}{4}\div\frac{3}{2}=\frac{1}{2}\,\text{m}^2\). 2. Positive multiples of \(\frac{1}{4}\,\text{m}\) that sum to \(\frac{3}{2}\,\text{m}\) give these unordered pairs: \(\frac{1}{4}\,\text{m}\) and \(\frac{5}{4}\,\text{m}\), \(\frac{1}{2}\,\text{m}\) and \(1\,\text{m}\), or \(\frac{3}{4}\,\text{m}\) and \(\frac{3}{4}\,\text{m}\). 3. Their products are \(\frac{5}{16}\,\text{m}^2\), \(\frac{1}{2}\,\text{m}^2\), and \(\frac{9}{16}\,\text{m}^2\), respectively. Only \(\frac{1}{2}\,\text{m}\times1\,\text{m}\) has the required product. 4. Since the width is greater than the height, the width is \(1\,\text{m}\) and the height is \(\frac{1}{2}\,\text{m}\).

Width: \(1\,\text{m}\) Height: \(\frac{1}{2}\,\text{m}\)