Free math worksheets

- 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: \(V=4\times3\times0.5=6\,\text{cm}^3\).

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

- For part a), use all three dimensions shown on the prism. - For part b), keep the other two dimensions unchanged while splitting only the \(2.5\,\text{cm}\) edge. - How should volumes of nonoverlapping pieces relate to the volume of the whole prism?

1. For a), \(V=2.5\times2\times3=15\,\text{cm}^3\). 2. For b), the \(2\,\text{cm}\)-long part has volume \(2\times2\times3=12\,\text{cm}^3\). 3. The \(0.5\,\text{cm}\)-long part has volume \(0.5\times2\times3=3\,\text{cm}^3\). 4. Their sum is \(12+3=15\,\text{cm}^3\), matching the volume of the whole prism because the two parts exactly partition it without overlap.

a) \(15\,\text{cm}^3\) b) \(12\,\text{cm}^3\) and \(3\,\text{cm}^3\); together they make \(15\,\text{cm}^3\).

- 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 decimal edge length with an equivalent fraction. - 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) \(\frac{9}{4}\,\text{ft}^3=2.25\,\text{ft}^3\)

- 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 a), \(V=2.5\times1.2\times3=9\,\text{cm}^3\). 2. For prism b), \(V=1.5\times2\times3=9\,\text{cm}^3\). 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 a): \(9\,\text{cm}^3\); prism b): \(9\,\text{cm}^3\) 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.

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