Free math worksheets

- Look at whether the rate is based on area or volume. - Match the denominator's dimension to the geometric quantity being measured.

1. Mass per unit area uses \(\text{kg/m}^2\), so that unit fits the floor covering. 2. Mass per unit volume uses \(\text{kg/m}^3\), so that unit fits the solid foam block.

Floor covering: \(\text{kg/m}^2\) Foam block: \(\text{kg/m}^3\)

- Area density tells you the mass for each square meter. - Multiply the number of square meters by the mass per square meter.

1. Multiply area by area density: \(m=(5)(0.6)=3\,\text{kg}\).

The mass is \(3\,\text{kg}\).

- Area density is mass divided by area. - Check that the denominator of the final unit is square meters.

1. Divide mass by area: \(\rho_A=\frac{2.4}{4}=0.6\,\text{kg/m}^2\).

The area density is \(0.6\,\text{kg/m}^2\).

- Find the volume of each solid. - Use \(m=\rho V\) to find mass from density and volume. - Compare the two masses.

1. Solid A's volume is \(12\cdot5\cdot10=600\,\text{cm}^3\). Its mass is \(600\cdot0.8=480\,\text{g}\). 2. Solid B's volume is \(8^3=512\,\text{cm}^3\). Its mass is \(512\cdot0.5=256\,\text{g}\). 3. Since \(480>256\), Solid A has the greater mass.

Solid A has the greater mass: \(480\,\text{g}\), compared with \(256\,\text{g}\) for Solid B.

- Determine the height from its relationship to the width before finding volume. - Use the rectangular-prism dimensions to calculate the bar's volume. - Apply \(m=\rho V\), then convert grams to kilograms.

1. The height is \(5-2=3\,\text{cm}\). 2. The bar's volume is \(V=10\cdot5\cdot3=150\,\text{cm}^3\). 3. Its mass is \(m=\rho V=19.3\cdot150=2895\,\text{g}\). 4. Convert to kilograms: \(2895\,\text{g}=2.895\,\text{kg}\).

The bar's mass is \(2.895\,\text{kg}\).

- Convert all dimensions to decimeters. - Find the rectangular-prism volume. - Multiply the volume by the mass per cubic decimeter.

1. Convert the dimensions to decimeters: \(1.2\,\text{dm}\), \(1.2\,\text{dm}\), and \(25\,\text{dm}\). 2. The volume is \(1.2\cdot1.2\cdot25=36\,\text{dm}^3\). 3. The mass is \(36\cdot0.8=28.8\,\text{kg}\).

a) The volume is \(36\,\text{dm}^3\). b) The beam has a mass of about \(28.8\,\text{kg}\).

- Treat the gravel layer as a rectangular prism. - Convert cubic centimeters to liters. - Multiply liters by the mass per liter.

1. The gravel volume is \(80\cdot40\cdot5=16{,}000\,\text{cm}^3=16\,\text{L}\). 2. Its mass is \(16\cdot1.6=25.6\,\text{kg}\).

a) \(16\,\text{L}\) of gravel are needed. b) The gravel has a mass of \(25.6\,\text{kg}\).

- Determine which geometric measurement matches the denominator in \(\text{kg/m}^2\). - Find the rectangular area before applying the density. - Check that square meters cancel from the units.

1. The panel's area is \(A=(2.4)(1.5)=3.6\,\text{m}^2\). 2. Since area density is mass divided by area, \(m=(0.80)(3.6)=2.88\,\text{kg}\).

\(2.88\,\text{kg}\)

- Read the two side lengths from the diagram and find the panel's area. - Area density compares mass with two-dimensional area, not with volume. - Divide the total mass by the total area and keep the units attached.

1. The displayed dimensions give area \(A=(5)(3)=15\,\text{ft}^2\). 2. Area density is mass divided by area, so \(\frac{30\,\text{lb}}{15\,\text{ft}^2}=2\,\text{lb/ft}^2\).

\(2\,\text{lb/ft}^2\)

- Inspect the units in the density before deciding whether to calculate area or volume. - Convert the thickness to meters so all three dimensions use compatible units. - Use unit cancellation to check whether your calculation ends in kilograms.

1. The given density is mass per unit volume, so multiplying it by an area leaves units of \(\text{kg/m}\), not kilograms. Maya has mixed an area measurement with a volume density. 2. Convert the thickness: \(5\,\text{cm}=0.05\,\text{m}\). 3. The board's volume is \(V=(2)(1)(0.05)=0.10\,\text{m}^3\). 4. The correct mass is \(m=(30)(0.10)=3\,\text{kg}\).

Maya used area with a volume density, so the units do not produce mass. The correct mass is \(3\,\text{kg}\).

- The copper's volume stays constant when it is reshaped. - Convert the block's volume to cubic millimeters. - Divide volume by cross-sectional area to find length.

1. The block's volume is \(12\cdot10\cdot6=720\,\text{cm}^3\). 2. Convert the volume: \(720\,\text{cm}^3=720{,}000\,\text{mm}^3\). 3. The wire's cross-sectional area is \(2\cdot2=4\,\text{mm}^2\). 4. Since volume is preserved, the wire's length is \(720{,}000\div4=180{,}000\,\text{mm}=180\,\text{m}\).

The wire is \(180\,\text{m}\) long.

- How do you find the volume of a rectangular prism? - What does using the same material tell you about mass and volume? - Find the mass of one cubic centimeter of the wood. - You can also compare the two volumes using a scale factor.

1. The first prism has volume \(V_1=10\cdot5\cdot4\,\text{cm}^3=200\,\text{cm}^3\). 2. The oak's density is \(\rho=\frac{140\,\text{g}}{200\,\text{cm}^3}=0.7\,\text{g/cm}^3\). 3. The second prism has volume \(V_2=15\cdot8\cdot5\,\text{cm}^3=600\,\text{cm}^3\). 4. Its mass is \(m_2=0.7\,\text{g/cm}^3\cdot600\,\text{cm}^3=420\,\text{g}\).

\(420\,\text{g}\)

- Use the cylinder volume formula and compare how each dimension changes. - What happens to the radius when the diameter is halved? - How does halving the radius affect the squared factor in the volume formula? - You can compare volumes without calculating the wax's density explicitly.

1. Candle \(1\) has \(r_1=3\,\text{cm}\) and \(h_1=10\,\text{cm}\). Candle \(2\) has \(r_2=1.5\,\text{cm}\) and \(h_2=20\,\text{cm}\). 2. The first volume is \(V_1=\pi\cdot3^2\cdot10\,\text{cm}^3=90\pi\,\text{cm}^3\). 3. The second volume is \(V_2=\pi\cdot1.5^2\cdot20\,\text{cm}^3=45\pi\,\text{cm}^3\). 4. Thus, \(V_2=\frac{1}{2}V_1\). Because the candles use the same wax, mass is proportional to volume. 5. Therefore, \(m_2=\frac{1}{2}\cdot250\,\text{g}=125\,\text{g}\).

\(125\,\text{g}\)

- Start with the rectangular-prism volume of the slab. - Rearrange the density formula for mass before substituting. - After converting pounds to short tons, decide how a non-whole number of truckloads affects the minimum trip count.

1. The slab volume is \(V=15\cdot12\cdot0.5=90\,\text{ft}^3\). 2. Rearrange the density formula to \(m=\rho V\). Then \(m=150\cdot90=13{,}500\,\text{lb}\). 3. Since \(2000\,\text{lb}=1\) short ton, the mass is \(6.75\) short tons. 4. The number of truckloads is \(6.75\div1.5=4.5\). Since a partial trip cannot carry the remaining concrete, at least \(5\) trips are needed.

a) \(90\,\text{ft}^3\) b) \(6.75\) short tons c) \(5\) trips

- Convert all measurements to a consistent system of centimeters and grams. - How are mass, volume, and density related? - Model the wire as a very long cylinder. - How can cylinder volume be used to find length when the cross-sectional area is known?

1. Convert the mass: \(5.2\,\text{kg}=5200\,\text{g}\). 2. The wire's volume is \(V=\frac{m}{\rho}=\frac{5200}{8.9}\,\text{cm}^3\). 3. The radius is \(0.4\,\text{mm}=0.04\,\text{cm}\), so the cross-sectional area is \(A=\pi(0.04)^2\,\text{cm}^2\). 4. Since \(V=AL\), \(L=\frac{V}{A}\approx116{,}236.76\,\text{cm}\). 5. Therefore, \(L\approx1162.4\,\text{m}\).

About \(1162.4\,\text{m}\)

- Use mass and density to find the bolt's volume first. - How can volume and length be used to find the circular base area? - How do you find a circle's diameter from its area?

1. The bolt's volume is \(V=\frac{850\,\text{g}}{7.85\,\text{g/cm}^3}\approx108.2803\,\text{cm}^3\). 2. Its circular base area is \(B=\frac{V}{h}\approx9.0234\,\text{cm}^2\). 3. The radius is \(r=\sqrt{\frac{B}{\pi}}\approx1.6948\,\text{cm}\). 4. The diameter is \(d=2r\approx3.3895\,\text{cm}=33.8953\,\text{mm}\), which rounds to \(33.9\,\text{mm}\).

About \(33.9\,\text{mm}\)

- Model the pipe as a large cylinder with a smaller cylinder removed. - Find the inside radius from the outside radius and wall thickness. - Which length unit matches density measured in \(\text{kg/dm}^3\)?

1. Use decimeters. The outside radius is \(R=30\,\text{cm}=3\,\text{dm}\), the inside radius is \(r=30\,\text{cm}-8\,\text{cm}=22\,\text{cm}=2.2\,\text{dm}\), and the length is \(25\,\text{dm}\). 2. The concrete volume is the difference of two cylinder volumes: \(V=\pi(R^2-r^2)h\). 3. Thus, \(V=\pi(3^2-2.2^2)\cdot25\,\text{dm}^3=104\pi\,\text{dm}^3\). 4. The mass is \(m=V\rho=104\pi\cdot2.4\,\text{kg}=249.6\pi\,\text{kg}\approx784.14\,\text{kg}\).

About \(784.14\,\text{kg}\)

- Which quantity relates mass and volume? - Write the cylinder volume formula. - Solve the formula for height.

1. The volume is \(V=\frac{m}{\rho}=\frac{1215\,\text{g}}{2.7\,\text{g/cm}^3}=450\,\text{cm}^3\). 2. From \(V=\pi r^2h\), \(h=\frac{V}{\pi r^2}=\frac{450}{9\pi}\,\text{cm}=\frac{50}{\pi}\,\text{cm}\approx15.92\,\text{cm}\).

a) \(450\,\text{cm}^3\) b) \(h\approx15.92\,\text{cm}\)

- Does the liquid volume change when it is poured? - Find the area of the right-triangle base. - Use \(V=Bh\) for the prism. - Reuse the exact volume from part a in part b.

1. The oil volume is \(V=\pi\cdot4^2\cdot10\,\text{cm}^3=160\pi\,\text{cm}^3\). 2. Its mass is \(m=V\rho=160\pi\cdot0.92\,\text{g}=147.2\pi\,\text{g}\approx462.4\,\text{g}\). 3. The triangular base area of the prism is \(B=\frac{1}{2}\cdot8\cdot10\,\text{cm}^2=40\,\text{cm}^2\). 4. The volume stays constant, so the oil depth is \(h=\frac{160\pi\,\text{cm}^3}{40\,\text{cm}^2}=4\pi\,\text{cm}\approx12.57\,\text{cm}\).

a) About \(462.4\,\text{g}\) b) About \(12.57\,\text{cm}\)

- Convert the mass to grams so the density units are compatible. - Use the density relationship to find volume. - Solve the sphere volume formula for the radius. - Double the radius to obtain the diameter.

1. Convert the mass: \(450\,\text{kg}=450{,}000\,\text{g}\). 2. Find the volume from density: \(V=\frac{m}{\rho}=\frac{450{,}000}{8.7}\approx51{,}724.1379\,\text{cm}^3\). 3. Solve the sphere volume formula for radius: \(r=\sqrt[3]{\frac{3V}{4\pi}}\approx23.1136\,\text{cm}\). 4. The diameter is \(d=2r\approx46.2273\,\text{cm}\), which rounds to \(46.2\,\text{cm}\).

The bronze sphere's diameter is approximately \(46.2\,\text{cm}\).

- The pipe's cross section is an annulus. - Use the difference-of-squares identity. - Convert every length measurement to centimeters before calculating mass. - Use \(m=\rho V\). - Identify how the length appears in the volume formula.

1. Subtract the inner cylinder volume from the outer cylinder volume: \(V=\frac{\pi D^2}{4}L-\frac{\pi d^2}{4}L=\frac{\pi L}{4}(D^2-d^2)\). 2. Factor the difference of squares: \(D^2-d^2=(D-d)(D+d)\). Thus, \(V=\frac{\pi L}{4}(D-d)(D+d)\). 3. Convert the length: \(2.5\,\text{m}=250\,\text{cm}\). 4. Using \(\pi\approx3.14\), the volume is \(V=\frac{3.14\cdot250}{4}(12^2-10^2)=8635\,\text{cm}^3\). 5. The mass is \(m=V\rho=8635\cdot7.85\,\text{g}=67{,}784.75\,\text{g}=67.78475\,\text{kg}\approx67.78\,\text{kg}\). 6. Volume is directly proportional to length, so doubling the length doubles both the volume and the mass.

a) \(V=\frac{\pi L}{4}(D^2-d^2)\) b) \(V=\frac{\pi L}{4}(D-d)(D+d)\) c) \(m\approx67.78\,\text{kg}\) d) The mass doubles.

- Convert cubic centimeters to liters after finding the volume. - Use \(m=\rho V\). - Compare the outside and inside radii to interpret \(R-r\). - Volume scales by the cube of a linear scale factor.

1. Using \(\pi\approx3.14\), the annular cross-sectional area is \(A=3.14\cdot(50^2-40^2)=2826\,\text{cm}^2\). 2. The concrete volume is \(V=2826\cdot80=226{,}080\,\text{cm}^3\). 3. Since \(1000\,\text{cm}^3=1\,\text{L}\), the volume is \(226.08\,\text{L}\). 4. The mass is \(m=226.08\cdot2.4\,\text{kg}=542.592\,\text{kg}\approx542.59\,\text{kg}\). 5. The difference \(R-r\) is the wall thickness. 6. Doubling every linear dimension multiplies volume by \(2^3=8\).

a) \(226.08\,\text{L}\) b) \(542.59\,\text{kg}\) c) \(R-r\) is the wall thickness. d) The volume is multiplied by \(8\).

- Substitute each outside diameter into the given formula. - Common factors cancel when you form a ratio. - For the same material, mass is proportional to volume.

1. For the first pipe, \(V_1=\pi L\cdot2\cdot(20-2)=36\pi L\,\text{mm}^3\). 2. For the second pipe, \(V_2=\pi L\cdot2\cdot(40-2)=76\pi L\,\text{mm}^3\). 3. Because the pipes are made of the same material, their masses are proportional to their metal volumes. 4. The mass ratio is \(\frac{V_2}{V_1}=\frac{76\pi L}{36\pi L}=\frac{19}{9}\approx2.11\). 5. Therefore, the second pipe is about \(2.11\) times as heavy, not exactly twice as heavy.

a) \(V_1=36\pi L\,\text{mm}^3\) and \(V_2=76\pi L\,\text{mm}^3\) b) No. The second pipe is \(\frac{19}{9}\approx2.11\) times as heavy.

- Convert all lengths to centimeters. - Change the diameter to a radius. - Find the cylinder's volume, then use \(m=\rho V\). - Convert grams to kilograms at the end.

1. The radius is \(2\,\text{cm}\), and the length is \(250\,\text{mm}=25\,\text{cm}\). 2. The rod's volume is \(V=\pi r^2h=\pi\cdot2^2\cdot25=100\pi\,\text{cm}^3\). 3. Its mass is \(m=\rho V=2.7\cdot100\pi\,\text{g}=270\pi\,\text{g}\approx848.23\,\text{g}\). 4. This is approximately \(0.84823\,\text{kg}\), which rounds to \(0.85\,\text{kg}\).

About \(0.85\,\text{kg}\)

- Convert the measurements to decimeters because the density uses cubic decimeters. - Find the area of the trapezoidal base. - Multiply the base area by the prism length. - Use \(m=\rho V\).

1. Convert the trapezoid dimensions to decimeters: \(6\,\text{dm}\), \(4\,\text{dm}\), and \(5\,\text{dm}\). 2. The base area is \(B=\frac{6+4}{2}\cdot5=25\,\text{dm}^2\). 3. The prism length is \(2.5\,\text{m}=25\,\text{dm}\), so the volume is \(V=25\cdot25=625\,\text{dm}^3\). 4. The mass is \(m=\rho V=2.5\cdot625=1562.5\,\text{kg}\).

The piece has a mass of \(1562.5\,\text{kg}\).

- Identify the two familiar solids that make up the silo. - Determine which measurements each volume formula requires. - Convert diameters to radii before substituting. - Add the component volumes.

1. The cylinder has radius \(2.00\,\text{m}\), so \(V_{\text{cylinder}}=\pi\cdot2.00^2\cdot5.50=22\pi\,\text{m}^3\). 2. The hopper has radii \(2.00\,\text{m}\) and \(0.30\,\text{m}\). Its volume is \(V_{\text{hopper}}=\frac{1}{3}\pi\cdot1.80\left(2.00^2+2.00\cdot0.30+0.30^2\right)=2.814\pi\,\text{m}^3\). 3. The total capacity is \(V_{\text{total}}=(22+2.814)\pi=24.814\pi\,\text{m}^3\approx77.96\,\text{m}^3\).

The silo's total capacity is approximately \(77.96\,\text{m}^3\).

- Convert each diameter to a radius before using a volume formula. - Recall how many cubic centimeters are in one liter. - For percent error, compare the absolute difference with the exact value.

1. The radii are \(9\,\text{cm}\) and \(6\,\text{cm}\). 2. The exact volume is \(V=\frac{1}{3}\pi\cdot32\left(9^2+9\cdot6+6^2\right)=1{,}824\pi\,\text{cm}^3\approx5{,}730.27\,\text{cm}^3\). 3. This is approximately \(5.73\,\text{L}\). 4. The average diameter is \(15\,\text{cm}\), so the estimated cylinder radius is \(7.5\,\text{cm}\). The estimate is \(V_{\text{estimate}}=\pi\cdot7.5^2\cdot32=1{,}800\pi\,\text{cm}^3\approx5{,}654.87\,\text{cm}^3\). 5. The percent error is \(\frac{1{,}824\pi-1{,}800\pi}{1{,}824\pi}\cdot100\%\approx1.32\%\). The estimate is smaller than the exact volume.

a) \(V=1{,}824\pi\,\text{cm}^3\approx5{,}730.27\,\text{cm}^3\) b) The vase holds approximately \(5.73\,\text{L}\). c) The estimate is \(1{,}800\pi\,\text{cm}^3\approx5{,}654.87\,\text{cm}^3\), which is approximately \(1.32\%\) less than the exact volume.

- Convert all measurements to compatible units before calculating. - Identify the solid that models the planter. - Find the two square areas from their side lengths. - Solve the frustum volume formula for height.

1. Convert the capacity: \(40\,\text{L}=40{,}000\,\text{cm}^3\). 2. The bottom and top base areas are \(B_1=30^2=900\,\text{cm}^2\) and \(B_2=45^2=2{,}025\,\text{cm}^2\). 3. Use \(V=\frac{h}{3}\left(B_1+\sqrt{B_1B_2}+B_2\right)\): \(40{,}000=\frac{h}{3}\left(900+\sqrt{900\cdot2{,}025}+2{,}025\right)\). 4. Since \(\sqrt{900\cdot2{,}025}=1{,}350\), \(40{,}000=\frac{h}{3}\cdot4{,}275\). 5. Therefore, \(h=\frac{3\cdot40{,}000}{4{,}275}\approx28.07\,\text{cm}\).

The planter needs an interior height of approximately \(28.07\,\text{cm}\).

- Decompose the silo into its two basic solids. - Read the shared radius and each height from the panels. - Find the cone's slant height from its radius and perpendicular height. - Include only the cylinder's lateral area and the cone's lateral area.

1. The panels show a common radius \(r=3\,\text{m}\), cylinder height \(8\,\text{m}\), and cone height \(4\,\text{m}\). 2. The cylinder volume is \(V_{\text{cylinder}}=\pi\cdot3^2\cdot8=72\pi\,\text{m}^3\). 3. The cone volume is \(V_{\text{cone}}=\frac{1}{3}\pi\cdot3^2\cdot4=12\pi\,\text{m}^3\). Thus, the total volume is \(84\pi\,\text{m}^3\approx263.89\,\text{m}^3\). 4. The cylinder's lateral area is \(A_{\text{cylinder}}=2\pi\cdot3\cdot8=48\pi\,\text{m}^2\). 5. The cone's slant height is \(\ell=\sqrt{3^2+4^2}=5\,\text{m}\), so its lateral area is \(A_{\text{cone}}=\pi\cdot3\cdot5=15\pi\,\text{m}^2\). 6. The total painted area is \(48\pi+15\pi=63\pi\,\text{m}^2\approx197.92\,\text{m}^2\).

a) The silo's total volume is approximately \(263.89\,\text{m}^3\). b) The total area to be painted is approximately \(197.92\,\text{m}^2\).

- Decompose the L-shaped cross section into rectangles. - Multiply the cross-sectional area by the prism length shown in the diagram. - Use \(m=\rho V\). - Convert grams to kilograms.

1. Split the L-shaped cross section into a \(2\,\text{cm}\times8\,\text{cm}\) rectangle and a \(4\,\text{cm}\times2\,\text{cm}\) rectangle. 2. The cross-sectional area is \(B=2\cdot8+4\cdot2=24\,\text{cm}^2\). 3. The volume is \(V=24\cdot15=360\,\text{cm}^3\). 4. The mass is \(m=\rho V=8.4\cdot360=3024\,\text{g}=3.024\,\text{kg}\).

The volume is \(360\,\text{cm}^3\), and the mass is \(3.024\,\text{kg}\).

- Decompose the L-shaped cross section into rectangles. - Use the cross-sectional area and the prism length shown to find volume. - Use \(m=\rho V\). - For total surface area, use the cross-sectional perimeter and include both ends.

1. The L-shaped cross-sectional area is the sum of a \(5\,\text{dm}\times1\,\text{dm}\) rectangle and a \(1\,\text{dm}\times7\,\text{dm}\) rectangle: \(B=5+7=12\,\text{dm}^2\). Thus, \(V=12\cdot12=144\,\text{dm}^3\). 2. The mass is \(m=2.5\cdot144=360\,\text{kg}\). 3. The cross-sectional perimeter is \(5+1+4+7+1+8=26\,\text{dm}\). Therefore, \(SA=2B+PL=2\cdot12+26\cdot12=336\,\text{dm}^2\).

a) \(144\,\text{dm}^3\) b) \(360\,\text{kg}\) c) \(336\,\text{dm}^2\)

- Divide the cross-section into two rectangles. - Convert the beam length to centimeters before finding volume. - Convert cubic centimeters to cubic decimeters before using the density.

1. The top rectangle has area \(20\cdot5=100\,\text{cm}^2\), and the stem has area \(6\cdot15=90\,\text{cm}^2\). The cross-sectional area is \(190\,\text{cm}^2\). 2. Convert the length: \(3\,\text{m}=300\,\text{cm}\). The volume is \(190\cdot300=57{,}000\,\text{cm}^3\). 3. Since \(1000\,\text{cm}^3=1\,\text{dm}^3\), the volume is \(57\,\text{dm}^3\). 4. The mass is \(57\cdot2.4=136.8\,\text{kg}\).

a) The cross-sectional area is \(190\,\text{cm}^2\). b) The volume is \(57{,}000\,\text{cm}^3\). c) The mass is \(136.8\,\text{kg}\).

- Divide the cross-section into two flanges and one web. - Convert the beam length to centimeters. - Use density to find mass in grams, then convert to kilograms.

1. The two flanges have total area \(2(15\cdot3)=90\,\text{cm}^2\), and the web has area \(3\cdot10=30\,\text{cm}^2\). The cross-sectional area is \(120\,\text{cm}^2\). 2. Convert the length: \(5\,\text{m}=500\,\text{cm}\). The volume is \(120\cdot500=60{,}000\,\text{cm}^3\). 3. The mass is \(60{,}000\cdot7.85=471{,}000\,\text{g}=471\,\text{kg}\).

a) The cross-sectional area is \(120\,\text{cm}^2\). b) The volume is \(60{,}000\,\text{cm}^3\). c) The mass is \(471\,\text{kg}\).

- Convert every length to decimeters first. - Try both adding two prism volumes and subtracting a missing prism from a larger prism. - Multiply volume by density to find mass.

1. In decimeters, the base measures \(5\,\text{dm}\times3\,\text{dm}\times1\,\text{dm}\), and the upper section measures \(2\,\text{dm}\times3\,\text{dm}\times2\,\text{dm}\). 2. By decomposition, \(V=5\cdot3\cdot1+2\cdot3\cdot2=15+12=27\,\text{dm}^3\). 3. By subtraction, the enclosing prism has volume \(5\cdot3\cdot3=45\,\text{dm}^3\), and the missing prism has volume \(3\cdot3\cdot2=18\,\text{dm}^3\). Thus \(V=45-18=27\,\text{dm}^3\). 4. The mass is \(27\cdot2.5=67.5\,\text{kg}\).

a) The volume is \(27\,\text{dm}^3\). b) The mass is \(67.5\,\text{kg}\).

- Convert all dimensions to decimeters. - Subtract the opening's volume from the outer prism's volume. - Multiply the frame volume by the mass per cubic decimeter.

1. In decimeters, the outer prism measures \(3\,\text{dm}\times3\,\text{dm}\times1\,\text{dm}\), and the opening measures \(1\,\text{dm}\times1\,\text{dm}\times1\,\text{dm}\). 2. The frame's volume is \(3\cdot3\cdot1-1\cdot1\cdot1=9-1=8\,\text{dm}^3\). 3. The mass is \(8\cdot7.8=62.4\,\text{kg}\).

a) The volume is \(8\,\text{dm}^3\). b) The mass is \(62.4\,\text{kg}\).

- Subtract the rectangular opening from the outer rectangle. - Multiply the cross-sectional area by the prism length shown in the diagram. - Multiply volume by density, then convert grams to kilograms.

1. The outer rectangle has area \(40\cdot30=1200\,\text{cm}^2\). 2. The opening is \(40-2(10)=20\,\text{cm}\) wide and \(30-10=20\,\text{cm}\) high, so its area is \(400\,\text{cm}^2\). 3. The concrete cross-sectional area is \(1200-400=800\,\text{cm}^2\). 4. The volume is \(800\cdot100=80{,}000\,\text{cm}^3\). 5. The mass is \(80{,}000\cdot2.4=192{,}000\,\text{g}=192\,\text{kg}\).

The section has mass \(192\,\text{kg}\).

- Calculate the volumes of the component solids separately. - Account for both pyramids. - Add the volumes before using the density.

1. The rectangular prism's volume is \(V_1=4\cdot4\cdot2=32\,\text{cm}^3\). 2. One pyramid has volume \(V_2=\frac{1}{3}\cdot4\cdot4\cdot3=16\,\text{cm}^3\). 3. The total volume is \(V_{\text{total}}=32+2\cdot16=64\,\text{cm}^3\). 4. The mass is \(m=64\cdot8.4=537.6\,\text{g}\).

The paperweight's mass is \(537.6\,\text{g}\).

- Decompose the workpiece into two rectangular prisms. - Use the relationship among volume, density, and mass. - From above, determine which horizontal regions are exposed. - Include the lateral faces of both prisms, but exclude the bottom face.

1. The component volumes are \(V_1=10\cdot10\cdot4=400\,\text{cm}^3\) and \(V_2=6\cdot6\cdot8=288\,\text{cm}^3\). Thus, \(V=400+288=688\,\text{cm}^3\). 2. The mass is \(m=688\cdot2.7=1857.6\,\text{g}\). 3. The lower prism's lateral area is \(4\cdot10\cdot4=160\,\text{cm}^2\). The visible portion of its top is \(10^2-6^2=64\,\text{cm}^2\). 4. The upper prism's lateral area is \(4\cdot6\cdot8=192\,\text{cm}^2\), and its top area is \(6^2=36\,\text{cm}^2\). 5. Therefore, \(SA=160+64+192+36=452\,\text{cm}^2\).

a) \(V=688\,\text{cm}^3\) b) \(m=1857.6\,\text{g}\) c) \(SA=452\,\text{cm}^2\)

- Model the drilled hole as a cylinder removed from the plate. - Determine the cylinder's radius and height. - Subtract the removed volume from the original plate volume.

1. The undrilled plate has volume \(V_{\text{plate}}=15\cdot15\cdot4=900\,\text{cm}^3\). 2. The hole is a cylinder with radius \(3\,\text{cm}\) and height \(4\,\text{cm}\), so \(V_{\text{hole}}=\pi\cdot3^2\cdot4=36\pi\,\text{cm}^3\). 3. The remaining volume is \(V_{\text{remaining}}=(900-36\pi)\,\text{cm}^3\approx786.90\,\text{cm}^3\).

The remaining metal volume is approximately \(786.90\,\text{cm}^3\).

- In panel a), subtract the hollow area from the outer square’s area. - Use the common height to relate volume and cross-sectional area. - Select the positive solution for a side length.

1. Because the columns have equal heights and equal material volumes, their material cross-sectional areas are equal. 2. Thus, \(10^2-x^2=8^2\). Therefore, \(100-x^2=64\), so \(x^2=36\). 3. Since a length is positive, \(x=6\,\text{cm}\). 4. The material volume is \(8^2\cdot15=960\,\text{cm}^3\).

\(x=6\,\text{cm}\) Material volume: \(960\,\text{cm}^3\)

- Read the dimensions separately for the cylinder and rectangular-prism models. - Convert each modeled volume to mass using the same density. - Compare the difference in predictions with the cylinder prediction, because that is the reference specified in the criterion.

1. The cylinder model has radius \(5\,\text{cm}\) and height \(20\,\text{cm}\), so \(V_c=\pi(5^2)(20)=500\pi\,\text{cm}^3\). 2. Its predicted mass is \(m_c=0.040(500\pi)=20\pi\,\text{g}\approx62.8\,\text{g}\). 3. The rectangular-prism model has volume \(V_p=(10)(10)(20)=2000\,\text{cm}^3\), so its predicted mass is \(m_p=0.040(2000)=80.0\,\text{g}\). 4. The difference is \(80-20\pi\approx17.2\,\text{g}\). Relative to the cylinder prediction, \(\frac{80-20\pi}{20\pi}\cdot100\%\approx27.3\%\). 5. Since \(27.3\%>10\%\), the two models are not equivalent under the stated criterion.

Cylinder model: \(62.8\,\text{g}\) Rectangular-prism model: \(80.0\,\text{g}\) The predictions differ by about \(27.3\%\) of the cylinder prediction, so the models are not equivalent.

- Check the capacity constraint before comparing material use. - For each closed box, all six faces contribute to the cardboard area. - Do not choose only by volume; the decision criterion is least surface area among feasible designs.

1. The volumes are \(V_A=(8)(10)(13)=1040\,\text{cm}^3\), \(V_B=(9)(9)(13)=1053\,\text{cm}^3\), and \(V_C=(7)(12)(12)=1008\,\text{cm}^3\). All three meet the capacity requirement. 2. For a closed rectangular prism, \(SA=2(lw+lh+wh)\). 3. Design A has \(SA_A=2(80+104+130)=628\,\text{cm}^2\). 4. Design B has \(SA_B=2(81+117+117)=630\,\text{cm}^2\). 5. Design C has \(SA_C=2(84+84+144)=624\,\text{cm}^2\). 6. Design C meets the capacity requirement and has the least surface area, so it uses the least cardboard under the stated model.

All three designs meet the \(1000\,\text{cm}^3\) requirement. Design C uses the least cardboard because its total surface area is \(624\,\text{cm}^2\), compared with \(628\,\text{cm}^2\) for A and \(630\,\text{cm}^2\) for B.

- First find the volume of the triangular prism. - How are mass, volume, and density related? - Use the density to determine the cylinder's required volume. - Solve the cylinder volume formula for height.

1. The prism's base area is \(B=\frac{1}{2}\cdot4\,\text{cm}\cdot3\,\text{cm}=6\,\text{cm}^2\), so its volume is \(V_P=6\,\text{cm}^2\cdot10\,\text{cm}=60\,\text{cm}^3\). 2. The aluminum's density is \(\rho=\frac{162\,\text{g}}{60\,\text{cm}^3}=2.7\,\text{g/cm}^3\). 3. The cylinder must have volume \(V_C=\frac{270\,\text{g}}{2.7\,\text{g/cm}^3}=100\,\text{cm}^3\). 4. Use \(V_C=\pi r^2h\): \(100=\pi\cdot2^2h=4\pi h\). 5. Therefore, \(h=\frac{25}{\pi}\,\text{cm}\approx7.9577\,\text{cm}\), which rounds to \(8.0\,\text{cm}\).

About \(8.0\,\text{cm}\)

- Convert all lengths to centimeters first. - A quarter cylinder has quarter-circle bases, one curved face, and two rectangular faces. - Find mass by multiplying volume by density.

1. Convert the height: \(2.00\,\text{m}=200\,\text{cm}\). 2. Each quarter-circle base has area \(B=\frac{1}{4}\pi\cdot25^2\,\text{cm}^2=156.25\pi\,\text{cm}^2\). 3. The curved lateral area is \(\frac{1}{4}\cdot2\pi\cdot25\cdot200\,\text{cm}^2=2500\pi\,\text{cm}^2\). 4. The two rectangular side faces have combined area \(2\cdot25\cdot200\,\text{cm}^2=10{,}000\,\text{cm}^2\). 5. The total surface area is \(2\cdot156.25\pi+2500\pi+10{,}000\,\text{cm}^2=2812.5\pi+10{,}000\,\text{cm}^2\approx18{,}835.73\,\text{cm}^2\). 6. The volume is \(V=156.25\pi\cdot200\,\text{cm}^3=31{,}250\pi\,\text{cm}^3\). 7. The mass is \(m=31{,}250\pi\cdot2.7\,\text{g}\approx265{,}071.88\,\text{g}\approx265.07\,\text{kg}\).

a) About \(18{,}835.73\,\text{cm}^2\) b) About \(265.07\,\text{kg}\)

- Use the mass and density to determine how much volume of gold is present. - Convert the sheet thickness to centimeters before relating volume, area, and thickness. - For a thin sheet, \(V=A\cdot t\). - After finding the sheet area, use the fact that the sheet is square.

1. Use density to recover the gold volume: \(V=\frac{m}{\rho}=\frac{193}{19.3}=10\,\text{cm}^3\). 2. Convert the sheet thickness: \(0.0001\,\text{mm}=0.00001\,\text{cm}\). 3. Volume is conserved, so the sheet area is \(A=\frac{10\,\text{cm}^3}{0.00001\,\text{cm}}=1{,}000{,}000\,\text{cm}^2\). 4. Since \(1{,}000{,}000\,\text{cm}^2=100\,\text{m}^2\), the square's side length is \(\sqrt{100}=10\,\text{m}\).

The square sheet's side length is \(10\,\text{m}\).

- First find the areas of the two regular hexagonal bases. - Use the frustum volume formula without rounding intermediate radical values. - Relate mass, density, and volume. - Round only the final result.

1. The area of a regular hexagon with side length \(s\) is \(A=\frac{3\sqrt{3}}{2}s^2\). 2. The lower and upper base areas are \(B_1=\frac{3\sqrt{3}}{2}(3)^2=\frac{27\sqrt{3}}{2}\,\text{cm}^2\) and \(B_2=\frac{3\sqrt{3}}{2}(1)^2=\frac{3\sqrt{3}}{2}\,\text{cm}^2\). 3. The frustum volume is \(V=\frac{h}{3}\left(B_1+\sqrt{B_1B_2}+B_2\right)\). Here, \(\sqrt{B_1B_2}=\frac{9\sqrt{3}}{2}\), so \(V=\frac{4}{3}\left(\frac{27\sqrt{3}}{2}+\frac{9\sqrt{3}}{2}+\frac{3\sqrt{3}}{2}\right)=26\sqrt{3}\approx45.0333\,\text{cm}^3\). 4. The mass is \(m=\rho V=2.5\cdot26\sqrt{3}=65\sqrt{3}\approx112.6\,\text{g}\).

The paperweight's mass is approximately \(112.6\,\text{g}\).

- Identify the shape of the water-filled part of the vase. - First determine the radius of the water surface halfway up the vase. - Use the linear change in radius from bottom to top. - Convert cubic centimeters to liters at the end.

1. The bottom radius is \(6\,\text{cm}\), and the top radius is \(10\,\text{cm}\). 2. Because the radius changes linearly with height, the radius at \(15\,\text{cm}\) is \(6+\frac{10-6}{30}\cdot15=8\,\text{cm}\). 3. The water forms a conical frustum with height \(15\,\text{cm}\), bottom radius \(6\,\text{cm}\), and top radius \(8\,\text{cm}\). Its volume is \(V=\frac{1}{3}\pi\cdot15\left(8^2+8\cdot6+6^2\right)=740\pi\,\text{cm}^3\approx2{,}324.78\,\text{cm}^3\). 4. Since \(1{,}000\,\text{cm}^3=1\,\text{L}\), the water volume is approximately \(2.32\,\text{L}\).

The vase contains approximately \(2.32\,\text{L}\) of water.

- Convert measurements to decimeters because the density uses cubic decimeters. - Use the relationship between a circle's circumference and radius. - The approximation uses the arithmetic mean of the two end areas.

1. In decimeters, \(h=12\,\text{dm}\), \(r_1=\frac{20}{2\pi}=\frac{10}{\pi}\,\text{dm}\), and \(r_2=\frac{14}{2\pi}=\frac{7}{\pi}\,\text{dm}\). 2. The exact frustum volume is \(V=\frac{12\pi}{3}\left(\frac{100}{\pi^2}+\frac{70}{\pi^2}+\frac{49}{\pi^2}\right)=\frac{876}{\pi}\,\text{dm}^3\approx278.84\,\text{dm}^3=0.279\,\text{m}^3\). 3. The mass is \(m=\frac{876}{\pi}\cdot2.4\approx669.21\,\text{kg}\). 4. The end-base areas are \(B_1=\frac{100}{\pi}\,\text{dm}^2\) and \(B_2=\frac{49}{\pi}\,\text{dm}^2\). The approximation gives \(V_{\text{approx}}=\frac{B_1+B_2}{2}\cdot12=\frac{894}{\pi}\,\text{dm}^3\approx284.57\,\text{dm}^3=0.285\,\text{m}^3\). 5. The approximation is greater by approximately \(5.73\,\text{dm}^3\), or \(0.006\,\text{m}^3\), which is about \(2.05\%\) of the exact volume.

a) \(V\approx0.279\,\text{m}^3\) b) The pillar's mass is approximately \(669.21\,\text{kg}\). c) The approximation is \(0.285\,\text{m}^3\), about \(0.006\,\text{m}^3\) or \(2.05\%\) greater than the exact volume.

- Decompose the composite solid into its two parts. - Subtract the cylinder volume from the total volume. - Convert liters to cubic centimeters. - Convert each diameter to a radius before using the frustum formula.

1. Convert the total volume: \(70\,\text{L}=70{,}000\,\text{cm}^3\). 2. The cylinder has radius \(20\,\text{cm}\), so \(V_{\text{cylinder}}=\pi\cdot20^2\cdot10=4{,}000\pi\,\text{cm}^3\). 3. The frustum volume is \(V_{\text{frustum}}=70{,}000-4{,}000\pi\,\text{cm}^3\). 4. With radii \(20\,\text{cm}\) and \(30\,\text{cm}\), \(V_{\text{frustum}}=\frac{1}{3}\pi h\left(20^2+20\cdot30+30^2\right)=\frac{1{,}900\pi h}{3}\). 5. Therefore, \(h=\frac{3\left(70{,}000-4{,}000\pi\right)}{1{,}900\pi}\approx28.87\,\text{cm}\).

The lower frustum is approximately \(28.87\,\text{cm}\) tall.

- View the cross section as a large rectangle with pieces removed. - Identify the areas of the bevel and the notch from the diagram. - Multiply the net cross-sectional area by the component's length.

1. Enclose the cross section in a \(10\,\text{mm}\times8\,\text{mm}\) rectangle with area \(80\,\text{mm}^2\). 2. The triangular region removed by the bevel has area \(\frac{1}{2}\cdot3\cdot(8-5)=4.5\,\text{mm}^2\). 3. The rectangular notch has area \(2\cdot2=4\,\text{mm}^2\). 4. The net cross-sectional area is \(A=80-4.5-4=71.5\,\text{mm}^2\). 5. The volume is \(V=71.5\cdot20=1430\,\text{mm}^3\).

The component's volume is \(1430\,\text{mm}^3\).

- Use the listed coordinates to identify the cross-section boundaries and dimensions. - Decompose the region into rectangles and a triangle. - Distinguish added regions from the notch that must be subtracted.

1. Decompose the cross section into a lower rectangle, an upper-left rectangle, and an upper-right triangle, then subtract the rectangular notch. 2. The lower rectangle has area \(12\cdot4=48\,\text{m}^2\). 3. The upper-left rectangle has area \(8\cdot4=32\,\text{m}^2\), and the upper-right triangle has area \(\frac{1}{2}\cdot4\cdot4=8\,\text{m}^2\). 4. The notch has area \(2\cdot2=4\,\text{m}^2\), so the total cross-sectional area is \(A=48+32+8-4=84\,\text{m}^2\). 5. The volume is \(V=84\cdot25=2{,}100\,\text{m}^3\).

The foundation's volume is \(2{,}100\,\text{m}^3\).

- For each allowed radius, translate the capacity requirement into a lower bound on height. - The height must satisfy both the capacity condition and the whole-number, maximum-height constraints. - Compare material only after removing infeasible designs. - An open-top cylinder uses one circular base plus its lateral surface.

1. The capacity condition is \(\pi r^2h\ge1200\). 2. For \(r=4\), \(h\ge\frac{1200}{16\pi}\approx23.87\), so the smallest whole-number height is \(24\,\text{cm}\). This violates the \(20\,\text{cm}\) limit. 3. For \(r=5\), \(h\ge\frac{1200}{25\pi}\approx15.28\), so the smallest feasible whole-number height is \(16\,\text{cm}\). Its material area is \(SA=\pi(5^2)+2\pi(5)(16)=185\pi\,\text{cm}^2\). 4. For \(r=6\), \(h\ge\frac{1200}{36\pi}\approx10.61\), so the smallest feasible whole-number height is \(11\,\text{cm}\). Its material area is \(SA=\pi(6^2)+2\pi(6)(11)=168\pi\,\text{cm}^2\). 5. Since \(168\pi<185\pi\), the \(6\,\text{cm}\)-radius, \(11\,\text{cm}\)-high canister is the feasible design that uses the least sheet metal.

Radius \(4\,\text{cm}\): minimum height \(24\,\text{cm}\), so it is not feasible. Radius \(5\,\text{cm}\): minimum height \(16\,\text{cm}\), material area \(185\pi\,\text{cm}^2\). Radius \(6\,\text{cm}\): minimum height \(11\,\text{cm}\), material area \(168\pi\,\text{cm}^2\). The best design has radius \(6\,\text{cm}\) and height \(11\,\text{cm}\).

- Use the marked central angle to identify the sector and triangle that remain in the circular cross section. - The flat seat is the chord joining the two marked points on the circle. - Use the cross-sectional area and log length to find volume before applying density. - Compare the calculated mass with the group's total lifting limit.

1. The diagram shows radius \(30\,\text{cm}\), and the length is \(200\,\text{cm}\). 2. The remaining cross-sectional area is a \(270^\circ\) sector plus a right triangle with legs \(30\,\text{cm}\): \(B=\frac{270}{360}\pi\cdot30^2+\frac{1}{2}\cdot30\cdot30\,\text{cm}^2=675\pi+450\,\text{cm}^2\). 3. The volume is \(V=(675\pi+450)\cdot200\,\text{cm}^3\approx514{,}115.01\,\text{cm}^3\). 4. The seat width is the chord of the marked \(90^\circ\) central angle. It is the hypotenuse of a right triangle with legs \(30\,\text{cm}\), so \(w=\sqrt{30^2+30^2}\,\text{cm}=30\sqrt{2}\,\text{cm}\approx42.43\,\text{cm}\). 5. Using the unrounded volume, the bench mass is \(0.6\,\text{g/cm}^3\cdot V\approx308{,}469.00\,\text{g}=308.47\,\text{kg}\). 6. Three people can lift at most \(3\cdot50\,\text{kg}=150\,\text{kg}\). Since \(308.47>150\), they cannot carry it under the stated limit.

a) About \(514{,}115.01\,\text{cm}^3\) b) \(30\sqrt{2}\,\text{cm}\approx42.43\,\text{cm}\) c) No. The bench has mass about \(308.47\,\text{kg}\), which exceeds their combined \(150\,\text{kg}\) limit.