- Identify the volume formula for a sphere. - Substitute the given radius and keep \(\pi\) exact.

1. Use \(V=\frac{4}{3}\pi r^3\): \(V=\frac{4}{3}\pi(3^3)=36\pi\,\text{cm}^3\).

The volume is \(36\pi\,\text{cm}^3\).

- Identify the volume formula for a cone. - Use the perpendicular height shown in the problem.

1. Use \(V=\frac{1}{3}\pi r^2h\): \(V=\frac{1}{3}\pi(3^2)(4)=12\pi\,\text{cm}^3\).

The volume is \(12\pi\,\text{cm}^3\).

- The base area is already given, so no base-area calculation is needed. - Use the perpendicular height in the pyramid volume formula.

1. Use \(V=\frac{1}{3}Bh\): \(V=\frac{1}{3}(25)(6)=50\,\text{cm}^3\).

The volume is \(50\,\text{cm}^3\).

- Does the labeled segment represent a radius or a diameter? - How are the radius and diameter related? - The sphere volume formula uses the radius, not the diameter.

1. The diameter is \(8\,\text{cm}\), so the radius is half as long: \(r=4\,\text{cm}\). 2. Substitute the radius in \(V=\frac{4}{3}\pi r^3\) to get \(V=\frac{4}{3}\pi(4)^3\,\text{cm}^3\).

Radius: \(4\,\text{cm}\). Volume expression: \(\frac{4}{3}\pi(4)^3\,\text{cm}^3\).

- Imagine a line from the tip to the plane of the base that makes a right angle. - Which labeled length follows that line, rather than a sloping edge?

1. The perpendicular height is \(5\,\text{cm}\), measured at a right angle from the tip to the base plane. 2. A sloping edge is not the perpendicular distance, so it cannot replace the height in the volume formula.

\(5\,\text{cm}\); a sloping edge is not perpendicular to the base.

- Compare the figures' bases and perpendicular heights, not their slanted edges. - How does the pyramid volume formula differ from the prism formula?

1. The volume of a pyramid with the same base area and perpendicular height is one-third the volume of the matching prism. 2. Its volume is \(72\div3=24\,\text{cm}^3\).

\(24\,\text{cm}^3\), one-third of the cuboid's volume.

- Read the radius and height from the diagram. - Use the cone volume formula. - Write the exact answer with \(\pi\) before calculating a decimal approximation. - Round only in the final step.

1. The diagram shows radius \(r=5\,\text{cm}\) and height \(h=12\,\text{cm}\). 2. Use \(V=\frac{1}{3}\pi r^2h\): \(V=\frac{1}{3}\pi\cdot5^2\cdot12=100\pi\,\text{cm}^3\). 3. Numerically, \(100\pi\approx314.2\,\text{cm}^3\).

Exact value: \(100\pi\,\text{cm}^3\) Rounded value: \(314.2\,\text{cm}^3\)

- Determine whether the labeled measurement is a radius or diameter. - Use the sphere volume formula. - Simplify the exact expression before finding a decimal approximation. - Round only the decimal value.

1. The diagram shows radius \(r=6\,\text{cm}\). 2. Use \(V=\frac{4}{3}\pi r^3\): \(V=\frac{4}{3}\pi\cdot6^3=288\pi\,\text{cm}^3\). 3. Numerically, \(288\pi\approx904.8\,\text{cm}^3\).

Exact value: \(288\pi\,\text{cm}^3\) Rounded value: \(904.8\,\text{cm}^3\)

- Identify the solid as a cone. - Use the radius and height in the cone volume formula. - Convert cubic centimeters to milliliters using the given equivalence. - Round only the final value.

1. Use the cone volume formula: \(V=\frac{1}{3}\pi r^2h\). 2. Substitute the measurements: \(V=\frac{1}{3}\pi\cdot4^2\cdot9=48\pi\,\text{cm}^3\). 3. Since \(48\pi\approx150.7964\), the capacity is approximately \(150.8\,\text{mL}\).

The maximum capacity is approximately \(150.8\,\text{mL}\).

- Determine whether the given circular measurement is a radius or a diameter. - Convert the diameter to a radius. - Substitute the radius and height into the cone volume formula.

1. The radius is half the diameter: \(r=5\div2=2.5\,\text{cm}\). 2. Use the cone volume formula: \(V=\frac{1}{3}\pi r^2h\). 3. Substitute: \(V=\frac{1}{3}\pi\cdot2.5^2\cdot12=25\pi\,\text{cm}^3\). 4. Since \(25\pi\approx78.5398\), the volume is approximately \(78.5\,\text{cm}^3\).

The container volume is approximately \(78.5\,\text{cm}^3\).

- Use the cone volume formula. - Compare the cone formula with the cylinder formula for the same base and height.

1. Use \(V=\frac{1}{3}\pi r^2h\). 2. Substitute: \(V=\frac{1}{3}\pi\cdot7^2\cdot12=196\pi\,\text{m}^3\). 3. Since \(196\pi\approx615.7522\), the volume is approximately \(615.8\,\text{m}^3\).

The sand pile volume is approximately \(615.8\,\text{m}^3\).

- Find the area of the square base first. - A pyramid has one-third the volume of a prism with the same base and height.

1. The square base area is \(B=12^2=144\,\text{m}^2\). 2. The pyramid volume is \(V=\frac{1}{3}Bh=\frac{1}{3}\cdot144\cdot9=432\,\text{m}^3\).

The monument volume is \(432\,\text{m}^3\).

- Identify whether the pictured measurement runs across the whole sphere. - The formula takes a radius, not a diameter. - Keep \(\pi\) until the final rounding step.

1. The displayed diameter is \(12\,\text{cm}\), so the radius is \(r=6\,\text{cm}\). 2. Use the sphere-volume formula: \(V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi(6)^3=288\pi\,\text{cm}^3\). 3. Therefore, \(V\approx904.8\,\text{cm}^3\) to the nearest tenth.

\(V=288\pi\,\text{cm}^3\approx904.8\,\text{cm}^3\)

- Find the area of the triangular base before using the solid's height. - Distinguish the dimensions inside the base triangle from the perpendicular height of the pyramid. - Check that the final unit is cubic centimeters.

1. The base area is \(B=\frac{1}{2}(8)(15)=60\,\text{cm}^2\). 2. The volume of a pyramid is \(V=\frac{1}{3}Bh\). 3. Thus, \(V=\frac{1}{3}(60)(9)=180\,\text{cm}^3\).

\(180\,\text{cm}^3\)

- Compare the base areas first. - The heights stay fixed. What does that mean for the volume ratio? - A doubled length does not mean a doubled square area.

1. Doubling both sides of the square base multiplies its area by \(2^2=4\). 2. Both pyramids have the same height and the same one-third factor. 3. Therefore the larger pyramid has four times the smaller pyramid's volume.

\(4\) times as great.

- What does each solid's circular base contribute to volume? - Compare the labeled radii and perpendicular heights. - Which two volume formulas apply?

1. Both have base area \(\pi(2)^2=4\pi\,\text{cm}^2\). 2. The cylinder volume is \(4\pi\cdot6=24\pi\,\text{cm}^3\). 3. The cone volume is \(\frac{1}{3}\cdot24\pi=8\pi\,\text{cm}^3\).

The cone occupies \(\frac{1}{3}\) of the cylinder volume. Cone: \(8\pi\,\text{cm}^3\); cylinder: \(24\pi\,\text{cm}^3\).

- Find the distance that is perpendicular to the base plane. - Where does the slant height lie compared with the vertical segment? - Check that you use the radius for the circular base.

1. Volume uses the perpendicular distance from the tip to the base, which is \(4\,\text{cm}\), not the slant height along the side. 2. The base radius is \(3\,\text{cm}\). 3. Thus \(V=\frac{1}{3}\pi(3)^2(4)=12\pi\,\text{cm}^3\).

The perpendicular height is \(4\,\text{cm}\); the slant height does not belong in the volume formula. \(V=12\pi\,\text{cm}^3\).

- Is the marked length a radius or a diameter? - How would you find the distance from the sphere's center to its surface? - Check which length belongs in the sphere-volume formula.

1. The diameter is \(6\,\text{cm}\), so \(r=3\,\text{cm}\). 2. Use \(V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi(3)^3=36\pi\,\text{cm}^3\).

\(36\pi\,\text{cm}^3\)

- Which part of the sphere volume formula changes? - A length factor affects volume in three dimensions. - Compare the two cubes of the radius as a ratio.

1. The new radius is three times the old radius. 2. Sphere volume is proportional to the cube of its radius. 3. The new volume is \(3^3=27\) times the old volume.

The volume becomes \(27\) times as great.

- Start with the cone volume formula, leaving height unknown. - Cancel a factor common to both sides of your equation. - Check the recovered height by substituting it back.

1. Use \(V=\frac{1}{3}\pi r^2h\). 2. Substituting the known values gives \(18\pi=\frac{1}{3}\pi(3)^2h=3\pi h\). 3. Divide by \(3\pi\): \(h=6\,\text{cm}\).

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

- The total amount of clay, and therefore the total volume, stays the same. - Write the sphere volume formula for each original sphere. - Factor out the common constants before solving for the new radius.

1. Volume is conserved. The total volume is \(\frac{4}{3}\pi\left(1^3+6^3+8^3\right)\). 2. The sum of the cubes is \(1+216+512=729\). 3. If \(R\) is the new radius, then \(\frac{4}{3}\pi R^3=\frac{4}{3}\pi\cdot729\). 4. Therefore, \(R^3=729\), so \(R=\sqrt[3]{729}=9\,\text{cm}\).

The new sphere has radius \(9\,\text{cm}\).

- Isolate \(r^3\) in the volume formula, then take a cube root. - Keep the calculator value of the radius available for the surface-area calculation instead of reusing the rounded radius. - Check that the first answer has length units and the second has square units.

1. Solve the volume formula for the radius: \(r=\sqrt[3]{\frac{3V}{4\pi}}\). Thus \(r=\sqrt[3]{\frac{3\cdot14.5}{4\pi}}\approx1.51\,\text{m}\). 2. Use the unrounded radius in the surface-area formula: \(SA=4\pi\left(\sqrt[3]{\frac{3\cdot14.5}{4\pi}}\right)^2\approx28.76\,\text{m}^2\).

1. \(r\approx1.51\,\text{m}\) 2. \(SA\approx28.76\,\text{m}^2\)

- Rearrange the sphere surface-area formula to isolate the radius. - Keep the calculator value of the radius for the volume calculation rather than reusing the rounded radius. - Check that radius uses linear units and volume uses cubic units.

1. From \(SA=4\pi r^2\), \(r=\sqrt{\frac{800}{4\pi}}\approx7.98\,\text{cm}\). 2. Use the unrounded radius in the volume formula: \(V=\frac{4}{3}\pi\left(\sqrt{\frac{800}{4\pi}}\right)^3\approx2127.69\,\text{cm}^3\).

\(r\approx7.98\,\text{cm}\) \(V\approx2127.69\,\text{cm}^3\)

- Convert liters to cubic centimeters. - A hemisphere has half the volume of a sphere. - Rearrange the volume formula to solve for the radius. - Double the radius because the problem asks for diameter.

1. Convert the capacity: \(4\,\text{L}=4{,}000\,\text{cm}^3\). 2. A hemisphere has volume \(V=\frac{2}{3}\pi r^3\). 3. Solve for the radius: \(r=\sqrt[3]{\frac{3V}{2\pi}}=\sqrt[3]{\frac{3\cdot4{,}000}{2\pi}}\approx12.4070\,\text{cm}\). 4. The diameter is \(d=2r\approx24.8140\,\text{cm}\), which rounds to \(24.8\,\text{cm}\).

The interior diameter is approximately \(24.8\,\text{cm}\).

- Find the volume of each solid and compare them. - Volume changes by the cube of a linear scale factor. - A hemisphere has half the volume of a sphere with the same radius. - Divide the new volume by the original volume.

1. The original sphere has radius \(1.5\,\text{cm}\), so its volume is \(V_1=\frac{4}{3}\pi\cdot1.5^3=4.5\pi\,\text{cm}^3\). 2. The new hemisphere has radius \(3\,\text{cm}\), so its volume is \(V_2=\frac{2}{3}\pi\cdot3^3=18\pi\,\text{cm}^3\). 3. The volume factor is \(\frac{V_2}{V_1}=\frac{18\pi}{4.5\pi}=4\). 4. Equivalently, doubling all linear dimensions multiplies a sphere's volume by \(2^3=8\), and taking half of that sphere gives a factor of \(4\).

The new hemisphere has \(4\) times the volume of the original sphere.

- Convert liters to cubic decimeters. - Rearrange the sphere volume formula and take a cube root. - Use the calculated radius in the sphere surface-area formula. - Keep the unrounded radius for the second calculation.

1. The volume is \(12\,\text{dm}^3\). 2. Solve \(V=\frac{4}{3}\pi r^3\) for the radius: \(r=\sqrt[3]{\frac{3V}{4\pi}}=\sqrt[3]{\frac{36}{4\pi}}=\sqrt[3]{\frac{9}{\pi}}\approx1.420248\,\text{dm}\). 3. Therefore, \(r\approx1.42\,\text{dm}\). 4. Using the unrounded radius, \(SA=4\pi r^2\approx25.3477\,\text{dm}^2\), so \(SA\approx25.35\,\text{dm}^2\).

a) \(r\approx1.42\,\text{dm}\) b) \(SA\approx25.35\,\text{dm}^2\)

- Use similarity to determine the small pyramid's base edge. - The cut divides the original height into two equal parts. - Find the volume of the original pyramid and subtract the volume of the smaller pyramid.

1. The cut is halfway up the original height, so the small removed pyramid has height \(9\,\text{cm}\) and linear scale factor \(\frac{1}{2}\). 2. Its base edge is \(\frac{1}{2}\cdot12=6\,\text{cm}\). 3. The original pyramid volume is \(V_L=\frac{1}{3}\cdot12^2\cdot18=864\,\text{cm}^3\). 4. The small pyramid volume is \(V_s=\frac{1}{3}\cdot6^2\cdot9=108\,\text{cm}^3\). 5. The frustum volume is \(864-108=756\,\text{cm}^3\).

The frustum volume is \(756\,\text{cm}^3\).

- Compare corresponding edges of the two rectangular bases. - Imagine extending the sloping edges until they meet at a tip. - How do the base-side ratios compare with the heights of the two similar pyramids? - Subtract the smaller pyramid's volume from the full pyramid's volume.

1. The corresponding side-length ratios are \(5.0\div12.5=3.2\div8.0=0.4\), so the removed pyramid has a length scale factor of \(0.4\). 2. Let \(H\) be the full pyramid height. The removed pyramid's height is \(0.4H\), so \(H-0.4H=9.0\). Thus \(H=15\,\text{cm}\) and the removed height is \(6\,\text{cm}\). 3. The lower and upper base areas are \(12.5\cdot8.0=100\,\text{cm}^2\) and \(5.0\cdot3.2=16\,\text{cm}^2\). 4. Subtract the pyramid volumes: \(V=\frac{1}{3}\cdot100\cdot15-\frac{1}{3}\cdot16\cdot6=468\,\text{cm}^3\).

\(468\,\text{cm}^3\)

- Convert all lengths to meters first. - Check the similarity ratio between the two rectangular bases. - Express the frustum as the difference of two similar pyramids. - Keep track of square and cubic units.

1. Convert the upper base dimensions: \(80\,\text{cm}=0.80\,\text{m}\) and \(50\,\text{cm}=0.50\,\text{m}\). 2. The upper base dimensions are one-third of the corresponding lower base dimensions. If the full pyramid height is \(H\), the removed pyramid height is \(\frac{H}{3}\). 3. The frustum height is \(H-\frac{H}{3}=\frac{2H}{3}=1.80\), so \(H=2.70\,\text{m}\), and the removed height is \(0.90\,\text{m}\). 4. The lower and upper base areas are \(2.40\cdot1.50=3.60\,\text{m}^2\) and \(0.80\cdot0.50=0.40\,\text{m}^2\). 5. Therefore, \(V=\frac{1}{3}\cdot3.60\cdot2.70-\frac{1}{3}\cdot0.40\cdot0.90=3.24-0.12=3.12\,\text{m}^3\).

\(V=3.12\,\text{m}^3\)

- Compare the side lengths of the top and bottom squares. - Extend the sides to create two similar pyramids sharing an apex. - Use the similarity ratio to determine the two pyramid heights. - Convert the final cubic-centimeter volume to liters.

1. The opening side is \(1.5\) times the bottom side. Extend the slanted sides downward to a common apex below the planter. 2. Let \(x\) be the height from the apex to the smaller bottom square. Then \(\frac{x+45}{x}=1.5\), so \(x=90\,\text{cm}\). The larger pyramid height is \(135\,\text{cm}\). 3. The large and small square base areas are \(60^2=3{,}600\,\text{cm}^2\) and \(40^2=1{,}600\,\text{cm}^2\). 4. The planter volume is \(V=\frac{1}{3}\cdot3{,}600\cdot135-\frac{1}{3}\cdot1{,}600\cdot90=162{,}000-48{,}000=114{,}000\,\text{cm}^3\). 5. Since \(1{,}000\,\text{cm}^3=1\,\text{L}\), the planter holds \(114\,\text{L}\).

The planter requires \(114\,\text{L}\) of soil.

- Compare corresponding dimensions of the two rectangular bases. - Model the frustum as a large pyramid minus a smaller similar pyramid. - Use the scale factor to determine the full and removed pyramid heights. - Convert cubic centimeters to cubic meters only after finding the volume.

1. Each upper-base dimension is half the corresponding lower-base dimension, so the removed top pyramid has scale factor \(\frac{1}{2}\) relative to the full pyramid. 2. Let \(H\) be the full pyramid height. The removed pyramid height is \(\frac{H}{2}\), and the frustum height is also \(\frac{H}{2}=50\). Thus, \(H=100\,\text{cm}\). 3. The lower and upper base areas are \(120\cdot80=9{,}600\,\text{cm}^2\) and \(60\cdot40=2{,}400\,\text{cm}^2\). 4. The frustum volume is \(V=\frac{1}{3}\cdot9{,}600\cdot100-\frac{1}{3}\cdot2{,}400\cdot50=320{,}000-40{,}000=280{,}000\,\text{cm}^3\). 5. Since \(1\,\text{m}^3=1{,}000{,}000\,\text{cm}^3\), \(V=0.28\,\text{m}^3\).

The required concrete volume is \(0.28\,\text{m}^3\).

- Convert the diameters to radii. - Extend the sides to model the frustum as the difference of two similar cones. - Use the radius ratio to determine the two cone heights. - Convert cubic centimeters to liters at the end.

1. The top and bottom radii are \(16\,\text{cm}\) and \(12\,\text{cm}\). 2. Extend the slanted sides below the bucket to form two similar cones. Let \(x\) be the height of the smaller cone. Since the radius ratio is \(16\div12=\frac{4}{3}\), \(\frac{x+28}{x}=\frac{4}{3}\), so \(x=84\,\text{cm}\). The larger cone height is \(112\,\text{cm}\). 3. The bucket volume is the difference of the two cone volumes: \(V=\frac{1}{3}\pi\cdot16^2\cdot112-\frac{1}{3}\pi\cdot12^2\cdot84=\frac{16{,}576\pi}{3}\approx17{,}358.35\,\text{cm}^3\). 4. Since \(1{,}000\,\text{cm}^3=1\,\text{L}\), the capacity is approximately \(17.358\,\text{L}\), or \(17.4\,\text{L}\) to the nearest tenth.

The bucket capacity is approximately \(17.4\,\text{L}\).

- Compare the two square edge lengths. - Extend the lateral edges to form the original full pyramid. - Use the similarity ratio to relate the full and removed pyramid heights. - Subtract the smaller pyramid volume from the larger one.

1. The upper base edge is half the lower base edge, so the small removed pyramid is a \(\frac{1}{2}\)-scale copy of the full pyramid. 2. If the full pyramid height is \(H\), the small pyramid height is \(\frac{H}{2}\). The frustum height is \(H-\frac{H}{2}=6\), so \(H=12\,\text{cm}\), and the small pyramid height is \(6\,\text{cm}\). 3. The lower and upper base areas are \(100\,\text{cm}^2\) and \(25\,\text{cm}^2\). 4. Subtract the small pyramid volume from the full pyramid volume: \(V=\frac{1}{3}\cdot100\cdot12-\frac{1}{3}\cdot25\cdot6=400-50=350\,\text{cm}^3\).

The paperweight volume is \(350\,\text{cm}^3\).

- A cross section parallel to a pyramid's base has the same shape as the base. - Find the lower volume by subtracting the small top pyramid from the original pyramid. - Use similarity to determine the cross-section edge length.

1. A plane parallel to a square pyramid's base creates a square cross section. 2. The original pyramid volume is \(V_L=\frac{1}{3}\cdot6^2\cdot9=108\,\text{cm}^3\). 3. The small pyramid above the cut has height \(9-3=6\,\text{cm}\), so its linear scale factor is \(6\div9=\frac{2}{3}\). Its base edge is \(\frac{2}{3}\cdot6=4\,\text{cm}\), and its volume is \(V_s=\frac{1}{3}\cdot4^2\cdot6=32\,\text{cm}^3\). 4. Therefore, the frustum volume is \(108-32=76\,\text{cm}^3\).

a) The cross section is a square. b) The frustum volume is \(76\,\text{cm}^3\).

- Find the volume of each solid separately. - The sphere's diameter equals the cube's side length. - Divide the sphere's volume by the cube's volume and convert to a percent.

1. The sphere has radius \(6\,\text{cm}\), so \(V_s=\frac{4}{3}\pi\cdot6^3=288\pi\,\text{cm}^3\approx904.78\,\text{cm}^3\). 2. The cube's volume is \(V_c=12^3=1728\,\text{cm}^3\). 3. The percentage is \(\frac{288\pi}{1728}\cdot100\%=\frac{\pi}{6}\cdot100\%\approx52.4\%\).

Sphere volume: \(288\pi\,\text{cm}^3\approx904.78\,\text{cm}^3\) Percent of cube volume: approximately \(52.4\%\)

- Compare corresponding top and bottom dimensions. - Extend the sides to model the pan as the difference of two similar pyramids. - Use the scale factor and the known depth to find the two pyramid heights. - Subtract the smaller pyramid volume from the larger one.

1. The top dimensions are \(\frac{4}{3}\) times the corresponding bottom dimensions. Extend the slanted sides below the pan to form two similar pyramids. 2. Let \(x\) be the height from the common apex to the smaller bottom rectangle. Then \(\frac{x+5}{x}=\frac{4}{3}\), so \(x=15\,\text{cm}\). The larger pyramid height is \(20\,\text{cm}\). 3. The top and bottom areas are \(32\cdot24=768\,\text{cm}^2\) and \(24\cdot18=432\,\text{cm}^2\). 4. The capacity is \(V=\frac{1}{3}\cdot768\cdot20-\frac{1}{3}\cdot432\cdot15=5120-2160=2960\,\text{cm}^3\).

The baking pan has a capacity of \(2960\,\text{cm}^3\).

- Imagine extending the sloping edges until they meet at one tip. - Compare the upper and lower square edges to find the scale factor for the removed pyramid. - Express the two pyramid heights using the frustum's perpendicular height. - Subtract the top pyramid's volume from the complete pyramid's volume.

1. The lower square edge is \(12\,\text{cm}\), the upper edge is \(4\,\text{cm}\), and the perpendicular frustum height is \(9\,\text{cm}\). 2. The upper edge is one-third the lower edge, so the small removed pyramid has height \(\frac{H}{3}\) if the complete pyramid has height \(H\). 3. The frustum height is \(H-\frac{H}{3}=\frac{2H}{3}=9\), giving \(H=13.5\,\text{cm}\) and a removed height of \(4.5\,\text{cm}\). 4. The complete pyramid has volume \(\frac{1}{3}\cdot12^2\cdot13.5=648\,\text{cm}^3\). 5. The small pyramid has volume \(\frac{1}{3}\cdot4^2\cdot4.5=24\,\text{cm}^3\). 6. Subtracting gives \(648-24=624\,\text{cm}^3\).

\(624\,\text{cm}^3\)

- Use the ratio of the upper and lower base edges as a similarity scale factor. - Relate the full pyramid height, removed pyramid height, and frustum height. - Find the two pyramid volumes before subtracting.

1. The linear scale factor from the full pyramid to the removed pyramid is \(3\div5=0.6\). 2. If the full pyramid height is \(H\), the removed height is \(0.6H\). Therefore, \(H-0.6H=0.4H=4.5\), so \(H=11.25\,\text{cm}\), and the removed height is \(6.75\,\text{cm}\). 3. The full pyramid volume is \(\frac{1}{3}\cdot5^2\cdot11.25=93.75\,\text{cm}^3\). 4. The removed pyramid volume is \(\frac{1}{3}\cdot3^2\cdot6.75=20.25\,\text{cm}^3\). 5. The frustum volume is \(93.75-20.25=73.5\,\text{cm}^3\).

The glass solid has volume \(73.5\,\text{cm}^3\).

- The volume formula needs the perpendicular height rather than the slant height. - Use the right triangle formed by the radius, perpendicular height, and slant height. - Keep \(\pi\) symbolic because an exact volume is requested.

1. In an axial right triangle, the radius, perpendicular height, and slant height satisfy \(6^2+h^2=10^2\). 2. Thus, \(h^2=64\), so \(h=8\,\text{cm}\). 3. The cone's volume is \(V=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi(6^2)(8)=96\pi\,\text{cm}^3\).

\(96\pi\,\text{cm}^3\)

- Work backward from the pyramid-volume formula to the base area. - Express the area of a square base in terms of one side length. - A geometric length must be positive.

1. Let the base side length be \(x\), so the base area is \(B=x^2\). 2. From \(V=\frac{1}{3}Bh\), \(588=\frac{1}{3}x^2(9)=3x^2\). 3. Therefore, \(x^2=196\), and the positive length is \(x=14\,\text{cm}\).

\(14\,\text{cm}\)

- Write the sphere volume formula for both spheres and form a ratio. - Cancel constants that are common to both volumes. - A volume scale factor corresponds to the cube of the linear scale factor. - Radius and diameter have the same scale factor.

1. The volume relationship is \(V_B=6V_A\). 2. Using \(V=\frac{4}{3}\pi r^3\), \(\frac{4}{3}\pi r_B^3=6\left(\frac{4}{3}\pi r_A^3\right)\). 3. Cancel the common factor to get \(r_B^3=6r_A^3\). 4. Take cube roots: \(r_B=\sqrt[3]{6}\,r_A\). 5. Diameter is proportional to radius, so \(k=\frac{d_B}{d_A}=\sqrt[3]{6}\approx1.82\).

\(k=\sqrt[3]{6}\approx1.82\)

- Compare the two rectangular end faces; corresponding lengths have the same ratio. - Extend the sloping sides in your mind to make a complete pyramid below the planter. - Similar pyramids give a relationship between the two distances from their shared tip. - Subtract the smaller pyramid's volume, then convert cubic centimeters to liters.

1. The upper rectangle is a scale copy of the lower rectangle, with scale factor \(\frac{60}{40}=\frac{45}{30}=1.5\). 2. Extend the sloping sides downward to complete a larger pyramid. Let \(x\) be the height from the common tip to the lower base. Then \(\frac{x+50}{x}=1.5\), so \(x=100\,\text{cm}\) and the complete height is \(150\,\text{cm}\). 3. The larger pyramid's base area is \(60\cdot45=2700\,\text{cm}^2\); the removed smaller one's base area is \(40\cdot30=1200\,\text{cm}^2\). 4. Subtract their volumes: \(V=\frac{1}{3}\cdot2700\cdot150-\frac{1}{3}\cdot1200\cdot100=95000\,\text{cm}^3\). 5. Convert to liters: \(95000\,\text{cm}^3=95\,\text{L}\).

\(95\,\text{L}\)

- Compare the upper and lower base edge lengths to find the similarity scale factor. - Model the frustum as a large pyramid with a smaller similar pyramid removed. - Express both pyramid heights in terms of one variable, then use the volume difference.

1. The upper base edge is half the lower base edge, so the small removed pyramid has linear scale factor \(\frac{1}{2}\) relative to the full pyramid. 2. Let \(H\) be the full pyramid height. Then the removed pyramid height is \(\frac{H}{2}\), so the frustum height is also \(\frac{H}{2}\). 3. The lower and upper base areas are \(100\,\text{cm}^2\) and \(25\,\text{cm}^2\). 4. Subtract the smaller pyramid's volume from the larger pyramid's volume: \(700=\frac{1}{3}\cdot100H-\frac{1}{3}\cdot25\left(\frac{H}{2}\right)=\frac{175H}{6}\). 5. Solving gives \(H=24\,\text{cm}\), so the frustum height is \(h=\frac{H}{2}=12\,\text{cm}\).

The frustum height is \(12\,\text{cm}\).

- Convert the diameters to radii and liters to cubic centimeters. - Extend the sides to form two similar cones. - Use the radius ratio to express both cone heights in terms of the bowl depth. - Set the difference of the cone volumes equal to the required capacity.

1. The top and bottom radii are \(11\,\text{cm}\) and \(7\,\text{cm}\), and \(1.5\,\text{L}=1{,}500\,\text{cm}^3\). 2. Extend the slanted sides below the bowl to form two similar cones. If the bowl depth is \(h\), similarity gives the smaller cone height \(\frac{7h}{4}\) and the larger cone height \(\frac{11h}{4}\). 3. The volume difference is \(1{,}500=\frac{1}{3}\pi\cdot11^2\left(\frac{11h}{4}\right)-\frac{1}{3}\pi\cdot7^2\left(\frac{7h}{4}\right)=\frac{247\pi h}{3}\). 4. Solving gives \(h=\frac{4{,}500}{247\pi}\approx5.7992\,\text{cm}\), so the required depth is \(5.8\,\text{cm}\).

The bowl must be approximately \(5.8\,\text{cm}\) deep.

- Model the frustum as a large cone with a smaller similar cone removed. - Use the radius ratio to relate the cone heights to the frustum height. - Set the difference of the two cone volumes equal to the given volume.

1. Extend the sides to form a large cone and a smaller similar cone. Let the frustum height be \(h\). 2. Since the radii are in the ratio \(3\) to \(2\), the smaller cone height is \(2h\), and the larger cone height is \(3h\). 3. The frustum volume is \(150=\frac{1}{3}\pi\cdot3^2\cdot3h-\frac{1}{3}\pi\cdot2^2\cdot2h=\frac{19\pi h}{3}\). 4. Therefore, \(h=\frac{450}{19\pi}\approx7.5389\,\text{dm}\), which rounds to \(7.54\,\text{dm}\).

The column height is approximately \(7.54\,\text{dm}\).

- Use a right triangle inside the pyramid that contains the perpendicular height. - Find the diagonal of the square base. - Apply the Pythagorean theorem to find the height. - Substitute that height into the pyramid volume formula.

1. The square base has diagonal \(s\sqrt{2}\), so the distance from its center to a vertex is \(\frac{s\sqrt{2}}{2}\). 2. The perpendicular height, the center-to-vertex segment, and a lateral edge form a right triangle. Thus, \(h^2+\left(\frac{s\sqrt{2}}{2}\right)^2=s^2\). 3. Therefore, \(h^2=s^2-\frac{1}{2}s^2=\frac{1}{2}s^2\), so \(h=\frac{s\sqrt{2}}{2}\). 4. Since the base area is \(s^2\), \(V=\frac{1}{3}s^2\cdot\frac{s\sqrt{2}}{2}=\frac{\sqrt{2}}{6}s^3\). 5. For \(s=4.5\,\text{cm}\), \(V=\frac{\sqrt{2}}{6}(4.5)^3\approx21.48\,\text{cm}^3\).

a) \(h=\frac{s\sqrt{2}}{2}\) b) \(V=\frac{\sqrt{2}}{6}s^3\) c) \(V\approx21.48\,\text{cm}^3\)

- Convert the capacity to cubic centimeters before setting up a volume equation. - Imagine completing the original cone and removing the smaller cone below the cup. - The two radii tell you how the full and smaller cone heights compare.

1. Convert the capacity: \(1.8\,\text{L}=1800\,\text{cm}^3\). 2. Imagine extending the sloping sides downward to complete a large cone, with a smaller cone below the cup. The radii are in the ratio \(9:6=3:2\), so the large and small cone heights are \(3h\) and \(2h\). 3. Subtract their volumes: \(1800=\frac{1}{3}\pi\cdot9^2\cdot3h-\frac{1}{3}\pi\cdot6^2\cdot2h=57\pi h\). 4. Therefore, \(h=\frac{1800}{57\pi}\approx10.05189\,\text{cm}\), or \(10.1\,\text{cm}\) to the nearest tenth.

The measuring cup is about \(10.1\,\text{cm}\) tall.

- Substitute the known values into the cone-frustum volume formula. - Rearrange the equation into quadratic form in the unknown radius. - Solve the quadratic equation using an appropriate method. - Reject any solution that is not meaningful as a radius.

1. Substitute into the cone-frustum volume formula: \(703.7=\frac{1}{3}\pi\cdot6\left(4^2+4r+r^2\right)=2\pi(r^2+4r+16)\). 2. Divide by \(2\pi\): \(r^2+4r+16=\frac{703.7}{2\pi}\). 3. Write the quadratic equation as \(r^2+4r+16-\frac{703.7}{2\pi}=0\). 4. The quadratic formula gives \(r=\frac{-4\pm\sqrt{16-4\left(16-\frac{703.7}{2\pi}\right)}}{2}\). 5. The two solutions are approximately \(7.9999\) and \(-11.9999\). A radius must be positive, so \(r\approx8.0\,\text{cm}\).

The upper-base radius is \(8.0\,\text{cm}\).

- Compare the filled cone with the entire container using similar triangles. - Express the changing liquid radius in terms of its height. - For half the volume, use half the full capacity, not half the funnel's height. - Check that the solution is between the point and the rim.

1. The full cone has radius \(6\,\text{cm}\) and height \(18\,\text{cm}\). 2. By similarity, \(\frac{r}{h}=\frac{6}{18}=\frac{1}{3}\), so \(r(h)=\frac{h}{3}\). 3. Thus \(V(h)=\frac{1}{3}\pi(\frac{h}{3})^2h=\frac{\pi}{27}h^3\,\text{cm}^3\) on \(0\le h\le18\). 4. At \(h=9\), \(V(9)=27\pi\,\text{cm}^3\approx84.8\,\text{cm}^3\). 5. Full capacity is \(V(18)=216\pi\,\text{cm}^3\). Half capacity is \(108\pi\,\text{cm}^3\). 6. Solving \(\frac{\pi}{27}h^3=108\pi\) gives \(h=\sqrt[3]{2916}\,\text{cm}\approx14.3\,\text{cm}\), within the physical range.

a) \(V(h)=\frac{\pi}{27}h^3\,\text{cm}^3\), for \(0\le h\le18\). b) \(V(9)=27\pi\,\text{cm}^3\approx84.8\,\text{cm}^3\). c) \(h\approx14.3\,\text{cm}\).

- Convert the volume to cubic centimeters. - Use the ratio of the two square edge lengths to relate the full and removed pyramid heights. - Write the frustum volume as the difference of two pyramid volumes and solve for height.

1. Convert the volume: \(76\,\text{dm}^3=76{,}000\,\text{cm}^3\). 2. The upper-to-lower edge ratio is \(40\div60=\frac{2}{3}\). If the full pyramid height is \(H\), the removed pyramid height is \(\frac{2H}{3}\), so the frustum height is \(\frac{H}{3}\). 3. The lower and upper base areas are \(3600\,\text{cm}^2\) and \(1600\,\text{cm}^2\). 4. The volume difference is \(76{,}000=\frac{1}{3}\cdot3600H-\frac{1}{3}\cdot1600\left(\frac{2H}{3}\right)=\frac{7600H}{9}\). 5. Thus, \(H=90\,\text{cm}\), and the frustum height is \(\frac{H}{3}=30\,\text{cm}\).

The pedestal is \(30\,\text{cm}\) tall.