Free Math Worksheets

- Convert the diameter to a radius first. - Use the sphere volume and surface-area formulas. - Check the exponents carefully when entering the calculations.

1. The radius is \(r=12.4\div2=6.2\,\text{cm}\). 2. The volume is \(V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi\cdot6.2^3\approx998.3060\,\text{cm}^3\). 3. The surface area is \(S=4\pi r^2=4\pi\cdot6.2^2\approx483.0513\,\text{cm}^2\).

\(V\approx998.3\,\text{cm}^3\); \(S\approx483.1\,\text{cm}^2\)

- Replace every \(B_2\) in the formula with \(B_1\). - Combine the terms inside the parentheses. - Identify the familiar volume formula that results. - Consider what happens to the lateral faces when the two bases are the same size.

1. Substitute \(B_2=B_1\): \(V=\frac{h}{3}\left(B_1+\sqrt{B_1^2}+B_1\right)\). 2. Because an area is nonnegative, \(\sqrt{B_1^2}=B_1\). Thus, \(V=\frac{h}{3}\left(3B_1\right)=B_1h\). 3. The formula \(V=B_1h\) is the volume formula for a prism. When the parallel, similar bases have equal areas, the solid does not taper, so it becomes a prism.

1. \(V=B_1h\) 2. The solid is a prism because its parallel, similar bases are the same size, so it does not taper.

- 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 \(432\,\text{m}^3\).

- 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. - Use the requested rounded radius in the surface-area formula. - Track the units for length, area, and volume.

1. Solve the volume formula for the radius: \(r=\sqrt[3]{\frac{3V}{4\pi}}\). Then \(r=\sqrt[3]{\frac{3\cdot14.5}{4\pi}}\approx1.51\,\text{m}\). 2. Using \(r=1.51\,\text{m}\), \(S=4\pi(1.51)^2\approx28.65\,\text{m}^2\).

1. \(r\approx1.51\,\text{m}\) 2. \(S\approx28.65\,\text{m}^2\)

- Rearrange the sphere surface-area formula to isolate the radius. - Substitute the rounded radius into the volume formula. - Follow the problem’s rounding instruction at each stage.

1. From \(S=4\pi r^2\), \(r=\sqrt{\frac{800}{4\pi}}\approx7.9788\,\text{cm}\). Following the instruction, round to \(r\approx7.98\,\text{cm}\). 2. Use the rounded radius: \(V=\frac{4}{3}\pi(7.98)^3\approx2128.62\,\text{cm}^3\).

\(r\approx7.98\,\text{cm}\) \(V\approx2128.62\,\text{cm}^3\)

- 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, \(S=4\pi r^2\approx25.3477\,\text{dm}^2\), so \(S\approx25.35\,\text{dm}^2\).

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

- Verify that the upper and lower rectangles are similar. - Extend the slanted sides to model the frustum as the difference of two similar pyramids. - Use the scale factor to relate the two pyramid heights. - Convert cubic centimeters to liters at the end.

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 slanted sides downward to form a smaller and a larger similar rectangular pyramid. Let \(x\) be the height from the common apex to the lower base. Then \(\frac{x+50}{x}=1.5\), so \(x=100\,\text{cm}\). The larger pyramid has height \(150\,\text{cm}\). 3. The upper and lower base areas are \(2{,}700\,\text{cm}^2\) and \(1{,}200\,\text{cm}^2\), respectively. 4. Subtract the smaller pyramid's volume from the larger pyramid's volume: \(V=\frac{1}{3}\cdot2{,}700\cdot150-\frac{1}{3}\cdot1{,}200\cdot100=135{,}000-40{,}000=95{,}000\,\text{cm}^3\). 5. Since \(1{,}000\,\text{cm}^3=1\,\text{L}\), the planter holds \(95\,\text{L}\).

The planter holds \(95\,\text{L}\) of soil.

- 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 side lengths of the two rectangular bases. - Extend the sides to model the frustum as a full pyramid minus a smaller similar pyramid. - Use the scale factor to relate the two pyramid heights. - Subtract the two pyramid volumes.

1. The corresponding side-length ratios are \(5.0\div12.5=3.2\div8.0=0.4\), so the small removed pyramid has scale factor \(0.4\) relative to the full pyramid. 2. Let \(H\) be the full pyramid height. The removed pyramid height is \(0.4H\), so the frustum height is \(H-0.4H=0.6H=9\). 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: \(V=\frac{1}{3}\cdot100\cdot15-\frac{1}{3}\cdot16\cdot6=500-32=468\,\text{cm}^3\).

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

- 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, \(V=\frac{1}{3}\cdot3.60\cdot2.70-\frac{1}{3}\cdot0.40\cdot0.90=3.24-0.12=3.12\,\text{m}^3\).

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

- First calculate \(B_1\) and \(B_2\) from the square edge lengths. - Consider what shape results when the top face shrinks to a single point. - Substitute zero for \(B_2\) in the general formula.

1. When \(b=4\,\text{cm}\), \(B_1=8^2=64\,\text{cm}^2\) and \(B_2=4^2=16\,\text{cm}^2\). Therefore, \(V=\frac{12}{3}\left(64+\sqrt{64\cdot16}+16\right)=4\left(64+32+16\right)=448\,\text{cm}^3\). 2. When \(b=0\,\text{cm}\), \(B_2=0\). Then \(V=\frac{12}{3}\left(64+\sqrt{64\cdot0}+0\right)=4\cdot64=256\,\text{cm}^3\). 3. When the top area shrinks to zero, the solid ends at an apex and becomes a pyramid. In general, setting \(B_2=0\) gives \(V=\frac{h}{3}(B_1+0+0)=\frac{1}{3}B_1h\).

1. \(V=448\,\text{cm}^3\) 2. \(V=256\,\text{cm}^3\) 3. The solid becomes a pyramid, and the formula becomes \(V=\frac{1}{3}B_1h\).

- 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 \(17.4\,\text{L}\).

- Use the proportional relationship between the radii and heights of the similar cones. - The large cone's height includes both the frustum and the small cone. - Use the cone volume formula for each cone before subtracting.

1. Similar triangles give \(\frac{x}{4}=\frac{x+6}{8}\). 2. Solving, \(8x=4x+24\), so \(x=6\,\text{cm}\). The large cone's height is \(12\,\text{cm}\). 3. The cone volumes are \(V_{\text{large}}=\frac{1}{3}\pi\cdot8^2\cdot12=256\pi\,\text{cm}^3\) and \(V_{\text{small}}=\frac{1}{3}\pi\cdot4^2\cdot6=32\pi\,\text{cm}^3\). 4. Therefore, the frustum's volume is \(256\pi-32\pi=224\pi\,\text{cm}^3\approx703.72\,\text{cm}^3\). 5. Direct substitution gives \(V=\frac{6\pi}{3}\left(8^2+8\cdot4+4^2\right)=2\pi\left(64+32+16\right)=224\pi\,\text{cm}^3\), confirming the result.

a) \(x=6\,\text{cm}\) b) \(V=224\pi\,\text{cm}^3\approx703.72\,\text{cm}^3\) c) Direct substitution also gives \(224\pi\,\text{cm}^3\).

- 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) \(V\approx21.48\,\text{cm}^3\)

- Compare the dimensions of the upper and lower rectangles. - Treat the frustum as a full pyramid with a smaller similar pyramid removed. - Use similarity to determine the two pyramid heights. - Convert cubic centimeters to liters at the end.

1. The upper base dimensions are half the corresponding lower base dimensions. Therefore, the removed 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 \(\frac{H}{2}=50\). Thus, \(H=100\,\text{cm}\). 3. The lower and upper base areas are \(80\cdot60=4{,}800\,\text{cm}^2\) and \(40\cdot30=1{,}200\,\text{cm}^2\). 4. The frustum volume is \(V=\frac{1}{3}\cdot4{,}800\cdot100-\frac{1}{3}\cdot1{,}200\cdot50=160{,}000-20{,}000=140{,}000\,\text{cm}^3\). 5. Since \(1{,}000\,\text{cm}^3=1\,\text{L}\), the volume is \(140\,\text{L}\).

The pedestal volume is \(140\,\text{L}\).

- 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 \(6\div9=\frac{2}{3}\). 4. Its base edge is \(\frac{2}{3}\cdot6=4\,\text{cm}\), and its volume is \(V_s=\frac{1}{3}\cdot4^2\cdot6=32\,\text{cm}^3\). 5. 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 \(V_s=\frac{4}{3}\pi\cdot6^3=288\pi\approx904.78\,\text{cm}^3\). 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 \(V=\frac{1}{3}\cdot768\cdot20-\frac{1}{3}\cdot432\cdot15=5{,}120-2{,}160=2{,}960\,\text{cm}^3\).

The baking pan has a capacity of \(2{,}960\,\text{cm}^3\).

- Compare the two square edge lengths. - Use similarity to relate the removed pyramid height to the full pyramid height. - Subtract the smaller pyramid volume from the larger one.

1. The upper base edge is one-third of the lower base edge. Therefore, the small removed pyramid has height \(\frac{H}{3}\) if the full pyramid height is \(H\). 2. The frustum height is \(H-\frac{H}{3}=\frac{2H}{3}=9\), so \(H=13.5\,\text{cm}\), and the removed pyramid height is \(4.5\,\text{cm}\). 3. The full pyramid volume is \(\frac{1}{3}\cdot12^2\cdot13.5=648\,\text{cm}^3\). 4. The removed pyramid volume is \(\frac{1}{3}\cdot4^2\cdot4.5=24\,\text{cm}^3\). 5. The frustum volume is \(648-24=624\,\text{cm}^3\).

The volume is \(624\,\text{cm}^3\).

- 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 \(73.5\,\text{cm}^3\).

- 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 \(k=\frac{d_B}{d_A}=\sqrt[3]{6}\approx1.8171\).

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

- 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 \(7.54\,\text{dm}\).

- Set the area of each square equal to the area of the corresponding circle. - Simplify the square root by using the squared factors inside it. - Look for a common factor in all three terms.

1. For equal base areas, \(a_1^2=\pi r_1^2\), so \(r_1=\frac{a_1}{\sqrt{\pi}}\). For equal top areas, \(a_2^2=\pi r_2^2\), so \(r_2=\frac{a_2}{\sqrt{\pi}}\). 2. For the conical frustum, substitute \(B_1=\pi r_1^2\) and \(B_2=\pi r_2^2\): \(V=\frac{h}{3}\left(\pi r_1^2+\sqrt{(\pi r_1^2)(\pi r_2^2)}+\pi r_2^2\right)\). 3. Because radii are nonnegative, \(\sqrt{\pi^2r_1^2r_2^2}=\pi r_1r_2\). 4. Factoring out \(\pi\) gives \(V=\frac{\pi h}{3}\left(r_1^2+r_1r_2+r_2^2\right)\).

a) \(r_1=\frac{a_1}{\sqrt{\pi}}\) and \(r_2=\frac{a_2}{\sqrt{\pi}}\) b) Substitution and simplification give \(V=\frac{\pi h}{3}\left(r_1^2+r_1r_2+r_2^2\right)\).

- Use the binomial expansion when squaring the average radius. - Put the two expressions over a common denominator. - Factor the difference and determine when it equals zero.

1. Substituting \(r_1=r_2=r\) gives \(V=\frac{1}{3}\pi h(r^2+r^2+r^2)=\pi r^2h\), the cylinder volume formula. 2. The proposed formula becomes \(V_{\text{estimate}}=\pi h\left(\frac{r_1+r_2}{2}\right)^2=\frac{\pi h}{4}(r_1^2+2r_1r_2+r_2^2)\). 3. Subtracting the estimate from the exact formula gives \(V-V_{\text{estimate}}=\frac{\pi h}{12}\left(4r_1^2+4r_1r_2+4r_2^2-3r_1^2-6r_1r_2-3r_2^2\right)\). 4. Therefore, \(V-V_{\text{estimate}}=\frac{\pi h}{12}(r_1-r_2)^2\). This difference is positive when \(r_1\ne r_2\), so the proposed cylinder formula is not exact and underestimates the frustum volume.

1. The formula simplifies to \(V=\pi r^2h\). 2. The proposed formula is not exact when \(r_1\ne r_2\). It underestimates the volume by \(\frac{\pi h}{12}(r_1-r_2)^2\).

- Identify the segment in the cube that becomes a tetrahedron edge. - Use the general volume formula for a pyramid. - Determine the shape and area of a corner pyramid's base. - Solve the edge-length relationship for \(a\) before substituting.

1. Each tetrahedron edge is a face diagonal of the cube. By the Pythagorean theorem, \(b^2=a^2+a^2=2a^2\), so \(b=a\sqrt{2}\). 2. A corner pyramid has a right-triangular base with area \(\frac{1}{2}a^2\) and perpendicular height \(a\). Thus, \(V_{\text{corner}}=\frac{1}{3}\cdot\frac{1}{2}a^2\cdot a=\frac{1}{6}a^3\). 3. Subtracting the four corner pyramids from the cube gives \(V_{\text{tetrahedron}}=a^3-4\cdot\frac{1}{6}a^3=\frac{1}{3}a^3\). 4. From \(b=a\sqrt{2}\), \(a=\frac{b}{\sqrt{2}}\). Therefore, \(V=\frac{1}{3}\left(\frac{b}{\sqrt{2}}\right)^3=\frac{b^3}{6\sqrt{2}}=\frac{\sqrt{2}}{12}b^3\).

a) \(b=a\sqrt{2}\) b) \(V_{\text{corner}}=\frac{1}{6}a^3\) and \(V_{\text{tetrahedron}}=\frac{1}{3}a^3\) c) \(V=\frac{\sqrt{2}}{12}b^3\)

- Model the cup as the difference of two similar cones. - Use the radius ratio to express the cone heights in terms of the cup height. - Set the volume difference equal to the given capacity and solve.

1. Extend the sides below the cup to form two similar cones. Let the cup height be \(h\). 2. Since the radii are in the ratio \(9\) to \(6\), or \(3\) to \(2\), the smaller cone height is \(2h\), and the larger cone height is \(3h\). 3. The volume difference is \(1{,}790.7=\frac{1}{3}\pi\cdot9^2\cdot3h-\frac{1}{3}\pi\cdot6^2\cdot2h=57\pi h\). 4. Thus, \(h=\frac{1{,}790.7}{57\pi}\approx9.99996\,\text{cm}\), which rounds to \(10.0\,\text{cm}\).

The measuring cup height is \(10.0\,\text{cm}\).

- 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 \(8.0\,\text{cm}\).

- Use a vertical cross section and similarity to relate the liquid-surface radius to the fill height. - Substitute that radius expression into the cone volume formula. - For part c, first find the funnel's full capacity. - Set the volume function equal to half the maximum volume and solve for \(h\).

1. The full cone has radius \(6\,\text{cm}\) and height \(18\,\text{cm}\). By similarity, \(\frac{r}{h}=\frac{6}{18}=\frac{1}{3}\), so the liquid-surface radius is \(r(h)=\frac{h}{3}\). 2. Substitute into the cone volume formula: \(V(h)=\frac{1}{3}\pi\left(\frac{h}{3}\right)^2h=\frac{\pi}{27}h^3\). 3. At \(h=9\), \(V(9)=\frac{\pi}{27}\cdot9^3=27\pi\approx84.8\,\text{cm}^3\). 4. The maximum volume is \(V(18)=216\pi\,\text{cm}^3\), so half the capacity is \(108\pi\,\text{cm}^3\). 5. Solve \(\frac{\pi}{27}h^3=108\pi\): \(h^3=2{,}916\), so \(h=\sqrt[3]{2{,}916}\approx14.3\,\text{cm}\).

a) \(V(h)=\frac{\pi}{27}h^3\) b) \(V(9)=27\pi\,\text{cm}^3\approx84.8\,\text{cm}^3\) c) \(h=\sqrt[3]{2{,}916}\,\text{cm}\approx14.3\,\text{cm}\)

- 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 \(3{,}600\,\text{cm}^2\) and \(1{,}600\,\text{cm}^2\). 4. The volume difference is \(76{,}000=\frac{1}{3}\cdot3{,}600H-\frac{1}{3}\cdot1{,}600\left(\frac{2H}{3}\right)=\frac{7{,}600H}{9}\). 5. Thus, \(H=90\,\text{cm}\), and the frustum height is \(\frac{H}{3}=30\,\text{cm}\).

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