- Read the radius and height from the diagram. - A cone has one-third the volume of a cylinder with the same base and height. - Keep \(\pi\) in the exact answer.

From the diagram, \(r=3\,\text{cm}\) and \(h=4\,\text{cm}\). Using \(V=\frac{1}{3}\pi r^2h\), \(V=\frac{1}{3}\pi\cdot3^2\cdot4=12\pi\,\text{cm}^3\).

\(12\pi\,\text{cm}^3\)

- The base area is already given. - A cone has one-third the volume of a prism or cylinder with the same base area and height.

1. Use \(V=\frac{1}{3}Bh\). 2. \(V=\frac{1}{3}\cdot27\cdot6=54\,\text{cm}^3\).

\(54\,\text{cm}^3\)

- Check which segment forms a right angle with the base. - A sloping edge and the shortest distance from the tip to the base are different lengths.

1. Segment \(TH\) meets \(AB\), the line of the base diameter, at a right angle and reaches the base's center. 2. Thus \(TH\) is the perpendicular height of the cone. Segment \(TA\) is its slant height.

\(TH\) is the perpendicular height because it meets the base diameter at \(H\) at a right angle. \(TA\) is a slant height.

- Try choosing a simple value for the common product \(Bh\). - Does each formula give the same volume for that value? How could you compare the results?

1. The two formulas use the same base area \(B\) and height \(h\). 2. The cylinder has volume \(Bh\), three times the cone's \(\frac{1}{3}Bh\).

The cylinder holds three times as much as the cone, because the cone has one-third of its volume for the same base and height.

- Convert the diameter to a radius before using the formula. - Identify the radius and perpendicular height. - Leave \(\pi\) unevaluated for an exact answer.

The radius is \(10\div2=5\,\text{cm}\). Then \(V=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi\cdot5^2\cdot12=100\pi\,\text{cm}^3\).

\(100\pi\,\text{cm}^3\)

- Decide whether the marked base measurement goes across the entire circle or from center to edge. - Convert a diameter to a radius if needed. - Use the perpendicular height in the cone formula.

1. The diagram labels the base diameter \(8\,\text{cm}\), so the radius is \(4\,\text{cm}\). 2. The perpendicular height is \(9\,\text{cm}\). 3. \(V=\frac{1}{3}\pi\cdot4^2\cdot9=48\pi\,\text{cm}^3\).

\(48\pi\,\text{cm}^3\).

- Treat the radius as the unknown in the cone volume formula. - Isolate \(r^2\) before taking a square root. - Use the positive root for a length.

1. Substitute into \(V=\frac{1}{3}\pi r^2h\): \(48\pi=\frac{1}{3}\pi r^2(9)=3\pi r^2\). 2. Divide by \(3\pi\): \(r^2=16\). 3. Since a radius is positive, \(r=4\,\text{cm}\).

\(4\,\text{cm}\)

- What stays the same about the two solids? - Write each volume formula using the same base area and height. - Compare the formulas before using the given volume.

1. For the same base area \(B\) and height \(h\), the cylinder volume is \(Bh\) and the cone volume is \(\frac{1}{3}Bh\). 2. Thus the cone's volume is one-third of \(54\pi\,\text{cm}^3\), or \(18\pi\,\text{cm}^3\).

\(18\pi\,\text{cm}^3\). A cone has one-third the volume of a cylinder with the same base area and height.

- Does the base measurement travel from the center to the rim, or across the whole circle? - Would your answer be reasonable if it exceeded the volume of a cylinder with the same base and height?

1. Read the radius \(r=4\,\text{cm}\) and perpendicular height \(h=6\,\text{cm}\). 2. Substitute: \(V=\frac{1}{3}\pi\cdot4^2\cdot6\). 3. Simplify to \(V=32\pi\,\text{cm}^3\).

\(V=\frac{1}{3}\pi\cdot4^2\cdot6=32\pi\,\text{cm}^3\).

- If their base areas were equal, how would the cone and cylinder volumes compare? - How might a change in base area affect that comparison?

1. A cone's volume is one-third its base area times height: \(V_{\text{cone}}=\frac{1}{3}\cdot30\cdot4=40\,\text{cm}^3\). 2. The cylinder's volume is its base area times height: \(V_{\text{cylinder}}=10\cdot4=40\,\text{cm}^3\). 3. The volumes are equal even though the cone has a larger base area.

Yes. \(V_{\text{cone}}=\frac{1}{3}\cdot30\cdot4=40\,\text{cm}^3\) and \(V_{\text{cylinder}}=10\cdot4=40\,\text{cm}^3\).

- Does the question ask for the radius, height, or base area? - Put the known height and volume into the base-area form of the cone formula. - Check that the unit for base area is squared, not cubed.

1. Substitute the volume and height: \(60=\frac{1}{3}B\cdot5\). 2. Multiply by \(3\) and divide by \(5\): \(B=36\,\text{cm}^2\).

\(36\,\text{cm}^2\).

- Substitute the known volume and radius into the cone formula. - Treat the height as the unknown. - Cancel common factors before solving the equation.

Substitute into \(V=\frac{1}{3}\pi r^2h\): \(150\pi=\frac{1}{3}\pi\cdot25h\). Cancel \(\pi\) and multiply by \(3\): \(450=25h\). Therefore, \(h=18\,\text{cm}\).

\(18\,\text{cm}\)

- Write the cone and cylinder volume formulas side by side. - Notice which factors are identical in the two formulas. - Compare the coefficients after substituting the same radius and height.

The cone volume is \(V_c=\frac{1}{3}\pi\cdot4^2\cdot9=48\pi\,\text{cm}^3\). The cylinder volume is \(V_y=\pi\cdot4^2\cdot9=144\pi\,\text{cm}^3\). Therefore, the cone has one-third the volume of the cylinder, or the cylinder has three times the cone's volume.

Cone: \(48\pi\,\text{cm}^3\); cylinder: \(144\pi\,\text{cm}^3\). The cone's volume is one-third of the cylinder's.

- Read the radius and height from the diagram. - Find the cone's volume in cubic centimeters. - Use the equivalence between cubic centimeters and milliliters. - Round only at the end.

From the diagram, \(r=6\,\text{cm}\) and \(h=12\,\text{cm}\). The volume is \(V=\frac{1}{3}\pi\cdot6^2\cdot12=144\pi\,\text{cm}^3\approx452.4\,\text{cm}^3\). Since \(1\,\text{cm}^3=1\,\text{mL}\), the cup holds about \(452.4\,\text{mL}\).

About \(452.4\,\text{mL}\)

- Find the volume of sand in the cone first. - The amount of sand does not change when it is transferred. - Use the receiving cylinder's radius with an unknown depth.

1. From the diagram, the cone has radius \(3\,\text{cm}\) and height \(12\,\text{cm}\), so its volume is \(V=\frac{1}{3}\pi\cdot3^2\cdot12=36\pi\,\text{cm}^3\). 2. The sand volume is unchanged when it is transferred. If the cylinder depth is \(h\), then \(36\pi=\pi\cdot3^2h=9\pi h\). 3. Therefore, \(h=4\,\text{cm}\).

\(4\,\text{cm}\)

- Which segment in the cross-section meets the base at a right angle? - Relate the radius, the unknown perpendicular height, and the labeled slanted edge in the right triangle. - Use the cone volume formula only after identifying the perpendicular height.

1. The height in the cone volume formula is the perpendicular distance from the apex to the base plane, not the slanted edge from the apex to the rim. 2. In the cross-section, the radius, perpendicular height, and slanted edge form a right triangle. With radius \(5\,\text{cm}\) and slanted edge \(13\,\text{cm}\), \(h^2+5^2=13^2\). 3. Thus \(h^2=144\), so \(h=12\,\text{cm}\). 4. The cone volume is \(V=\frac{1}{3}\pi(5)^2(12)=100\pi\,\text{cm}^3\).

a) The slanted edge is not perpendicular to the base, so it is not the height used in the volume formula. b) \(12\,\text{cm}\) c) \(100\pi\,\text{cm}^3\)

- Read both radii and heights from the diagrams. - Compare how the radius and height change separately. - In the cone formula, which dimension is squared? - Combine the scale factors before deciding whether they cancel.

From the diagram, cone A has \(r=3\) and \(h=8\), while cone B has \(r=6\) and \(h=4\). Thus \(V_A=\frac{1}{3}\pi\cdot3^2\cdot8=24\pi\) and \(V_B=\frac{1}{3}\pi\cdot6^2\cdot4=48\pi\). Therefore, \(V_B=2V_A\). The error is treating radius as a linear factor in the volume formula; radius is squared, so doubling it multiplies the radius-squared factor by \(4\), while halving height multiplies volume by \(\frac{1}{2}\). The combined factor is \(4\cdot\frac{1}{2}=2\).

Cone B has twice the volume of cone A. The claim overlooks the radius-squared factor.