Free math worksheets

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

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

- 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 base measurement shown is a radius or a diameter. - Convert the diameter to a radius before substituting. - Use the perpendicular height shown in the diagram.

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

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

- 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, \(r=4\,\text{cm}\).

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

- 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.