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, \(h=4\,\text{cm}\).

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

- 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 student's 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 radius-squared factor was overlooked.