Free math worksheets

- Read the radius from the diagram. - Use the sphere volume formula. - Keep \(\pi\) in the exact answer.

From the diagram, the radius is \(3\,\text{cm}\). Using \(V=\frac{4}{3}\pi r^3\), \(V=\frac{4}{3}\pi\cdot3^3=36\pi\,\text{cm}^3\).

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

- Think about the standard volume relationship for a sphere rather than a surface-area formula. - Check that the radius appears with the power expected for a three-dimensional measurement.

1. The volume formula for a sphere is \(V=\frac{4}{3}\pi r^3\).

\(V=\frac{4}{3}\pi r^3\)

- Convert the diameter to a radius first. - The radius is cubed in the sphere volume formula. - Leave \(\pi\) in the exact answer.

The radius is \(10\div2=5\,\text{cm}\). Then \(V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi\cdot5^3=\frac{500\pi}{3}\,\text{cm}^3\).

\(\frac{500\pi}{3}\,\text{cm}^3\)

- Decide whether the displayed measurement is a radius or a diameter. - Convert the diameter to a radius before substituting. - The radius is cubed in the sphere volume formula.

From the diagram, the diameter is \(12\,\text{cm}\), so the radius is \(6\,\text{cm}\). Therefore, \(V=\frac{4}{3}\pi\cdot6^3=288\pi\,\text{cm}^3\).

\(288\pi\,\text{cm}^3\)

- Find the volume of the corresponding full sphere first. - Use the fact that a hemisphere is one of two equal halves of a sphere.

1. A full sphere of radius \(6\,\text{cm}\) has volume \(V=\frac{4}{3}\pi\cdot6^3=288\pi\,\text{cm}^3\). 2. A hemisphere has half that volume, so \(V=144\pi\,\text{cm}^3\).

\(144\pi\,\text{cm}^3\)

- Substitute the known volume into the sphere formula. - Isolate \(r^3\) before taking a cube root. - Once the radius is known, relate it to the diameter.

Substitute into \(V=\frac{4}{3}\pi r^3\): \(288\pi=\frac{4}{3}\pi r^3\). Cancel \(\pi\) and multiply by \(\frac{3}{4}\): \(r^3=216\). Thus \(r=6\,\text{cm}\), so the diameter is \(12\,\text{cm}\).

Radius: \(6\,\text{cm}\); diameter: \(12\,\text{cm}\)

- Keep the original radius as a variable. - Replace the new radius by \(3r\) in the sphere formula. - Pay attention to the exponent on the radius.

Let sphere A have radius \(r\), so \(V_A=\frac{4}{3}\pi r^3\). Sphere B has radius \(3r\), so \(V_B=\frac{4}{3}\pi(3r)^3=27\left(\frac{4}{3}\pi r^3\right)=27V_A\). Therefore, tripling the radius multiplies the volume by \(27\).

Sphere B has \(27\) times the volume of sphere A.

- Read the sphere's radius from the diagram. - Cube the radius before multiplying by the remaining factors. - Round only the final volume.

From the diagram, the radius is \(4.5\,\text{cm}\). Then \(V=\frac{4}{3}\pi(4.5)^3=121.5\pi\,\text{cm}^3\approx381.7\,\text{cm}^3\).

About \(381.7\,\text{cm}^3\)

- Represent the new radius as a scale factor times the original radius. - Track how that scale factor is affected by the exponent in the sphere volume formula. - Work backward from the given volume factor.

1. Sphere volume is proportional to the cube of the radius, so the volume scale factor is \(k^3\). 2. Since the volume factor is \(8\), \(k^3=8\). 3. Therefore, \(k=2\).

\(k=2\)

- Read the dimensions from the two solids before computing either volume. - Use the sphere and cylinder volume formulas separately. - For part b, set the two general formulas equal. - Cancel only factors that are nonzero for a genuine solid.

1. From the diagram, both solids have radius \(3\,\text{cm}\), and the cylinder has height \(4\,\text{cm}\). The sphere volume is \(\frac{4}{3}\pi\cdot3^3=36\pi\,\text{cm}^3\). The cylinder volume is \(\pi\cdot3^2\cdot4=36\pi\,\text{cm}^3\), so they are equal. 2. Set the general volumes equal: \(\frac{4}{3}\pi r^3=\pi r^2h\). For \(r>0\), cancel \(\pi r^2\) to get \(h=\frac{4}{3}r\).

a) Both volumes are \(36\pi\,\text{cm}^3\). b) \(h=\frac{4}{3}r\)