Free Math Worksheets

- Which formula relates cylinder volume, radius, and height? - Rearrange the formula so the height is isolated. - Check that all units are compatible.

1. Use \(V=\pi r^2h\) and solve for height: \(h=\frac{V}{\pi r^2}\). 2. Substitute: \(h=\frac{850}{\pi\cdot4.5^2}\,\text{cm}\approx13.3617\,\text{cm}\). 3. Rounded to the nearest tenth, \(h\approx13.4\,\text{cm}\).

About \(13.4\,\text{cm}\)

- Convert the diameter to a radius. - Use the cylinder volume formula. - Leave \(\pi\) in the exact answer.

1. The radius is \(r=12\div2=6\,\text{cm}\). 2. The volume is \(V=\pi r^2h=\pi\cdot6^2\cdot20=720\pi\,\text{cm}^3\).

\(720\pi\,\text{cm}^3\)

- Use the cylinder volume formula. - Use the \(\pi\) key on a calculator. - Round only the final value.

1. Use \(V=\pi r^2h\). 2. Substitute the dimensions: \(V=\pi\cdot2^2\cdot50=200\pi\,\text{cm}^3\). 3. Therefore, \(V\approx628.32\,\text{cm}^3\).

\(628.32\,\text{cm}^3\)

- Use \(1\,\text{L}=1\,\text{dm}^3\) and convert the height to decimeters. - Which formula gives the volume of a cylinder? - After finding the radius, how do you find the diameter?

1. Convert to compatible units: \(450\,\text{L}=450\,\text{dm}^3\) and \(1.10\,\text{m}=11\,\text{dm}\). 2. Use \(V=\pi r^2h\): \(450=11\pi r^2\). 3. Then \(r^2=\frac{450}{11\pi}\,\text{dm}^2\), so \(r=\sqrt{\frac{450}{11\pi}}\,\text{dm}\approx3.6086\,\text{dm}\). 4. The diameter is \(d=2r\approx7.2171\,\text{dm}=72.171\,\text{cm}\). 5. Rounded to the nearest tenth, \(d\approx72.2\,\text{cm}\).

About \(72.2\,\text{cm}\)

- Use the area formulas for a square and a circle. - How are base area, height, and volume related? - What happens when two different positive values are multiplied by the same positive number?

1. The prism's base area is \(B_P=(5\,\text{cm})^2=25\,\text{cm}^2\). 2. The cylinder's base area is \(B_C=\pi\cdot3^2\,\text{cm}^2=9\pi\,\text{cm}^2\approx28.27\,\text{cm}^2\). 3. The prism's volume is \(V_P=25\,\text{cm}^2\cdot10\,\text{cm}=250\,\text{cm}^3\). 4. The cylinder's volume is \(V_C=9\pi\,\text{cm}^2\cdot10\,\text{cm}=90\pi\,\text{cm}^3\approx282.7\,\text{cm}^3\). 5. The claim is true for positive equal heights because \(V=Bh\). Multiplying two base areas by the same positive height preserves their order.

a) \(B_P=25\,\text{cm}^2\); \(B_C=9\pi\,\text{cm}^2\approx28.27\,\text{cm}^2\) b) \(V_P=250\,\text{cm}^3\); \(V_C=90\pi\,\text{cm}^3\approx282.7\,\text{cm}^3\) c) The claim is true for positive equal heights because volume is proportional to base area when \(h\) is fixed.

- Write the cylinder volume formula first. - Substitute the changed dimension into the formula for each option. - Pay attention to which variable is squared. - Compare each new expression with the original volume.

1. The original volume is \(V=\pi r^2h\). 2. For option A, \(V_A=\pi r^2(2h)=2V\), so the volume doubles. 3. For option B, \(V_B=\pi(2r)^2h=4\pi r^2h=4V\), so the volume is multiplied by \(4\). 4. Only option A produces exactly twice the original volume.

Option A gives \(V_A=2V\) and meets the goal. Option B gives \(V_B=4V\) and does not meet the goal.

- Convert all measurements to compatible units before calculating. - Which length unit is convenient when the volume must be in liters? - Convert minutes to hours at the end.

1. Use decimeters so the volume will be in liters. The radius is \(1.20\,\text{m}=12\,\text{dm}\), and the water depth is \(50\,\text{cm}=5\,\text{dm}\). 2. The volume is \(V=\pi\cdot12^2\cdot5\,\text{dm}^3=720\pi\,\text{dm}^3\approx2261.95\,\text{dm}^3\). 3. Since \(1\,\text{dm}^3=1\,\text{L}\), the pool contains about \(2262\,\text{L}\). 4. The fill time is \(\frac{2261.95\,\text{L}}{15\,\text{L/min}}\approx150.80\,\text{min}\). 5. In hours, \(150.80\div60\approx2.51\,\text{h}\), or about \(2.5\) hours.

a) About \(2262\,\text{L}\) b) About \(2.5\) hours

- Recall the relationship among cubic meters, cubic decimeters, and liters. - Does the water volume change when it is poured into another container? - How does a smaller base area affect the depth for the same volume?

1. The first barrel has radius \(40\,\text{cm}=0.40\,\text{m}\). 2. Its maximum volume is \(V_{\max}=\pi\cdot0.40^2\cdot1.20\,\text{m}^3\approx0.60319\,\text{m}^3\approx603.2\,\text{L}\). 3. At a depth of \(0.45\,\text{m}\), the water volume is \(V=\pi\cdot0.40^2\cdot0.45\,\text{m}^3\approx0.22619\,\text{m}^3\approx226.2\,\text{L}\). 4. The volume stays constant when poured into the second container. With radius \(0.30\,\text{m}\), \(h=\frac{\pi\cdot0.40^2\cdot0.45}{\pi\cdot0.30^2}\,\text{m}=0.80\,\text{m}\). 5. The new water depth is \(80\,\text{cm}\).

a) About \(603.2\,\text{L}\) b) About \(226.2\,\text{L}\) c) \(80\,\text{cm}\)

- How are volume, base area, and height related? - The cylinder's base is a circle. - In \(V=\pi r^2h\), what happens to \(r^2\) when \(r\) is halved?

1. Convert the volume: \(1.5\,\text{L}=1500\,\text{cm}^3\). 2. From \(V=Gh\), \(h=\frac{1500\,\text{cm}^3}{120\,\text{cm}^2}=12.5\,\text{cm}\). 3. Since \(G=\pi r^2\), \(r=\sqrt{\frac{120}{\pi}}\,\text{cm}\approx6.18\,\text{cm}\). 4. Halving the radius multiplies \(r^2\) by \(\left(\frac{1}{2}\right)^2=\frac{1}{4}\), while doubling the height multiplies \(h\) by \(2\). 5. The total volume factor is \(\frac{1}{4}\cdot2=\frac{1}{2}\), so the volume is halved.

a) \(12.5\,\text{cm}\) b) About \(6.18\,\text{cm}\) c) The volume is multiplied by \(\frac{1}{2}\), so it is halved.

- The copper volume does not change when the rod is drawn into wire. - Model the wire as a very long, thin cylinder. - Convert square millimeters to square centimeters carefully. - How many centimeters are in one kilometer?

1. The rod's radius is \(1\,\text{cm}\), so its volume is \(V=\pi\cdot1^2\cdot20\,\text{cm}^3=20\pi\,\text{cm}^3\). 2. Convert the wire's cross-sectional area: \(0.1\,\text{mm}^2=0.001\,\text{cm}^2\). 3. Since \(V=A\ell\), \(\ell=\frac{20\pi\,\text{cm}^3}{0.001\,\text{cm}^2}=20{,}000\pi\,\text{cm}\). 4. This is \(200\pi\,\text{m}\approx628.32\,\text{m}\approx0.628\,\text{km}\).

About \(0.628\,\text{km}\)

- Convert every length to meters. - Find the inside radius by subtracting the wall thickness from the outside radius. - The cross section is an annulus. - Multiply the annular area by the shaft height.

1. The outside radius is \(R=5.4\,\text{m}\). The wall thickness is \(0.45\,\text{m}\), so the inside radius is \(r=5.4-0.45=4.95\,\text{m}\). 2. The annular cross-sectional area is \(G=\pi(R^2-r^2)=\pi(5.4^2-4.95^2)\,\text{m}^2\approx14.632\,\text{m}^2\). 3. The concrete volume is \(V=Gh\approx14.632\cdot165\,\text{m}^3\approx2414.275\,\text{m}^3\). 4. Rounded to the nearest cubic meter, the volume is \(2414\,\text{m}^3\).

About \(2414\,\text{m}^3\)

- Treat each pipe as an outer cylinder with an inner cylinder removed. - Use \(R^2-r^2=(R-r)(R+r)\) in part b. - Which factor is equal for both pipes, and which is larger for pipe B?

1. Convert the length: \(2\,\text{m}=200\,\text{cm}\). 2. For pipe A, \(V_A=\pi(5^2-4^2)\cdot200\,\text{cm}^3=1800\pi\,\text{cm}^3\approx5654.87\,\text{cm}^3\). 3. For pipe B, \(V_B=\pi(10^2-9^2)\cdot200\,\text{cm}^3=3800\pi\,\text{cm}^3\approx11{,}938.05\,\text{cm}^3\). 4. Since \(R^2-r^2=(R-r)(R+r)\), equal wall thickness makes \(R-r\) equal for both pipes. Pipe B has a larger value of \(R+r\), so its annular area and material volume are larger.

a) \(V_A=1800\pi\,\text{cm}^3\approx5654.87\,\text{cm}^3\); \(V_B=3800\pi\,\text{cm}^3\approx11{,}938.05\,\text{cm}^3\) b) Equal wall thickness does not mean equal annular area. Pipe B has a larger average circumference, so it uses more plastic.

- Convert lengths to decimeters when the answer is in liters. - Find the current and target water depths first. - The difference in depths is the height of the additional cylindrical volume. - Convert diameter to radius.

1. Use decimeters: the radius is \(3.5\,\text{dm}\), and the barrel height is \(11\,\text{dm}\). 2. The current water depth is \(\frac{1}{3}\cdot11\,\text{dm}=\frac{11}{3}\,\text{dm}\). 3. The target depth is \(11-1=10\,\text{dm}\), so the increase in depth is \(10-\frac{11}{3}=\frac{19}{3}\,\text{dm}\). 4. The additional volume is \(V=\pi\cdot3.5^2\cdot\frac{19}{3}\,\text{dm}^3\approx243.74\,\text{dm}^3\). 5. Since \(1\,\text{dm}^3=1\,\text{L}\), about \(244\,\text{L}\) must be added.

About \(244\,\text{L}\)

- Use the hollow-cylinder volume formula. - First solve for the annular cross-sectional area or outside radius. - Subtract the inside radius from the outside radius.

1. Let \(R\) be the outside radius. Use \(V=\pi(R^2-r^2)h\): \(600=15\pi(R^2-3^2)\). 2. Solve for \(R^2\): \(R^2=\frac{600}{15\pi}+9\approx21.732\). 3. Thus, \(R\approx\sqrt{21.732}\,\text{cm}\approx4.6618\,\text{cm}\). 4. The wall thickness is \(R-r\approx4.6618-3=1.6618\,\text{cm}\), which rounds to \(1.66\,\text{cm}\).

About \(1.66\,\text{cm}\)

- Start with the cylinder volume formula. - Replace the radius with the given expression involving \(h\). - Notice how doubling a quantity affects its square.

1. Substitute \(r=\frac{1}{4}h\) into \(V=\pi r^2h\): \(V=\pi\left(\frac{1}{4}h\right)^2h=\frac{\pi h^3}{16}\). 2. If the radius is doubled while the height remains constant, then \(V_{\text{new}}=\pi(2r)^2h=4\pi r^2h=4V\). 3. Therefore, the volume is multiplied by \(4\).

a) \(V=\frac{\pi h^3}{16}\) b) The volume is multiplied by \(4\).

- Subtract the hole's volume from the original cylinder's volume. - Convert each diameter to a radius. - For the same material, the percent decrease in mass matches the percent decrease in volume. - Compare the removed volume with the original volume.

1. The outside radius is \(10\,\text{cm}\), and the hole radius is \(4\,\text{cm}\). 2. The original cylinder has volume \(V_{\text{original}}=\pi\cdot10^2\cdot10=1000\pi\,\text{cm}^3\). 3. The removed cylinder has volume \(V_{\text{hole}}=\pi\cdot4^2\cdot10=160\pi\,\text{cm}^3\). 4. The finished volume is \(V=1000\pi-160\pi=840\pi\,\text{cm}^3\approx2638.94\,\text{cm}^3\). 5. Because the material is unchanged, the percent decrease in mass equals the percent decrease in volume: \(\frac{160\pi}{1000\pi}\cdot100\%=16\%\).

a) \(840\pi\,\text{cm}^3\approx2638.94\,\text{cm}^3\) b) The finished part is \(16\%\) lighter.

- Choose the correct volume formula for each solid. - Find the triangular base area before finding the prism volume. - Use \(\pi\) when calculating the cylinder volume. - Compare the three numerical results.

1. Model A has volume \(V_A=10^3=1000\,\text{cm}^3\). 2. Model B has base area \(\frac{1}{2}\cdot20\cdot10=100\,\text{cm}^2\), so \(V_B=100\cdot10=1000\,\text{cm}^3\). 3. Model C has volume \(V_C=\pi\cdot5^2\cdot14=350\pi\,\text{cm}^3\approx1099.56\,\text{cm}^3\). 4. Therefore, \(V_C>V_A=V_B\).

\(C>A=B\)

- Use the cylinder volume formula. - Convert cubic centimeters to liters. - Round only after converting.

1. The cylinder volume is \(V=\pi r^2h=\pi\cdot50^2\cdot120=300{,}000\pi\,\text{cm}^3\). 2. Numerically, \(V\approx942{,}477.8\,\text{cm}^3\). 3. Since \(1000\,\text{cm}^3=1\,\text{L}\), the capacity is \(942.4778\,\text{L}\approx942.5\,\text{L}\).

\(942.5\,\text{L}\)

- Find the volume of each cylinder separately. - Add the two cylinder volumes. - Round the total only at the end.

1. Cylinder a) has volume \(V_a=\pi\cdot6^2\cdot3=108\pi\,\text{cm}^3\). 2. Cylinder b) has volume \(V_b=\pi\cdot3^2\cdot8=72\pi\,\text{cm}^3\). 3. The total volume is \(V=108\pi+72\pi=180\pi\,\text{cm}^3\approx565.5\,\text{cm}^3\).

\(565.5\,\text{cm}^3\)

- First find the volume the jar can hold when full. - Then calculate \(80\%\) of that total volume.

1. The full cylinder volume is \(V=\pi r^2h=\pi\cdot5^2\cdot15=375\pi\approx1{,}178.0972\,\text{cm}^3\). 2. The flour occupies \(80\%\) of the total volume: \(0.80\cdot375\pi=300\pi\approx942.4778\,\text{cm}^3\). 3. Rounded to the nearest tenth, the volume is \(942.5\,\text{cm}^3\).

The flour occupies approximately \(942.5\,\text{cm}^3\).

- Write a volume formula for each candle and identify what is equal. - Set the volumes equal. - Cancel common factors. - How must a squared factor change when another factor is cut in half but the product stays constant?

1. Let \(V_1=\pi r_1^2h_1\) and \(V_2=\pi r_2^2h_2\). 2. Since the volumes are equal and \(h_2=\frac{1}{2}h_1\), \(\pi r_1^2h_1=\pi r_2^2\left(\frac{1}{2}h_1\right)\). 3. Cancel \(\pi\) and \(h_1\): \(r_1^2=\frac{1}{2}r_2^2\). 4. Thus, \(r_2^2=2r_1^2\), so \(r_2=\sqrt{2}r_1\). 5. The radius scale factor is \(\sqrt{2}\approx1.41\).

The second radius is \(\sqrt{2}\) times the first radius, and \(\sqrt{2}\approx1.41\).

- Express a \(25\%\) increase as a scale factor. - Set the original and new cylinder volumes equal. - Remember that the radius is squared in the volume formula. - Compare the new height with \(100\%\) of the original height.

1. Let the original dimensions be \(r_1\) and \(h_1\). The new radius is \(r_2=1.25r_1\). 2. Equal volumes give \(\pi r_1^2h_1=\pi r_2^2h_2\). 3. Substitute \(r_2=1.25r_1\): \(\pi r_1^2h_1=\pi(1.25r_1)^2h_2=1.5625\pi r_1^2h_2\). 4. Divide by \(\pi r_1^2\): \(h_1=1.5625h_2\), so \(h_2=0.64h_1\). 5. The new height is \(64\%\) of the original height, so the decrease is \(100\%-64\%=36\%\).

The height must be decreased by \(36\%\).