Free math worksheets

- The base area is already given, so you do not need to find the radius. - Multiply the base area by the perpendicular height.

1. Use \(V=Bh\). 2. \(V=18\cdot5=90\,\text{cm}^3\).

\(90\,\text{cm}^3\)

- Identify which labeled dimension is the radius. - Substitute the labeled values into the cylinder volume formula.

1. From the diagram, \(r=2\,\text{cm}\) and \(h=7\,\text{cm}\). 2. Substituting into \(V=\pi r^2h\) gives \(V=\pi(2)^2(7)\,\text{cm}^3\).

\(\pi(2)^2(7)\,\text{cm}^3\)

- 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: \(h=\frac{850}{\pi\cdot4.5^2}\,\text{cm}\approx13.3612\,\text{cm}\). 3. Rounded to the nearest tenth, \(h\approx13.4\,\text{cm}\).

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

- Read the radius and height directly from the diagram. - Use the cylinder volume formula. - Leave \(\pi\) in the exact answer.

1. From the diagram, the radius is \(6\,\text{cm}\) and the height is \(20\,\text{cm}\). 2. The volume is \(V=\pi r^2h=\pi\cdot6^2\cdot20=720\pi\,\text{cm}^3\).

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

- Read the radius and height from the diagram. - Use the cylinder volume formula. - Round only the final value.

1. From the diagram, the radius is \(2\,\text{cm}\) and the height is \(50\,\text{cm}\). 2. Use \(V=\pi r^2h\): \(V=\pi\cdot2^2\cdot50=200\pi\,\text{cm}^3\). 3. Therefore, \(V\approx628.32\,\text{cm}^3\).

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

- Imagine pouring the same amount of water into a narrow container and a wide container. - Use the relationship \(V=Bh\) between volume, base area, and height. - For the same volume, a larger base area produces a proportionally smaller height increase.

1. The same volume of water spreads across a larger base in container A, so the water level rises faster in container B. 2. For equal volumes, \(V=Bh\). Since the base area of A is three times the base area of B, its height increase is one-third as great: \(15\div 3=5\). Container A rises \(5\,\text{cm}\) per minute. 3. The equations are \(h_A=5t\) and \(h_B=15t\).

a) Container B, because it has the smaller base area. b) \(5\,\text{cm}\) c) \(h_A=5t\) and \(h_B=15t\)

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

a) \(B_P=25\,\text{cm}^2\); \(B_C=9\pi\,\text{cm}^2\) b) \(V_P=250\,\text{cm}^3\); \(V_C=90\pi\,\text{cm}^3\) 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? - Use the hose rate after finding the water volume. - Convert minutes to hours at the end, and round only as requested.

1. Use decimeters so the volume is 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 water volume is \(V=\pi\cdot12^2\cdot5=720\pi\,\text{dm}^3\approx2261.95\,\text{L}\), which rounds to \(2262\,\text{L}\). 3. Using the unrounded volume, the fill time is \(\frac{720\pi}{15}=48\pi\) minutes. 4. In hours, this is \(\frac{48\pi}{60}=0.8\pi\approx2.513\) hours, which rounds to \(2.5\) hours.

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

- Convert the diameter to a radius before using the cylinder formula. - Recall the relationship between cubic meters 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 \(0.40\,\text{m}\). Its maximum volume is \(\pi(0.40)^2(1.20)\approx0.60319\,\text{m}^3\approx603.2\,\text{L}\). 2. At a depth of \(0.45\,\text{m}\), the water volume is \(\pi(0.40)^2(0.45)\approx0.22619\,\text{m}^3\approx226.2\,\text{L}\). 3. The water volume is unchanged when it is poured into the second container. With radius \(0.30\,\text{m}\), \(h=\frac{\pi(0.40)^2(0.45)}{\pi(0.30)^2}=0.80\,\text{m}=80\,\text{cm}\).

a) \(603.2\,\text{L}\) b) \(226.2\,\text{L}\) 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: \(1.5\,\text{L}=1500\,\text{cm}^3\). From \(V=Bh\), \(h=\frac{1500}{120}=12.5\,\text{cm}\). 2. Since \(B=\pi r^2\), \(r=\sqrt{\frac{120}{\pi}}\,\text{cm}\approx6.18\,\text{cm}\). 3. Halving the radius multiplies \(r^2\) by \(\frac{1}{4}\), while doubling the height multiplies \(h\) by \(2\). The total volume factor is \(\frac{1}{4}\cdot2=\frac{1}{2}\), so the volume is halved.

a) \(12.5\,\text{cm}\) b) \(6.18\,\text{cm}\) 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 \(200\pi\,\text{m}\approx628.32\,\text{m}\approx0.628\,\text{km}\).

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

- 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 \(A=\pi(R^2-r^2)=\pi(5.4^2-4.95^2)\,\text{m}^2\approx14.632\,\text{m}^2\). 3. The concrete volume is \(V=Ah\approx14.632\cdot165\,\text{m}^3\approx2414.275\,\text{m}^3\). 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: \(2\,\text{m}=200\,\text{cm}\). For pipe A, \(V_A=\pi(5^2-4^2)\cdot200=1800\pi\,\text{cm}^3\). 2. For pipe B, \(V_B=\pi(10^2-9^2)\cdot200=3800\pi\,\text{cm}^3\). 3. 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 cross-sectional area and material volume are larger.

a) \(V_A=1800\pi\,\text{cm}^3\); \(V_B=3800\pi\,\text{cm}^3\) 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, \(V_{\text{new}}=\pi(2r)^2h=4\pi r^2h=4V\). 3. Therefore, doubling the radius while keeping the height fixed multiplies the volume 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 \(4\,\text{cm}\). The original volume is \(1000\pi\,\text{cm}^3\), and the removed volume is \(160\pi\,\text{cm}^3\). 2. The finished volume is \(1000\pi-160\pi=840\pi\,\text{cm}^3\). 3. For the same material, the percent decrease in mass equals the percent decrease in volume. The removed fraction is \(\frac{160\pi}{1000\pi}=0.16\), so the finished part is \(16\%\) lighter.

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

- Read each model's dimensions from its diagram. - Choose the correct volume formula for each solid. - Find the triangular base area before finding the prism volume. - Compare exact or sufficiently precise volume values.

1. From the diagram, model A is a cube with side \(10\,\text{cm}\), so \(V_A=10^3=1000\,\text{cm}^3\). 2. Model B is a triangular prism with triangular base \(20\,\text{cm}\) by \(10\,\text{cm}\) and prism length \(10\,\text{cm}\). Its base area is \(\frac{1}{2}\cdot20\cdot10=100\,\text{cm}^2\), so \(V_B=1000\,\text{cm}^3\). 3. Model C is a cylinder with radius \(5\,\text{cm}\) and height \(14\,\text{cm}\), so \(V_C=\pi\cdot5^2\cdot14=350\pi\,\text{cm}^3\approx1099.56\,\text{cm}^3\). 4. Therefore, \(V_C>V_A=V_B\).

\(C>A=B\)

- Read the radius and height from the diagram. - Use the cylinder volume formula. - Convert cubic centimeters to liters. - Round only after converting.

1. From the diagram, the radius is \(50\,\text{cm}\) and the height is \(120\,\text{cm}\). 2. The volume is \(V=\pi\cdot50^2\cdot120=300{,}000\pi\,\text{cm}^3\approx942{,}477.8\,\text{cm}^3\). 3. Since \(1000\,\text{cm}^3=1\,\text{L}\), the capacity is about \(942.4778\,\text{L}\), which rounds to \(942.5\,\text{L}\).

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

- Read the dimensions of each cylinder from its panel. - Find each cylinder's volume separately. - Add the two volumes before rounding.

1. From the diagram, cylinder a) has radius \(6\,\text{cm}\) and height \(3\,\text{cm}\), so \(V_a=\pi\cdot6^2\cdot3=108\pi\,\text{cm}^3\). 2. Cylinder b) has radius \(3\,\text{cm}\) and height \(8\,\text{cm}\), so \(V_b=\pi\cdot3^2\cdot8=72\pi\,\text{cm}^3\). 3. The total volume is \(180\pi\,\text{cm}^3\approx565.5\,\text{cm}^3\).

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

- Read the cylinder's dimensions from the diagram. - First find the volume the jar can hold when full. - Then calculate \(80\%\) of that volume. - Round only the final result.

1. From the diagram, the jar has radius \(5\,\text{cm}\) and height \(15\,\text{cm}\). 2. The full volume is \(V=\pi\cdot5^2\cdot15=375\pi\,\text{cm}^3\). 3. The flour occupies \(80\%\) of the total: \(0.80\cdot375\pi=300\pi\,\text{cm}^3\approx942.4778\,\text{cm}^3\). 4. Rounded to the nearest tenth, the flour 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 \(36\%\).