Free Math Worksheets

- How are radius and diameter related? - Which equation relates circumference and diameter? - How can you rearrange the circumference equation to find diameter? - Keep the units within each part consistent.

1. For a), \(d = 2r = 2 \cdot 5.5\,\text{cm} = 11\,\text{cm}\). Then \(C = \pi d = 11\pi\,\text{cm} \approx 34.56\,\text{cm}\). 2. For b), \(r = \frac{d}{2} = \frac{18\,\text{m}}{2} = 9\,\text{m}\). Then \(C = \pi d = 18\pi\,\text{m} \approx 56.55\,\text{m}\). 3. For c), \(d = \frac{C}{\pi} = \frac{25\,\text{dm}}{\pi} \approx 7.96\,\text{dm}\). Then \(r = \frac{d}{2} \approx 3.98\,\text{dm}\).

a) \(d = 11\,\text{cm}\), \(C \approx 34.56\,\text{cm}\) b) \(r = 9\,\text{m}\), \(C \approx 56.55\,\text{m}\) c) \(d \approx 7.96\,\text{dm}\), \(r \approx 3.98\,\text{dm}\)

- Which formula relates a circle's circumference and diameter? - Rearrange the formula to isolate the diameter. - Which measurement is given, and which is unknown?

1. The edging is the circumference, so \(C=\pi d\). 2. Solve for diameter: \(d=\frac{C}{\pi}\). 3. Substitute: \(d=\frac{15\,\text{ft}}{\pi}\approx4.77\,\text{ft}\).

At most \(4.77\,\text{ft}\)

- Which equation relates circumference and diameter? - How do you find the radius from the circumference? - In \(C = 2\pi r\), what happens to \(r\) when \(C\) is multiplied by \(2\)?

1. Use \(C = \pi d\): \(d = \frac{15.70\,\text{m}}{\pi} \approx 5.00\,\text{m}\). 2. The new circumference is \(2 \cdot 15.70\,\text{m} = 31.40\,\text{m}\). 3. Use \(C = 2\pi r\): \(r = \frac{31.40\,\text{m}}{2\pi} \approx 5.00\,\text{m}\). 4. Since \(C = 2\pi r\), circumference and radius are proportional. Doubling the circumference doubles the radius.

a) Approximately \(5.00\,\text{m}\) b) Approximately \(5.00\,\text{m}\) c) The radius doubles.

- First find the original circumference. - Write the circumference formula for both the original and enlarged rings, then subtract. - Does the formula for the change in radius contain the original radius?

1. The original circumference is \(C = 2\pi r = 2\pi\cdot20\,\text{cm} = 40\pi\,\text{cm} \approx 125.66\,\text{cm}\). 2. For a circumference change of \(\Delta C = 10\,\text{cm}\), \(\Delta C = 2\pi\Delta r\). 3. Therefore, \(\Delta r = \frac{10\,\text{cm}}{2\pi} \approx 1.59\,\text{cm}\). 4. The formula \(\Delta r = \frac{\Delta C}{2\pi}\) does not contain the original radius. Therefore, the increase is still about \(1.59\,\text{cm}\), even when the original radius is \(200\,\text{m}\).

a) Approximately \(125.66\,\text{cm}\) b) Approximately \(1.59\,\text{cm}\) c) No. The increase remains approximately \(1.59\,\text{cm}\) because \(\Delta r = \frac{\Delta C}{2\pi}\) is independent of the original radius.

- What is the radius of the outer lane? - Can you find the difference between the circumferences without calculating both full circumferences? - How should the starting positions compensate for the longer outer lane?

1. The inner-lane circumference is \(C_i = 2\pi\cdot36\,\text{m} = 72\pi\,\text{m} \approx 226.19\,\text{m}\). 2. The outer radius is \(36\,\text{m} + 1.22\,\text{m} = 37.22\,\text{m}\). 3. The difference in lap lengths can be found directly from the radius difference: \(\Delta C = 2\pi\cdot1.22\,\text{m} \approx 7.67\,\text{m}\). 4. To travel the same distance to the common finish line, the outer-lane runner must start \(7.67\,\text{m}\) ahead.

a) Approximately \(226.19\,\text{m}\) b) Approximately \(7.67\,\text{m}\) c) Approximately \(7.67\,\text{m}\) ahead

- Which equation relates a wheel's radius and circumference? - Convert the travel distance to the same unit as the wheel circumferences. - One rotation moves a wheel forward by one circumference. - A complete-rotation count must be a whole number.

1. The front wheel's circumference is \(C_1 = 2\pi\cdot28\,\text{cm} = 56\pi\,\text{cm} \approx 175.93\,\text{cm}\). 2. The rear wheel's circumference is \(C_2 = 56\pi\,\text{cm} - 25\,\text{cm} \approx 150.93\,\text{cm}\). 3. Its radius is \(r_2 = \frac{C_2}{2\pi} \approx \frac{150.93}{2\pi}\,\text{cm} \approx 24.02\,\text{cm}\). 4. Convert \(100\,\text{m}\) to \(10{,}000\,\text{cm}\). The front wheel makes \(\frac{10000}{56\pi} \approx 56.84\) rotations, so it completes \(56\) full rotations. 5. The rear wheel makes \(\frac{10000}{56\pi - 25} \approx 66.26\) rotations, so it completes \(66\) full rotations. 6. The difference is \(66 - 56 = 10\) complete rotations.

a) Approximately \(175.93\,\text{cm}\) b) Approximately \(24.02\,\text{cm}\) c) Front wheel: \(56\) complete rotations Rear wheel: \(66\) complete rotations Difference: \(10\) complete rotations

- Think of the pool and walkway as two concentric circles. - How does the walkway width change the radius? - Subtract the two outer-circumference expressions before substituting numbers. - Does the original radius remain in the simplified difference?

1. The pool's diameter is \(d = \frac{18.85\,\text{m}}{\pi} \approx 6.00\,\text{m}\), so its radius is about \(3.00\,\text{m}\). 2. With a \(1.20\,\text{m}\) walkway, the outer radius is approximately \(3.00\,\text{m} + 1.20\,\text{m} = 4.20\,\text{m}\). 3. The outer circumference is \(C_2 = 2\pi\cdot4.20\,\text{m} \approx 26.39\,\text{m}\). 4. Doubling the walkway width adds another \(1.20\,\text{m}\) to the outer radius, so the circumference increases by \(2\pi\cdot1.20\,\text{m} \approx 7.54\,\text{m}\). 5. In general, \(2\pi(r+2w)-2\pi(r+w)=2\pi w\). The original radius cancels, so the increase does not depend on the pool's diameter.

a) Approximately \(6.00\,\text{m}\) b) Approximately \(26.39\,\text{m}\) c) Approximately \(7.54\,\text{m}\); the increase does not depend on the original pool diameter.

- How far does the robot travel in one complete wheel rotation? - Convert all lengths to the same unit before dividing. - When a minimum number of complete rotations is required, which way should you round?

1. One wheel rotation moves the robot one circumference: \(C = \pi d = 12\pi\,\text{cm} \approx 37.70\,\text{cm}\). 2. Convert the target distance: \(75\,\text{m} = 7500\,\text{cm}\). 3. The number of rotations needed is \(\frac{7500}{12\pi} \approx 198.94\). 4. Because the robot must travel at least the required distance using complete rotations, round up to \(199\).

At least \(199\) complete rotations

- Both wheels travel the same total distance. - First find the distance traveled from the front wheel's rotations. - How do you find a circle's circumference from its diameter?

1. The total distance is \(120 \cdot 4.20\,\text{m} = 504\,\text{m}\). 2. Convert the rear-wheel diameter: \(35\,\text{cm} = 0.35\,\text{m}\). 3. The rear-wheel circumference is \(C = \pi\cdot0.35\,\text{m} \approx 1.09956\,\text{m}\). 4. The number of rear-wheel rotations is \(\frac{504}{0.35\pi} \approx 458.37\), which rounds to \(458\).

Approximately \(458\) rotations

- Will the computer display too much or too little when the programmed diameter is smaller than the actual diameter? - The number of rotations is the same for the displayed and actual distances. - In a ratio of the two circumferences, does \(\pi\) cancel?

1. The displayed distance is based on the programmed circumference \(C_p = 68\pi\,\text{cm}\). The actual distance is based on \(C_a = 71\pi\,\text{cm}\). 2. The same number of rotations is used for both distances, so \(\frac{d_a}{d_p}=\frac{71\pi}{68\pi}=\frac{71}{68}\). 3. Therefore, \(d_a = 34.0\,\text{km} \cdot \frac{71}{68} = 35.5\,\text{km}\).

\(35.5\,\text{km}\)

- When a circular object makes one complete rotation, how far does it travel compared with its circumference? - How can you use several rotations to estimate one circumference? - What equation relates a circle's diameter and circumference?

1. Divide the total distance by the number of rotations: \(C = \frac{377\,\text{cm}}{10} = 37.7\,\text{cm}\). 2. Use \(C = \pi d\) and solve for \(\pi\): \(\pi = \frac{C}{d}\). 3. For the measured distance, \(\pi \approx \frac{37.7}{12} \approx 3.1417\). 4. With the corrected distance, the circumference would be \(C = \frac{378\,\text{cm}}{10} = 37.8\,\text{cm}\). 5. The corrected estimate is \(\pi \approx \frac{37.8}{12} = 3.15\).

a) \(37.7\,\text{cm}\) b) \(\pi \approx 3.1417\) c) \(\pi \approx 3.15\)

- Find the distance traveled in one full rotation. - Use the relationship among speed, distance, and time. - Convert the time to seconds for the calculation in part b).

1. The Ferris wheel's circumference is \(C=\pi\cdot160\,\text{m}=160\pi\,\text{m}\approx502.65\,\text{m}\). 2. One rotation takes \(t=\frac{160\pi\,\text{m}}{0.25\,\text{m/s}}\approx2010.62\,\text{s}\). 3. Converting to minutes gives \(\frac{2010.62}{60}\approx33.51\,\text{min}\). 4. For \(15\) minutes, \(t=15\cdot60\,\text{s}=900\,\text{s}\). 5. The distance is \(0.25\,\text{m/s}\cdot900\,\text{s}=225\,\text{m}=0.225\,\text{km}\).

a) Approximately \(33.51\,\text{min}\) b) \(0.225\,\text{km}\)

- How are radius and circumference related? - You can find the original radius first, or work directly with the change in radius. - How much circumference is added for each unit added to the radius?

1. A radius increase of \(0.8\,\text{cm}\) increases the circumference by \(\Delta C=2\pi\Delta r\). 2. Therefore, \(\Delta C=2\pi\cdot0.8\,\text{cm}\approx5.03\,\text{cm}\). 3. The new circumference is \(120\,\text{cm}+5.0265\ldots\,\text{cm}\approx125.03\,\text{cm}\).

Approximately \(125.03\,\text{cm}\)

- How far does the wheel travel in one rotation? - Convert all measurements to the same unit. - What does “complete rotations” mean when the quotient is not a whole number?

1. Convert the diameter: \(50\,\text{cm}=0.5\,\text{m}\). 2. One rotation covers one circumference: \(C=\pi(0.5\,\text{m})=0.5\pi\,\text{m}\approx1.5708\,\text{m}\). 3. The total number of rotations is \(\frac{250}{0.5\pi}\approx159.15\). 4. Therefore, the wheel completes \(159\) full rotations.

\(159\) complete rotations

- Find the circumference of each wheel type. - Divide the travel distance by each circumference. - Subtract the two rotation counts.

1. Convert the measurements: \(1\,\text{km}=1000\,\text{m}\) and \(80\,\text{cm}=0.80\,\text{m}\). 2. The front-wheel circumference is \(0.80\pi\,\text{m}\), so the front wheels make \(\frac{1000}{0.80\pi}\approx397.89\) rotations. 3. The rear-wheel circumference is \(1.60\pi\,\text{m}\), so the rear wheels make \(\frac{1000}{1.60\pi}\approx198.94\) rotations. 4. Using the unrounded rotation counts, the difference is approximately \(198.94\), or about \(199\) rotations.

Approximately \(199\) more rotations

- Imagine the roller creating one long rectangular strip. - Use area and roller width to find the travel distance. - Divide that distance by the drum's circumference.

1. The travel distance is the rolled area divided by the drum width: \(s=\frac{1200\,\text{m}^2}{2\,\text{m}}=600\,\text{m}\). 2. The drum circumference is \(C=\pi\cdot1.20\,\text{m}=1.20\pi\,\text{m}\approx3.7699\,\text{m}\). 3. The number of rotations is \(n=\frac{600}{1.20\pi}\approx159.15\).

Approximately \(159.15\) rotations

- How far does each seat travel in one rotation? - Both seats take the same time to complete a rotation. - Use distance divided by time to find each speed.

1. In one rotation, the inner seat travels \(2\pi\cdot3\,\text{m}=6\pi\,\text{m}\), so its speed is \(\frac{6\pi\,\text{m}}{10\,\text{s}}\approx1.88\,\text{m/s}\). 2. The outer seat travels \(2\pi\cdot5.50\,\text{m}=11\pi\,\text{m}\), so its speed is \(\frac{11\pi\,\text{m}}{10\,\text{s}}\approx3.46\,\text{m/s}\). 3. The speed difference is \(\frac{2\pi\cdot(5.50-3)}{10}\,\text{m/s}=\frac{\pi}{2}\,\text{m/s}\approx1.57\,\text{m/s}\).

Approximately \(1.57\,\text{m/s}\) faster

- Add the orbital altitude to Earth's radius. - How many minutes are in \(24\) hours? - Divide the total time by the time per orbit. - Multiply the number of orbits by one orbit's circumference.

1. The orbit radius is \(6370\,\text{km}+400\,\text{km}=6770\,\text{km}\). 2. One orbit has length \(2\pi\cdot6770\,\text{km}\approx42{,}537.16\,\text{km}\). 3. A \(24\)-hour day has \(1440\,\text{min}\), so the station completes \(\frac{1440}{93}\approx15.48387\) orbits. 4. The modeled distance is \(\frac{1440}{93}\cdot2\pi\cdot6770\,\text{km}\approx658{,}639.97\,\text{km}\). 5. Rounded to the nearest kilometer, the distance is \(658{,}640\,\text{km}\).

Approximately \(658{,}640\,\text{km}\)

- How many times does one wheel circumference fit into the total distance? - Keep track of meters and centimeters. - Round the number of rotations only at the end.

1. Since \(C=\pi d\), the diameter is \(d=\frac{C}{\pi}=\frac{100}{\pi}\,\text{cm}\approx31.83\,\text{cm}\). 2. Each rotation covers \(1.00\,\text{m}\), so the path length is \(235\cdot1.00\,\text{m}=235\,\text{m}\). 3. The second wheel has circumference \(C_2=\pi\cdot25\,\text{cm}\approx78.54\,\text{cm}=0.7854\,\text{m}\). 4. The number of rotations is \(n=\frac{235\,\text{m}}{0.25\pi\,\text{m}}\approx299.21\).

a) About \(31.83\,\text{cm}\) b) \(235\,\text{m}\) c) About \(299.21\) rotations

- Find the old and new circumference for each bed. - Expand the general new-circumference expression. - Which terms cancel when you subtract the old circumference?

1. For Bed A, the circumference increase is \(2\pi(3\,\text{ft})-2\pi(2\,\text{ft})=2\pi\,\text{ft}\approx6.28\,\text{ft}\). 2. For Bed B, the increase is \(2\pi(7\,\text{ft})-2\pi(6\,\text{ft})=2\pi\,\text{ft}\approx6.28\,\text{ft}\). 3. In general, \(2\pi(r+w)-2\pi r=2\pi r+2\pi w-2\pi r=2\pi w\). 4. The original radius cancels, so the circumference increase depends only on the added distance \(w\).

a) Each circumference increases by approximately \(6.28\,\text{ft}\). b) The increases are equal because \(2\pi(r+w)-2\pi r=2\pi w\), which does not depend on \(r\).