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Circumference problems

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5546367
A circle has diameter \(9\,\text{in}\). Find its circumference exactly in terms of \(\pi\).

Hints

- Which circumference formula uses the diameter directly? - Keep \(\pi\) in the answer because an exact value is requested.

Solution

1. Use the diameter form \(C=\pi d\). 2. Substitute \(d=9\,\text{in}\): \(C=9\pi\,\text{in}\).

Answer

\(9\pi\,\text{in}\)
5126647
Find the missing measurements for each circle. Round to the nearest hundredth when needed. a) \(r=5.5\,\text{cm}\). Find \(d\) and \(C\). b) \(d=18\,\text{m}\). Find \(r\) and \(C\). c) \(C=25\,\text{dm}\). Find \(d\) and \(r\).

Hints

- 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.

Solution

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 \(r = \frac{d}{2} \approx 3.98\,\text{dm}\).

Answer

a) \(d = 11\,\text{cm}\), \(C \approx 34.56\,\text{cm}\) b) \(r = 9\,\text{m}\), \(C \approx 56.55\,\text{m}\) c) \(d \approx 7.96\,\text{dm}\), \(r \approx 3.98\,\text{dm}\)
5138407
A landscaper has \(15\,\text{ft}\) of edging for a circular rose bed. What is the greatest possible diameter of the bed? Round to the nearest hundredth of a foot.

Hints

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

Solution

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

Answer

At most \(4.77\,\text{ft}\)
5512217
The diagram shows a wheel and its diameter. How far does a point on the rim travel in one full revolution? Give an exact answer in terms of \(\pi\) and an approximate answer to the nearest tenth of an inch.
Figure for problem 551221

Hints

- One full turn of the wheel corresponds to one circumference. - Read the diameter from the diagram and choose the circumference formula that uses diameter directly.

Solution

1. One full revolution covers one circumference of the wheel. 2. Use \(C=\pi d\) with the diagram’s diameter: \(C=24\pi\,\text{in.}\). 3. Numerically, \(24\pi\approx75.4\,\text{in.}\).

Answer

Exactly \(24\pi\,\text{in.}\), or approximately \(75.4\,\text{in.}\)
5126497
A circular flower bed has a circumference of \(15.70\,\text{m}\). a) Find the diameter of the flower bed to the nearest hundredth of a meter. b) A new circular flower bed will have twice the circumference. What will its radius be, to the nearest hundredth of a meter? c) In general, what happens to a circle's radius when its circumference is doubled?

Hints

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

Solution

1. Use \(C=\pi d\): \(d=\frac{15.70}{\pi}\,\text{m}\approx5.00\,\text{m}\). 2. The new circumference is \(31.40\,\text{m}\). 3. Use \(C=2\pi r\): \(r=\frac{31.40}{2\pi}\,\text{m}\approx5.00\,\text{m}\). 4. Since \(C=2\pi r\), circumference and radius are proportional. Doubling circumference doubles radius.

Answer

a) Approximately \(5.00\,\text{m}\) b) Approximately \(5.00\,\text{m}\) c) The radius doubles.
5126507
A metal ring has a radius of \(20\,\text{cm}\). a) Find the circumference exactly in terms of \(\pi\) and approximately to the nearest hundredth. b) The circumference is increased by exactly \(10\,\text{cm}\). By how many centimeters does the radius increase? Give an exact answer in terms of \(\pi\) and an approximation to the nearest hundredth. c) Would the radius increase by a different amount if the original ring had a radius of \(200\,\text{m}\), but its circumference still increased by exactly \(10\,\text{cm}\)? Justify your answer using a formula.

Hints

- Write the circumference formula before and after the increase. - Subtract the two equations to isolate the change in radius. - Does the simplified change formula contain the original radius?

Solution

1. The original circumference is \(C=40\pi\,\text{cm}\approx125.66\,\text{cm}\). 2. For a circumference change of \(\Delta C=10\,\text{cm}\), \(\Delta C=2\pi\Delta r\). 3. Therefore, \(\Delta r=\frac{5}{\pi}\,\text{cm}\approx1.59\,\text{cm}\). 4. The formula \(\Delta r=\frac{\Delta C}{2\pi}\) does not contain the original radius, so the increase is unchanged in part c.

Answer

a) \(40\pi\,\text{cm}\approx125.66\,\text{cm}\) b) \(\frac{5}{\pi}\,\text{cm}\approx1.59\,\text{cm}\) c) No. The increase is independent of the original radius.
5126517
A circular training track has an inner lane with radius \(36\,\text{m}\). The outer lane is a constant \(1.22\,\text{m}\) farther from the center. a) Find the length of one lap on the inner lane exactly in terms of \(\pi\) and approximately to the nearest hundredth. b) How much longer is one lap on the outer lane than one lap on the inner lane? Give an exact answer in terms of \(\pi\) and an approximation to the nearest hundredth. c) In a one-lap race, runners must travel the same distance and finish at the same line. How far ahead of the inner-lane starting line should the outer-lane runner start? Round to the nearest hundredth of a meter.

Hints

- You can compare the circumferences using only the difference in radii. - What is \(2\pi\) times the lane-width difference? - The starting stagger must compensate for the extra outer-lane distance.

Solution

1. The inner-lane circumference is \(72\pi\,\text{m}\approx226.19\,\text{m}\). 2. The radius difference is \(1.22\,\text{m}\), so the lap-length difference is \(2\pi(1.22\,\text{m})=2.44\pi\,\text{m}\approx7.67\,\text{m}\). 3. To travel the same distance to a common finish line, the outer-lane runner starts \(7.67\,\text{m}\) ahead.

Answer

a) \(72\pi\,\text{m}\approx226.19\,\text{m}\) b) \(2.44\pi\,\text{m}\approx7.67\,\text{m}\) c) Approximately \(7.67\,\text{m}\) ahead
5126557
A small cart has a front wheel with radius \(28\,\text{cm}\). a) Find the circumference of the front wheel exactly in terms of \(\pi\) and approximately to the nearest hundredth. b) The rear wheel is smaller, and its circumference is exactly \(25\,\text{cm}\) less than the front wheel's circumference. Find the rear wheel's radius to the nearest hundredth of a centimeter. c) Over a distance of \(100\,\text{m}\), how many complete rotations does each wheel make? How many more complete rotations does the rear wheel make than the front wheel?

Hints

- One rotation moves a wheel forward by one circumference. - Convert the travel distance to centimeters before dividing. - A complete-rotation count must be a whole number less than or equal to the total rotation count.

Solution

1. The front wheel's circumference is \(56\pi\,\text{cm}\approx175.93\,\text{cm}\). 2. The rear wheel's circumference is \(56\pi\,\text{cm}-25\,\text{cm}\). 3. Its radius is \(r_2=\frac{56\pi-25}{2\pi}\,\text{cm}\approx24.02\,\text{cm}\). 4. Convert \(100\,\text{m}\) to \(10{,}000\,\text{cm}\). The front wheel makes about \(56.84\) rotations, so it completes \(56\) full rotations. 5. The rear wheel makes about \(66.26\) rotations, so it completes \(66\) full rotations. 6. The difference is \(10\) complete rotations.

Answer

a) \(56\pi\,\text{cm}\approx175.93\,\text{cm}\) b) Approximately \(24.02\,\text{cm}\) c) Front wheel: \(56\) complete rotations Rear wheel: \(66\) complete rotations Difference: \(10\) complete rotations
5126567
A circular swimming pool has circumference \(C_1 = 18.85\,\text{m}\). a) Find the pool's diameter to the nearest hundredth of a meter. b) A paved walkway \(1.20\,\text{m}\) wide will be built around the pool. Find the circumference of the walkway's outer edge to the nearest hundredth of a meter. c) The walkway width is doubled from \(1.20\,\text{m}\) to \(2.40\,\text{m}\). By how many meters does the outer circumference increase? Round to the nearest hundredth and explain whether this increase depends on the pool's original diameter.

Hints

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

Solution

1. The pool's diameter is \(d=\frac{18.85}{\pi}\,\text{m}\approx6.00\,\text{m}\), so its radius is about \(3.00\,\text{m}\). 2. With a \(1.20\,\text{m}\) walkway, the outer radius is about \(4.20\,\text{m}\), so the outer circumference is approximately \(26.39\,\text{m}\). 3. Doubling the walkway width adds another \(1.20\,\text{m}\) to the outer radius, so the circumference increases by \(2\pi(1.20\,\text{m})\approx7.54\,\text{m}\). 4. In general, \(2\pi(r+2w)-2\pi(r+w)=2\pi w\), so the original radius cancels.

Answer

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.
5126617
A small mobile robot has wheels with diameter \(12\,\text{cm}\). During a test, the robot must travel at least \(75\,\text{m}\) in a straight line. What is the least number of complete wheel rotations needed to reach that distance?

Hints

- 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?

Solution

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 \(199\).

Answer

At least \(199\) complete rotations
5126627
A historic penny-farthing has a large front wheel with circumference \(4.20\,\text{m}\) and a small rear wheel with diameter \(35\,\text{cm}\). During a ride, the front wheel makes exactly \(120\) rotations. About how many rotations does the rear wheel make over the same distance? Round to the nearest whole number.

Hints

- 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?

Solution

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 \(458\).

Answer

Approximately \(458\) rotations
5126637
A bicycle computer calculates distance by multiplying the number of wheel rotations by the programmed tire circumference. The rider accidentally enters a tire diameter of \(68\,\text{cm}\), but the actual tire diameter is \(71\,\text{cm}\). At the end of a ride, the computer displays exactly \(34.0\,\text{km}\). What distance did the rider actually travel?

Hints

- 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?

Solution

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, \(d_a = 34.0\,\text{km} \cdot \frac{71}{68} = 35.5\,\text{km}\).

Answer

\(35.5\,\text{km}\)
5126897
In an experiment to estimate \(\pi\), a cylindrical container with diameter \(d = 12\,\text{cm}\) is rolled along a long strip of paper. To reduce measurement error, the container makes exactly \(10\) complete rotations. The total marked distance is \(377\,\text{cm}\). a) Use the data to find the circumference \(C\) of the container. b) Use the circumference to estimate \(\pi\). Round to the nearest ten-thousandth. c) Suppose the measured distance was \(1\,\text{cm}\) too short, so the actual distance was \(378\,\text{cm}\). What estimate of \(\pi\) would result? Round to the nearest ten-thousandth.

Hints

- One complete rotation covers one circumference. - Divide the total travel distance by \(10\) before estimating \(\pi\). - Use the same requested precision for both estimates.

Solution

1. Divide the total distance by the number of rotations: \(C=\frac{377\,\text{cm}}{10}=37.7\,\text{cm}\). 2. Use \(C=\pi d\): \(\pi\approx\frac{37.7}{12}\approx3.1417\). 3. With the corrected distance, \(C=\frac{378\,\text{cm}}{10}=37.8\,\text{cm}\). 4. The corrected estimate is \(\pi\approx\frac{37.8}{12}=3.1500\).

Answer

a) \(37.7\,\text{cm}\) b) \(\pi\approx3.1417\) c) \(\pi\approx3.1500\)
5127177
A Ferris wheel has diameter \(160\,\text{m}\). Its passenger cars move at a constant speed of \(0.25\,\text{m/s}\). a) How many minutes does one complete rotation take? Round to the nearest hundredth of a minute. b) How far, in kilometers, does a passenger travel during a \(15\)-minute ride?

Hints

- Find the distance traveled in one full rotation. - Use the relationship among speed, distance, and time. - Convert between seconds and minutes only after setting up the time calculation.

Solution

1. The Ferris wheel's circumference is \(160\pi\,\text{m}\). 2. One rotation takes \(\frac{160\pi\,\text{m}}{0.25\,\text{m/s}}=640\pi\,\text{s}\approx2010.62\,\text{s}\). 3. In minutes, this is \(\frac{640\pi}{60}\,\text{min}\approx33.51\,\text{min}\). 4. A \(15\)-minute ride lasts \(900\,\text{s}\), so the distance is \(0.25\,\text{m/s}\cdot900\,\text{s}=225\,\text{m}=0.225\,\text{km}\).

Answer

a) Approximately \(33.51\,\text{min}\) b) \(0.225\,\text{km}\)
5138417
A young tree trunk has a circumference of exactly \(120\,\text{cm}\). During a very rainy year, its radius increases by \(0.8\,\text{cm}\). Find the trunk's new circumference. Round to the nearest hundredth of a centimeter.

Hints

- 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?

Solution

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 \(120\,\text{cm}+5.0265\ldots\,\text{cm}\approx125.03\,\text{cm}\).

Answer

Approximately \(125.03\,\text{cm}\)
5138527
A unicyclist travels \(250\,\text{m}\). The wheel has diameter \(50\,\text{cm}\). How many complete rotations does the wheel make over this distance?

Hints

- 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?

Solution

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.

Answer

\(159\) complete rotations
5138537
A tractor has front wheels with diameter \(80\,\text{cm}\) and rear wheels with diameter \(1.60\,\text{m}\). The tractor travels exactly \(1\,\text{km}\). About how many more rotations do the front wheels make than the rear wheels? Round to the nearest whole rotation.

Hints

- Find the circumference of each wheel type in meters. - Divide the travel distance by each circumference. - Subtract the unrounded rotation counts before rounding the final difference.

Solution

1. Convert \(80\,\text{cm}=0.80\,\text{m}\) and \(1\,\text{km}=1000\,\text{m}\). 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 counts, the difference is approximately \(198.94\), which rounds to \(199\) rotations.

Answer

Approximately \(199\) more rotations
5138547
A road roller has a cylindrical drum with diameter \(1.20\,\text{m}\) and width \(2\,\text{m}\). It rolls a rectangular area of \(1200\,\text{m}^2\) exactly once. About how many rotations does the drum make? Round to the nearest whole rotation.

Hints

- Use the rolled area and drum width to find the distance traveled. - One complete rotation covers one drum circumference. - Round only the final rotation count.

Solution

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 \(1.20\pi\,\text{m}\). 3. The number of rotations is \(n=\frac{600}{1.20\pi}\approx159.15\), which rounds to \(159\).

Answer

Approximately \(159\) rotations
5138737
At a swing ride, the inner seats are \(3\,\text{m}\) from the axis of rotation, and the outer seats are \(5.50\,\text{m}\) from the axis. One full rotation takes exactly \(10\,\text{s}\). How many meters per second faster does a rider in an outer seat move than a rider in an inner seat? Round to the nearest hundredth.

Hints

- 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.

Solution

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 \(\frac{2\pi\cdot(5.50-3)}{10}\,\text{m/s}=\frac{\pi}{2}\,\text{m/s}\approx1.57\,\text{m/s}\).

Answer

Approximately \(1.57\,\text{m/s}\) faster
5138747
The International Space Station is modeled as traveling in a circular orbit about \(400\,\text{km}\) above Earth. One orbit takes about \(93\,\text{min}\), and Earth's radius is taken to be \(6370\,\text{km}\). Based on this model, find the total distance the station travels in \(24\) hours. Round to the nearest kilometer.

Hints

- 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.

Solution

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

Answer

Approximately \(658{,}640\,\text{km}\)
5141147
A measuring wheel is used to measure distances at a construction site. The wheel has a circumference of exactly \(1.00\,\text{m}\). a) What is the wheel's diameter in centimeters? Give an exact answer in terms of \(\pi\) and an approximation to the nearest hundredth. b) The wheel makes exactly \(235\) complete rotations along a path. How long is the path in meters? c) A different measuring wheel has a diameter of \(25\,\text{cm}\). How many rotations would that wheel make along the same path? Round to the nearest hundredth.

Hints

- Use \(C=\pi d\) for the first wheel. - One complete rotation covers one circumference. - Keep meters and centimeters consistent in part c.

Solution

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

Answer

a) \(\frac{100}{\pi}\,\text{cm}\approx31.83\,\text{cm}\) b) \(235\,\text{m}\) c) Approximately \(299.21\) rotations
5138757
Two circular flower beds will receive new stone borders. Bed A has radius \(2\,\text{ft}\), and Bed B has radius \(6\,\text{ft}\). For each bed, the new border increases the radius by exactly \(1\,\text{ft}\). a) Find the increase in circumference for each bed. Round to the nearest hundredth. b) Compare the increases. Use a general formula with original radius \(r\) and added distance \(w\) to explain why the increase is independent of the original radius.

Hints

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

Solution

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 \(w\).

Answer

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 \(r\).

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