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Grade 7 math worksheets

Build your own math worksheets from 35,000+ problems for grades 3 to 12, from fractions to AP Calculus. Every problem comes with step-by-step solutions.

Circumference problems

Learning goals

  • Finds a circle’s circumference from its radius or diameter.
  • Selects the relevant boundary lengths or formula.
  • Expresses circumference in units of length.
  • Solves for missing dimensions when enough information is provided.

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Circumference problems 1
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

\(9\pi\,\text{in}\)
Circumference problems 2
Mia wants to put a ribbon around the edge of a round paper plate. Does she need the plate's area or its circumference? Explain.

Hints

- Is the ribbon covering the plate or going around it? - Which measurement names the distance around a circle?

Solution

1. The ribbon follows the outer boundary, not the region inside it. 2. The length around the boundary is the circumference.

Answer

Circumference, because the ribbon goes around the edge.
Circumference problems 3
Which segment is a diameter of the circle, \(OA\) or \(AB\)? Explain.
Figure for problem 559779

Hints

- Find the center first. - How many endpoints of a diameter lie on the circle?

Solution

1. Segment \(AB\) joins two points on the circle and passes through its center \(O\). 2. Therefore \(AB\) is a diameter; \(OA\) is a radius.

Answer

\(AB\), because it goes through the center and reaches the circle on both ends.
Circumference problems 4
The diameter of a circle is \(1\,\text{m}\). Write its circumference in terms of \(\pi\). What does \(\pi\) tell you about the distance around any circle compared with its diameter?

Hints

- Which circle measurement is multiplied by \(\pi\)? - Does the circumference-to-diameter ratio depend on circle size?

Solution

1. Circumference equals \(\pi\) times the diameter: \(C=\pi d\). 2. With \(d=1\,\text{m}\), \(C=\pi\,\text{m}\). 3. Every circle's circumference is \(\pi\) times its diameter.

Answer

\(\pi\,\text{m}\). Circumference is always \(\pi\) times diameter.
Circumference problems 5
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) \(d \approx 7.96\,\text{dm}\), \(r \approx 3.98\,\text{dm}\)
Circumference problems 6
A landscaper has \(15\,\text{ft}\) of edging for a circular rose bed. What is the greatest possible diameter of the bed? Give the exact value and round it to the nearest hundredth of a foot.

Hints

- Which formula relates circumference and diameter? - What circumference uses the entire available edging? - How can you express the diameter exactly before rounding?

Solution

1. The edging is the circumference, so \(C=\pi d\). 2. Solve for diameter: \(d=\frac{C}{\pi}\). 3. With all the edging used, \(d=\frac{15}{\pi}\,\text{ft}\approx4.77\,\text{ft}\). A smaller diameter is also possible, but a larger one needs more edging.

Answer

Greatest possible diameter: \(\frac{15}{\pi}\,\text{ft}\approx4.77\,\text{ft}\).
Circumference problems 7
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 \(75.4\,\text{in.}\)
Circumference problems 8
A circle has radius \(7\,\text{cm}\). Show that both \(C=2\pi r\) and \(C=\pi d\) give the same circumference.

Hints

- Express the diameter using the given radius. - Substitute into each formula separately and compare.

Solution

1. The diameter is \(d=2\cdot7=14\,\text{cm}\). 2. The radius formula gives \(C=2\pi\cdot7=14\pi\,\text{cm}\). 3. The diameter formula gives \(C=\pi\cdot14=14\pi\,\text{cm}\).

Answer

Both formulas give \(14\pi\,\text{cm}\).
Circumference problems 9
A circular track has circumference \(18\pi\,\text{m}\). Find its radius.

Hints

- What formula uses the radius to find circumference? - What must be multiplied by \(2\pi\) to get the given circumference?

Solution

1. Use \(C=2\pi r\). 2. Since \(2\pi r=18\pi\), divide by \(2\pi\) to get \(r=9\,\text{m}\).

Answer

\(9\,\text{m}\).
Circumference problems 10
A garden has the shape of a semicircle with radius \(4\,\text{m}\). Find (a) the length of its curved edge and (b) the full distance around its boundary. Give exact answers in terms of \(\pi\).

Hints

- A semicircle's curved part is what fraction of a full circle? - Does a semicircle's full boundary include a straight edge?

Solution

1. The curved edge is half a full circumference, so it is \(\frac{1}{2}(2\pi\cdot4)=4\pi\,\text{m}\). 2. The full boundary also includes the straight diameter, \(8\,\text{m}\). 3. The full perimeter is \((4\pi+8)\,\text{m}\).

Answer

(a) \(4\pi\,\text{m}\) (b) \((4\pi+8)\,\text{m}\)
Circumference problems 11
A bicycle wheel makes \(12\) complete turns without slipping. How far does the bicycle move? Use \(\pi\approx3.14\) and round to the nearest centimeter.
Figure for problem 559784

Hints

- What distance does the wheel cover in one full turn? - How can you combine the distances from the equal turns? - Check that your unit is a length.

Solution

1. The diameter in the drawing is \(50\,\text{cm}\), so each turn moves \(50\pi\,\text{cm}\). 2. For \(12\) turns, distance is \(12\cdot50\cdot3.14=1884\,\text{cm}\).

Answer

\(1884\,\text{cm}\).
Circumference problems 12
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.
Circumference problems 13
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.
Circumference problems 14
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
Circumference problems 15
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
Circumference problems 16
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.
Circumference problems 17
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
Circumference problems 18
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
Circumference problems 19
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

\(35.5\,\text{km}\)
Circumference problems 20
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) \(\pi\approx3.1500\)
Circumference problems 21
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) \(0.225\,\text{km}\)
Circumference problems 22
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 \(125.03\,\text{cm}\)
Circumference problems 23
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
Circumference problems 24
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
Circumference problems 25
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
Circumference problems 26
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
Circumference problems 27
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 \(658{,}640\,\text{km}\)
Circumference problems 28
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
Circumference problems 29
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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