Step-by-Step Solutions: Area and Circumference of Circles
This page provides detailed solutions for various problems involving the calculation of the area and circumference of circles. We will use the following key formulas and the approximation $\pi \approx 3.14$.
- Circumference ($C$) using diameter ($d$): $C = \pi d$
- Circumference ($C$) using radius ($r$): $C = 2 \pi r$
- Area ($A$) using radius ($r$): $A = \pi r^2$
- Relationship: $d = 2r$ or $r = d/2$
Problem 1: The Pizza Party
Given: A giant circular pizza with a diameter ($d$) of 60 cm.
a) What is the circumference of this delicious pizza?
Formula: $C = \pi d$
Substitute: $C = 3.14 \times 60 \text{ cm}$
Calculate: $C = 188.4 \text{ cm}$
Answer: The circumference of the pizza is 188.4 cm.
b) If each student needs at least 75 cm² of pizza, how many students can this one giant pizza feed?
Find the radius: $r = d/2 = 60 \text{ cm} / 2 = 30 \text{ cm}$
Find the area (Formula: $A = \pi r^2$):
Substitute: $A = 3.14 \times (30 \text{ cm})^2 = 3.14 \times 900 \text{ cm}^2$
Calculate Area: $A = 2826 \text{ cm}^2$
Calculate number of students: Number = Total Area / Area per student
Substitute: Number = $2826 \text{ cm}^2 / 75 \text{ cm}^2/\text{student}$
Calculate: Number $\approx 37.68$
Since you can't feed a fraction of a student, round down.
Answer: The pizza can feed 37 students.
Problem 2: The School's Circular Garden
Given: Garden radius ($r_{garden}$) = 3.5 meters.
a) What is the area of the garden that the sprinkler needs to cover?
Formula: $A = \pi r_{garden}^2$
Substitute: $A = 3.14 \times (3.5 \text{ m})^2 = 3.14 \times 12.25 \text{ m}^2$
Calculate: $A = 38.465 \text{ m}^2$
Answer: The area the sprinkler needs to cover is 38.465 m².
b) If the water from the sprinkler reaches a distance of 4 meters in all directions, will it cover the entire garden?
Sprinkler reach (radius): $r_{sprinkler} = 4$ meters.
Garden radius: $r_{garden} = 3.5$ meters.
Compare: Is $r_{sprinkler} \ge r_{garden}$? Yes, $4 \text{ m} > 3.5 \text{ m}$.
Answer: Yes, the sprinkler will cover the entire garden because its reach (4 m) is greater than the radius of the garden (3.5 m).
Problem 3: The Circular Running Track
Given: Circumference ($C$) = 400 meters.
a) What is the diameter of this running track?
Formula: $C = \pi d \implies d = C / \pi$
Substitute: $d = 400 \text{ m} / 3.14$
Calculate: $d \approx 127.39 \text{ m}$
Answer: The diameter is approximately 127.39 meters.
b) Sarah runs 5 laps around the track. How far does she run in total?
Distance per lap: $C = 400$ m
Total distance: Distance per lap $\times$ Number of laps
Substitute: Total distance = $400 \text{ m} \times 5$
Calculate: Total distance = $2000$ m
Answer: Sarah runs 2000 meters.
c) What is the area of the grassy section in the center?
Find the radius: $r = d/2 \approx 127.39 \text{ m} / 2 \approx 63.695$ m (or $r = C/(2\pi) \approx 63.69$ m)
Let's use $r \approx 63.70$ m.
Formula: $A = \pi r^2$
Substitute: $A = 3.14 \times (63.70 \text{ m})^2 = 3.14 \times 4057.69 \text{ m}^2$
Calculate: $A \approx 12742.3 \text{ m}^2$
Answer: The area of the grassy section is approximately 12742.3 m².
Problem 4: Canadian Quarter vs. Loonie
Given: Quarter diameter ($d_Q$) = 2.38 cm, Loonie diameter ($d_L$) = 2.65 cm.
a) What is the difference in circumference?
Quarter Circumference ($C_Q = \pi d_Q$): $C_Q = 3.14 \times 2.38 \text{ cm} \approx 7.4732$ cm
Loonie Circumference ($C_L = \pi d_L$): $C_L = 3.14 \times 2.65 \text{ cm} \approx 8.321$ cm
Difference: $C_L - C_Q = 8.321 \text{ cm} - 7.4732 \text{ cm} = 0.8478$ cm
Answer: The difference in circumference is approximately 0.85 cm.
b) What is the difference in area?
Quarter Radius ($r_Q = d_Q / 2$): $r_Q = 2.38 / 2 = 1.19$ cm
Quarter Area ($A_Q = \pi r_Q^2$): $A_Q = 3.14 \times (1.19 \text{ cm})^2 \approx 4.446$ cm²
Loonie Radius ($r_L = d_L / 2$): $r_L = 2.65 / 2 = 1.325$ cm
Loonie Area ($A_L = \pi r_L^2$): $A_L = 3.14 \times (1.325 \text{ cm})^2 \approx 5.513$ cm²
Difference: $A_L - A_Q = 5.513 \text{ cm}^2 - 4.446 \text{ cm}^2 = 1.067$ cm²
Answer: The difference in area is approximately 1.07 cm².
Problem 5: The Ferris Wheel
Given: Radius ($r$) = 15 meters.
a) How far does a person travel in one complete rotation?
This is the circumference.
Formula: $C = 2 \pi r$
Substitute: $C = 2 \times 3.14 \times 15 \text{ m}$
Calculate: $C = 94.2 \text{ m}$
Answer: A person travels 94.2 meters in one rotation.
b) Total distance for 10 rotations?
Total distance: Distance per rotation $\times$ Number of rotations
Substitute: Total distance = $94.2 \text{ m} \times 10$
Calculate: Total distance = $942 \text{ m}$
Answer: The total distance traveled is 942 meters.
c) Area of the circular face to be painted?
Formula: $A = \pi r^2$
Substitute: $A = 3.14 \times (15 \text{ m})^2 = 3.14 \times 225 \text{ m}^2$
Calculate: $A = 706.5 \text{ m}^2$
Answer: The area to be painted is 706.5 m².
Problem 6: The Delicious Donut
Given: Diameter ($d$) = 10 cm.
a) What is the circumference of this donut?
Formula: $C = \pi d$
Substitute: $C = 3.14 \times 10 \text{ cm}$
Calculate: $C = 31.4 \text{ cm}$
Answer: The circumference is 31.4 cm.
b) Length of remaining outer edge after a semicircular bite?
A semicircular bite removes half the outer edge.
Remaining edge: $C / 2$
Substitute: Remaining edge = $31.4 \text{ cm} / 2$
Calculate: Remaining edge = $15.7 \text{ cm}$
Answer: The length of the remaining outer edge is 15.7 cm.
Problem 7: Mrs. Green's Flower Bed
Given: Radius ($r$) = 1.5 meters.
a) What is the length of the stone border needed?
The border length is the circumference.
Formula: $C = 2 \pi r$
Substitute: $C = 2 \times 3.14 \times 1.5 \text{ m}$
Calculate: $C = 9.42 \text{ m}$
Answer: She needs a border 9.42 meters long.
b) Total cost if stones are $2.50 per meter?
Total cost: Length $\times$ Cost per meter
Substitute: Total cost = $9.42 \text{ m} \times \$2.50/\text{m}$
Calculate: Total cost = $\$23.55$
Answer: The total cost will be $23.55.
Problem 8: The Old Vinyl Record
Given: Circumference ($C$) = 94.2 cm.
a) What is the diameter of the record?
Formula: $C = \pi d \implies d = C / \pi$
Substitute: $d = 94.2 \text{ cm} / 3.14$
Calculate: $d = 30 \text{ cm}$
Answer: The diameter is 30 cm.
b) What is the area of the playing surface?
(Assuming this means the total area of the record face)
Find radius: $r = d/2 = 30 \text{ cm} / 2 = 15 \text{ cm}$
Formula: $A = \pi r^2$
Substitute: $A = 3.14 \times (15 \text{ cm})^2 = 3.14 \times 225 \text{ cm}^2$
Calculate: $A = 706.5 \text{ cm}^2$
Answer: The area is 706.5 cm².
Problem 9: The Hula Hoop
Given: Diameter ($d$) = 80 cm.
a) How many meters does it travel in one rotation?
This is the circumference in meters.
Convert diameter to meters: $d = 80 \text{ cm} = 0.80 \text{ m}$
Formula: $C = \pi d$
Substitute: $C = 3.14 \times 0.80 \text{ m}$
Calculate: $C = 2.512 \text{ m}$
Answer: It travels 2.512 meters in one rotation.
b) Total distance traveled in 25 spins?
Total distance: Distance per rotation $\times$ Number of spins
Substitute: Total distance = $2.512 \text{ m} \times 25$
Calculate: Total distance = $62.8 \text{ m}$
Answer: The total distance is 62.8 meters.
Problem 10: The Rain Puddle
Given: Radius ($r$) = 0.75 meters.
a) What is the area covered by the puddle?
Formula: $A = \pi r^2$
Substitute: $A = 3.14 \times (0.75 \text{ m})^2 = 3.14 \times 0.5625 \text{ m}^2$
Calculate: $A \approx 1.76625 \text{ m}^2$
Answer: The area is approximately 1.77 m².
b) How long until it evaporates at 0.1 m²/hour?
Time: Total Area / Rate of evaporation
Substitute: Time = $1.76625 \text{ m}^2 / (0.1 \text{ m}^2/\text{hour})$
Calculate: Time = $17.6625$ hours
Answer: It will take approximately 17.7 hours to evaporate.
Problem 11: The Circular Target
Given: Target diameter ($d_T$) = 60 cm, Bullseye diameter ($d_B$) = 10 cm.
a) What is the area of the entire target?
Target Radius ($r_T = d_T / 2$): $r_T = 60 / 2 = 30$ cm
Formula: $A_T = \pi r_T^2$
Substitute: $A_T = 3.14 \times (30 \text{ cm})^2 = 3.14 \times 900 \text{ cm}^2$
Calculate: $A_T = 2826 \text{ cm}^2$
Answer: The area of the entire target is 2826 cm².
b) What is the area of the bullseye?
Bullseye Radius ($r_B = d_B / 2$): $r_B = 10 / 2 = 5$ cm
Formula: $A_B = \pi r_B^2$
Substitute: $A_B = 3.14 \times (5 \text{ cm})^2 = 3.14 \times 25 \text{ cm}^2$
Calculate: $A_B = 78.5 \text{ cm}^2$
Answer: The area of the bullseye is 78.5 cm².
c) What is the area of the target outside the bullseye?
Area Outside: Total Area - Bullseye Area
Substitute: Area Outside = $A_T - A_B = 2826 \text{ cm}^2 - 78.5 \text{ cm}^2$
Calculate: Area Outside = $2747.5 \text{ cm}^2$
Answer: The area outside the bullseye is 2747.5 cm².
Problem 12: The Bicycle Wheel
Given: Diameter ($d$) = 66 cm.
a) What is the circumference of the bicycle wheel?
Formula: $C = \pi d$
Substitute: $C = 3.14 \times 66 \text{ cm}$
Calculate: $C = 207.24 \text{ cm}$
Answer: The circumference is 207.24 cm.
b) How many rotations will the wheel make if the bicycle travels 100 meters?
Convert units: Distance = 100 m = 10000 cm. Circumference = 207.24 cm.
(Alternatively: Distance = 100 m. Circumference = 2.0724 m)
Number of rotations: Total Distance / Circumference
Substitute (using cm): Rotations = $10000 \text{ cm} / 207.24 \text{ cm/rotation}$
Calculate: Rotations $\approx 48.25...$
Round to the nearest whole number.
Answer: The wheel will make approximately 48 rotations.
Problem 13: The Circular Table and Tablecloth
Given: Table diameter ($d_{table}$) = 1.2 meters. Overhang = 20 cm.
a) What is the diameter of the tablecloth?
Convert overhang to meters: 20 cm = 0.20 m.
The diameter includes the table plus overhang on *both* sides.
Diameter tablecloth ($d_{cloth}$): $d_{table} + 2 \times \text{overhang}$
Substitute: $d_{cloth} = 1.2 \text{ m} + 2 \times 0.20 \text{ m} = 1.2 \text{ m} + 0.40 \text{ m}$
Calculate: $d_{cloth} = 1.6 \text{ m}$
Answer: The diameter of the tablecloth is 1.6 meters.
b) What is the area of the tablecloth?
Tablecloth Radius ($r_{cloth}$): $d_{cloth} / 2 = 1.6 \text{ m} / 2 = 0.8 \text{ m}$
Formula: $A_{cloth} = \pi r_{cloth}^2$
Substitute: $A_{cloth} = 3.14 \times (0.8 \text{ m})^2 = 3.14 \times 0.64 \text{ m}^2$
Calculate: $A_{cloth} = 2.0096 \text{ m}^2$
Answer: The area of the tablecloth is approximately 2.01 m².
Problem 14: Nickel vs. Dime
Given: Nickel radius ($r_N$) = 1.06 cm, Dime radius ($r_D$) = 0.94 cm.
a) What is the circumference of the nickel?
Formula: $C_N = 2 \pi r_N$
Substitute: $C_N = 2 \times 3.14 \times 1.06 \text{ cm}$
Calculate: $C_N \approx 6.6568 \text{ cm}$
Answer: The circumference of the nickel is approximately 6.66 cm.
b) What is the circumference of the dime?
Formula: $C_D = 2 \pi r_D$
Substitute: $C_D = 2 \times 3.14 \times 0.94 \text{ cm}$
Calculate: $C_D \approx 5.9032 \text{ cm}$
Answer: The circumference of the dime is approximately 5.90 cm.
c) What is the difference in their circumferences?
Difference: $C_N - C_D$
Substitute: Difference = $6.6568 \text{ cm} - 5.9032 \text{ cm}$
Calculate: Difference = $0.7536 \text{ cm}$
Answer: The difference is approximately 0.75 cm.
Problem 15: Sprinkler on the Lawn
Given: Lawn radius ($r_{lawn}$) = 8 meters, Sprinkler water radius ($r_{water}$) = 5 meters.
a) What is the area of the lawn watered by the sprinkler?
This is the area the water covers.
Formula: $A_{water} = \pi r_{water}^2$
Substitute: $A_{water} = 3.14 \times (5 \text{ m})^2 = 3.14 \times 25 \text{ m}^2$
Calculate: $A_{water} = 78.5 \text{ m}^2$
Answer: The watered area is 78.5 m².
b) What is the area of the lawn that is not watered?
Calculate total lawn area ($A_{lawn} = \pi r_{lawn}^2$):
Substitute: $A_{lawn} = 3.14 \times (8 \text{ m})^2 = 3.14 \times 64 \text{ m}^2$
Calculate: $A_{lawn} = 200.96 \text{ m}^2$
Calculate unwatered area: $A_{lawn} - A_{water}$
Substitute: Unwatered Area = $200.96 \text{ m}^2 - 78.5 \text{ m}^2$
Calculate: Unwatered Area = $122.46 \text{ m}^2$
Answer: The area not watered is 122.46 m².