To fill in the values for the objects you've listed, let's first define the relationships among radius, diameter, and circumference.
- Radius (r) is half of the diameter (d): \( r = \frac{d}{2} \)
- Diameter (d) is twice the radius: \( d = 2r \)
- Circumference (C) is given by the formula: \( C = 2\pi r \) or \( C = \pi d \)
Now, let's calculate the measurements for each object listed:
Cookie
- Radius: Let’s assume a small cookie has a radius of about 5 cm.
- Diameter: \( d = 2r = 2(5) = 10 \) cm.
- Circumference: \( C = 2\pi r = 2\pi(5) = 10\pi \) cm.
Small Plate
- Radius: Let’s assume a small plate has a radius of about 10 cm.
- Diameter: \( d = 2r = 2(10) = 20 \) cm.
- Circumference: \( C = 2\pi r = 2\pi(10) = 20\pi \) cm.
Quarter
- Radius: Approximately 1.9 cm (typical U.S. quarter).
- Diameter: \( d = 2r = 2(1.9) \approx 3.8 \) cm.
- Circumference: \( C = 2\pi r \approx 2\pi(1.9) \approx 3.8\pi \) cm.
Frisbee
- Radius: Let’s say around 15 cm.
- Diameter: \( d = 2r = 2(15) = 30 \) cm.
- Circumference: \( C = 2\pi r = 2\pi(15) = 30\pi \) cm.
Tennis Ball Can
- Given:
- Diameter: 7 cm (let's find radius first).
- Radius: \( r = \frac{d}{2} = \frac{7}{2} = 3.5 \) cm.
- Circumference: \( C = \pi d = \pi(7) = 7\pi \) cm.
Vinyl Record
- Typically, let's consider a vinyl record to have a diameter of about 30 cm.
- Radius: \( r = \frac{d}{2} = \frac{30}{2} = 15 \) cm.
- Circumference: \( C = \pi d = \pi(30) = 30\pi \) cm.
Summary Table:
| Object | Radius (cm) | Diameter (cm) | Circumference (cm) | |------------------|-------------|----------------|---------------------| | Cookie | 5 | 10 | 10π | | Small Plate | 10 | 20 | 20π | | Quarter | 1.9 | 3.8 | 3.8π | | Frisbee | 15 | 30 | 30π | | Tennis Ball Can | 3.5 | 7 | 7π | | Vinyl Record | 15 | 30 | 30π |
Feel free to adjust the specific radius values depending on the actual sizes of the objects you're considering!