Question

find the radius of each circle. use your calculators value of π. round your answer to the nearest tenth.

1. circumference = 62.8 m
2. circumference = 88.1 yd
3. circumference = 12.6 yd
4. circumference = 25.1 ft

find the diameter of each circle. use your calculators value of π. round your answer to the nearest tenth.

1. area = 201.1 in²
2. area = 78.5 ft²

find the circumference of each circle.

1. area = 64π mi²
2. area = 16π in²

find the area of each.

1. circumference = 6π yd
2. circumference = 22π in

Explanation:

Step1: Apply the radius - circumference formula

\(r=\frac{C}{2\pi}\), substituting \(C = 62.8\) m and using the calculator's value of \(\pi\).
\(r=\frac{62.8}{2\pi}\)
## Step2: Calculate the value
\(r=\frac{62.8}{2\times3.14159}\approx10.0\) m
- # Answer:
\(10.0\) m
- **Problem 18: \(C = 89.1\) yd**
- # Explanation:
## Step1: Apply the radius - circumference formula
\(r = \frac{C}{2\pi}\), substituting \(C = 89.1\) yd.
\(r=\frac{89.1}{2\pi}\)
## Step2: Calculate the value
\(r=\frac{89.1}{2\times3.14159}\approx14.2\) yd
- # Answer:
\(14.2\) yd
- **Problem 19: \(C = 12.6\) yd**
- # Explanation:
## Step1: Apply the radius - circumference formula
\(r=\frac{C}{2\pi}\), substituting \(C = 12.6\) yd.
\(r=\frac{12.6}{2\pi}\)
## Step2: Calculate the value
\(r=\frac{12.6}{2\times3.14159}\approx2.0\) yd
- # Answer:
\(2.0\) yd
- **Problem 20: \(C = 25.1\) ft**
- # Explanation:
## Step1: Apply the radius - circumference formula
\(r=\frac{C}{2\pi}\), substituting \(C = 25.1\) ft.
\(r=\frac{25.1}{2\pi}\)
## Step2: Calculate the value
\(r=\frac{25.1}{2\times3.14159}\approx4.0\) ft
- # Answer:
\(4.0\) ft
2. **For finding diameter given area**:
- First, use the formula for the area of a circle \(A=\pi r^{2}\) to find the radius \(r = \sqrt{\frac{A}{\pi}}\), and then the diameter \(d = 2r\).
- **Problem 21: \(A = 201.1\) in²**
- # Explanation:
## Step1: Find the radius
\(r=\sqrt{\frac{A}{\pi}}\), substituting \(A = 201.1\) in².
\(r=\sqrt{\frac{201.1}{\pi}}\)
## Step2: Calculate the radius
\(r=\sqrt{\frac{201.1}{3.14159}}\approx8.0\) in
## Step3: Find the diameter
\(d = 2r\), so \(d=2\times8.0 = 16.0\) in
- # Answer:
\(16.0\) in
- **Problem 22: \(A = 78.5\) ft²**
- # Explanation:
## Step1: Find the radius
\(r=\sqrt{\frac{A}{\pi}}\), substituting \(A = 78.5\) ft².
\(r=\sqrt{\frac{78.5}{\pi}}\)
## Step2: Calculate the radius
\(r=\sqrt{\frac{78.5}{3.14159}}\approx5.0\) ft
## Step3: Find the diameter
\(d = 2r\), so \(d = 2\times5.0=10.0\) ft
- # Answer:
\(10.0\) ft
3. **For finding circumference given area**:
- First, find the radius from the area formula \(A=\pi r^{2}\) (\(r=\sqrt{\frac{A}{\pi}}\)), then find the circumference using \(C = 2\pi r\).
- **Problem 23: \(A = 64\pi\) mi²**
- # Explanation:
## Step1: Find the radius
Since \(A=\pi r^{2}=64\pi\), then \(r^{2}=64\), and \(r = 8\) mi.
## Step2: Find the circumference
\(C = 2\pi r\), substituting \(r = 8\) mi.
\(C=2\pi\times8=16\pi\approx50.3\) mi
- # Answer:
\(50.3\) mi
- **Problem 24: \(A = 16\pi\) in²**
- # Explanation:
## Step1: Find the radius
Since \(A=\pi r^{2}=16\pi\), then \(r^{2}=16\), and \(r = 4\) in.
## Step2: Find the circumference
\(C = 2\pi r\), substituting \(r = 4\) in.
\(C=2\pi\times4 = 8\pi\approx25.1\) in
- # Answer:
\(25.1\) in
4. **For finding area given circumference**:
- First, find the radius from the circumference formula \(C = 2\pi r\) (\(r=\frac{C}{2\pi}\)), then find the area using \(A=\pi r^{2}\).
- **Problem 25: \(C = 6\pi\) yd**
- # Explanation:
## Step1: Find the radius
\(r=\frac{C}{2\pi}\), substituting \(C = 6\pi\) yd. So \(r=\frac{6\pi}{2\pi}=3\) yd.
## Step2: Find the area
\(A=\pi r^{2}\), substituting \(r = 3\) yd.
\(A=\pi\times3^{2}=9\pi\approx28.3\) yd²
- # Answer:
\(28.3\) yd²
- **Problem 26: \(C = 22\pi\) in**
- # Explanation:
## Step1: Find the radius
\(r=\frac{C}{2\pi}\), substituting \(C = 22\pi\) in. So \(r=\frac{22\pi}{2\pi}=11\) in.
## Step2: Find the area
\(A=\pi r^{2}\), substituting \(r = 11\) in.
\(A=\pi\times11^{2}=121\pi\approx380.1\) in²
- # Answer:
\(380.1\) in²

Answer:

1. For finding radius given circumference:

• The formula for the circumference of a circle is \(C = 2\pi r\), where \(C\) is the circumference and \(r\) is the radius. We can solve for \(r\) as \(r=\frac{C}{2\pi}\).
• Problem 17: \(C = 62.8\) m
• # Explanation:

Step1: Apply the radius - circumference formula

\(r=\frac{C}{2\pi}\), substituting \(C = 62.8\) m and using the calculator's value of \(\pi\).
\(r=\frac{62.8}{2\pi}\)

Step2: Calculate the value

\(r=\frac{62.8}{2\times3.14159}\approx10.0\) m

• # Answer:

\(10.0\) m

• Problem 18: \(C = 89.1\) yd
• # Explanation:

Step1: Apply the radius - circumference formula

\(r = \frac{C}{2\pi}\), substituting \(C = 89.1\) yd.
\(r=\frac{89.1}{2\pi}\)

Step2: Calculate the value

\(r=\frac{89.1}{2\times3.14159}\approx14.2\) yd

• # Answer:

\(14.2\) yd

• Problem 19: \(C = 12.6\) yd
• # Explanation:

Step1: Apply the radius - circumference formula

\(r=\frac{C}{2\pi}\), substituting \(C = 12.6\) yd.
\(r=\frac{12.6}{2\pi}\)

Step2: Calculate the value

\(r=\frac{12.6}{2\times3.14159}\approx2.0\) yd

• # Answer:

\(2.0\) yd

• Problem 20: \(C = 25.1\) ft
• # Explanation:

Step1: Apply the radius - circumference formula

\(r=\frac{C}{2\pi}\), substituting \(C = 25.1\) ft.
\(r=\frac{25.1}{2\pi}\)

Step2: Calculate the value

\(r=\frac{25.1}{2\times3.14159}\approx4.0\) ft

• # Answer:

\(4.0\) ft

1. For finding diameter given area:

• First, use the formula for the area of a circle \(A=\pi r^{2}\) to find the radius \(r = \sqrt{\frac{A}{\pi}}\), and then the diameter \(d = 2r\).
• Problem 21: \(A = 201.1\) in²
• # Explanation:

Step1: Find the radius

\(r=\sqrt{\frac{A}{\pi}}\), substituting \(A = 201.1\) in².
\(r=\sqrt{\frac{201.1}{\pi}}\)

Step2: Calculate the radius

\(r=\sqrt{\frac{201.1}{3.14159}}\approx8.0\) in

Step3: Find the diameter

\(d = 2r\), so \(d=2\times8.0 = 16.0\) in

• # Answer:

\(16.0\) in

• Problem 22: \(A = 78.5\) ft²
• # Explanation:

Step1: Find the radius

\(r=\sqrt{\frac{A}{\pi}}\), substituting \(A = 78.5\) ft².
\(r=\sqrt{\frac{78.5}{\pi}}\)

Step2: Calculate the radius

\(r=\sqrt{\frac{78.5}{3.14159}}\approx5.0\) ft

Step3: Find the diameter

\(d = 2r\), so \(d = 2\times5.0=10.0\) ft

• # Answer:

\(10.0\) ft

1. For finding circumference given area:

• First, find the radius from the area formula \(A=\pi r^{2}\) (\(r=\sqrt{\frac{A}{\pi}}\)), then find the circumference using \(C = 2\pi r\).
• Problem 23: \(A = 64\pi\) mi²
• # Explanation:

Step1: Find the radius

Since \(A=\pi r^{2}=64\pi\), then \(r^{2}=64\), and \(r = 8\) mi.

Step2: Find the circumference

\(C = 2\pi r\), substituting \(r = 8\) mi.
\(C=2\pi\times8=16\pi\approx50.3\) mi

• # Answer:

\(50.3\) mi

• Problem 24: \(A = 16\pi\) in²
• # Explanation:

Step1: Find the radius

Since \(A=\pi r^{2}=16\pi\), then \(r^{2}=16\), and \(r = 4\) in.

Step2: Find the circumference

\(C = 2\pi r\), substituting \(r = 4\) in.
\(C=2\pi\times4 = 8\pi\approx25.1\) in

• # Answer:

\(25.1\) in

1. For finding area given circumference:

• First, find the radius from the circumference formula \(C = 2\pi r\) (\(r=\frac{C}{2\pi}\)), then find the area using \(A=\pi r^{2}\).
• Problem 25: \(C = 6\pi\) yd
• # Explanation:

Step1: Find the radius

\(r=\frac{C}{2\pi}\), substituting \(C = 6\pi\) yd. So \(r=\frac{6\pi}{2\pi}=3\) yd.

Step2: Find the area

\(A=\pi r^{2}\), substituting \(r = 3\) yd.
\(A=\pi\times3^{2}=9\pi\approx28.3\) yd²

• # Answer:

\(28.3\) yd²

• Problem 26: \(C = 22\pi\) in
• # Explanation:

Step1: Find the radius

\(r=\frac{C}{2\pi}\), substituting \(C = 22\pi\) in. So \(r=\frac{22\pi}{2\pi}=11\) in.

Step2: Find the area

\(A=\pi r^{2}\), substituting \(r = 11\) in.
\(A=\pi\times11^{2}=121\pi\approx380.1\) in²

• # Answer:

\(380.1\) in²