Answer :
To find the circumference of a circle given its area, you can follow these detailed steps:
1. Given Data:
- The area of the circle is 64π square inches.
2. Formulas to Use:
- The formula for the area of a circle is [tex]\( A = πr² \)[/tex].
- The formula for the circumference of a circle is [tex]\( C = 2πr \)[/tex].
3. Solve for the Radius (r):
- Start with the area formula: [tex]\( 64π = πr² \)[/tex].
- To isolate [tex]\( r² \)[/tex], divide both sides of the equation by [tex]\( π \)[/tex]:
[tex]\[ 64 = r² \][/tex]
- Next, solve for [tex]\( r \)[/tex] by taking the square root of both sides:
[tex]\[ r = \sqrt{64} = 8 \][/tex]
So, the radius [tex]\( r \)[/tex] is 8 inches.
4. Calculate the Circumference (C):
- Use the circumference formula: [tex]\( C = 2πr \)[/tex].
- Substitute the radius [tex]\( r = 8 \)[/tex] into the formula:
[tex]\[ C = 2π(8) = 16π \][/tex]
5. Final Answer:
- The circumference of the circle is [tex]\( 16π \)[/tex] inches.
So, the circumference [tex]\( C \)[/tex] in inches, expressed in terms of [tex]\( π \)[/tex], is [tex]\( 16π \)[/tex].
1. Given Data:
- The area of the circle is 64π square inches.
2. Formulas to Use:
- The formula for the area of a circle is [tex]\( A = πr² \)[/tex].
- The formula for the circumference of a circle is [tex]\( C = 2πr \)[/tex].
3. Solve for the Radius (r):
- Start with the area formula: [tex]\( 64π = πr² \)[/tex].
- To isolate [tex]\( r² \)[/tex], divide both sides of the equation by [tex]\( π \)[/tex]:
[tex]\[ 64 = r² \][/tex]
- Next, solve for [tex]\( r \)[/tex] by taking the square root of both sides:
[tex]\[ r = \sqrt{64} = 8 \][/tex]
So, the radius [tex]\( r \)[/tex] is 8 inches.
4. Calculate the Circumference (C):
- Use the circumference formula: [tex]\( C = 2πr \)[/tex].
- Substitute the radius [tex]\( r = 8 \)[/tex] into the formula:
[tex]\[ C = 2π(8) = 16π \][/tex]
5. Final Answer:
- The circumference of the circle is [tex]\( 16π \)[/tex] inches.
So, the circumference [tex]\( C \)[/tex] in inches, expressed in terms of [tex]\( π \)[/tex], is [tex]\( 16π \)[/tex].