Answer :
Certainly! Let's solve the problem step-by-step.
Given:
- The area of the circle, [tex]\( A = \pi r^2 \)[/tex]
- The circumference of the circle, [tex]\( C = 2\pi r \)[/tex]
We need to find the radius [tex]\( r \)[/tex] such that the area and the circumference are equal.
So we set up the equation:
[tex]\[ \pi r^2 = 2\pi r \][/tex]
Let's solve for [tex]\( r \)[/tex].
1. Remove the common factor [tex]\( \pi \)[/tex] from both sides:
[tex]\[ r^2 = 2r \][/tex]
2. Rearrange the equation to get all terms on one side:
[tex]\[ r^2 - 2r = 0 \][/tex]
3. Factor the quadratic equation:
[tex]\[ r(r - 2) = 0 \][/tex]
4. Solve for [tex]\( r \)[/tex] by setting each factor equal to zero:
- First factor:
[tex]\[ r = 0 \][/tex]
- Second factor:
[tex]\[ r - 2 = 0 \][/tex]
[tex]\[ r = 2 \][/tex]
Therefore, the solutions to the equation are [tex]\( r = 0 \)[/tex] and [tex]\( r = 2 \)[/tex].
Since a radius of [tex]\( 0 \)[/tex] does not make sense in the context of a real circle, we discard [tex]\( r = 0 \)[/tex].
Thus, the radius of the circle such that its area and its circumference are equal is:
[tex]\[ r = 2 \][/tex]
This completes the solution.
Given:
- The area of the circle, [tex]\( A = \pi r^2 \)[/tex]
- The circumference of the circle, [tex]\( C = 2\pi r \)[/tex]
We need to find the radius [tex]\( r \)[/tex] such that the area and the circumference are equal.
So we set up the equation:
[tex]\[ \pi r^2 = 2\pi r \][/tex]
Let's solve for [tex]\( r \)[/tex].
1. Remove the common factor [tex]\( \pi \)[/tex] from both sides:
[tex]\[ r^2 = 2r \][/tex]
2. Rearrange the equation to get all terms on one side:
[tex]\[ r^2 - 2r = 0 \][/tex]
3. Factor the quadratic equation:
[tex]\[ r(r - 2) = 0 \][/tex]
4. Solve for [tex]\( r \)[/tex] by setting each factor equal to zero:
- First factor:
[tex]\[ r = 0 \][/tex]
- Second factor:
[tex]\[ r - 2 = 0 \][/tex]
[tex]\[ r = 2 \][/tex]
Therefore, the solutions to the equation are [tex]\( r = 0 \)[/tex] and [tex]\( r = 2 \)[/tex].
Since a radius of [tex]\( 0 \)[/tex] does not make sense in the context of a real circle, we discard [tex]\( r = 0 \)[/tex].
Thus, the radius of the circle such that its area and its circumference are equal is:
[tex]\[ r = 2 \][/tex]
This completes the solution.
