Question

4.) two circles have circumferences that add up to 12π centimeters, and their area adds to 20π centimeters squared. find the radius of each.

Explanation:

Step1: Set up equations

Let the radii of the two circles be $r_1$ and $r_2$. The circumference of a circle is $C = 2\pi r$ and the area is $A=\pi r^{2}$. We have the equations: $2\pi r_1 + 2\pi r_2=12\pi$ and $\pi r_1^{2}+\pi r_2^{2}=20\pi$.
Simplify the first - equation:
$2\pi r_1 + 2\pi r_2=12\pi$ can be divided by $2\pi$ to get $r_1 + r_2 = 6$, so $r_2=6 - r_1$.
Simplify the second - equation:
$\pi r_1^{2}+\pi r_2^{2}=20\pi$ can be divided by $\pi$ to get $r_1^{2}+r_2^{2}=20$.

Step2: Substitute $r_2$ into the second - simplified equation

Substitute $r_2 = 6 - r_1$ into $r_1^{2}+r_2^{2}=20$:
$r_1^{2}+(6 - r_1)^{2}=20$.
Expand $(6 - r_1)^{2}$ using the formula $(a - b)^{2}=a^{2}-2ab + b^{2}$, where $a = 6$ and $b = r_1$.
$r_1^{2}+36-12r_1+r_1^{2}=20$.
Combine like - terms:
$2r_1^{2}-12r_1 + 36-20=0$.
$2r_1^{2}-12r_1 + 16 = 0$.
Divide the entire equation by 2:
$r_1^{2}-6r_1 + 8 = 0$.

Step3: Solve the quadratic equation

Factor the quadratic equation $r_1^{2}-6r_1 + 8 = 0$.
$(r_1 - 2)(r_1 - 4)=0$.
Set each factor equal to zero:
If $r_1 - 2=0$, then $r_1 = 2$.
If $r_1 - 4=0$, then $r_1 = 4$.
When $r_1 = 2$, $r_2=6 - 2 = 4$.
When $r_1 = 4$, $r_2=6 - 4 = 2$.

Answer:

The radii of the two circles are 2 cm and 4 cm.