we use the formula of the area
![A=\pi* r^2](https://img.qammunity.org/2023/formulas/mathematics/high-school/8wbdxz67tl4r7vg1yhkaxc4upw6qhkrx91.png)
where A is the area and r the radious
Radious
we can replace the value of the area and solve for r
![\begin{gathered} 36\pi=\pi* r^2 \\ \\ r^2=(36\pi)/(\pi) \\ \\ r^2=36 \\ r=6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ypunj7wesnhv2gyvwyg3wtyuzmodp4w56y.png)
Value of the radious is 6inches
Diameter
diameter is twice radious
![\begin{gathered} d=6*2 \\ d=12 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7nxyl8ps8u73lac7i5ry20nn6fk8ayr7ku.png)
Value of the diamter is 12 inches
Circumference
we use the formula of the periemter
![P=2\pi* r](https://img.qammunity.org/2023/formulas/mathematics/college/7jdbiit2j29qv8gxgbt0m468z3w9bhe4q5.png)
and replace the value of radious
![\begin{gathered} P=2\pi*6 \\ P=12\pi \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/vgpqr1m07zbnzj1fklg355vkfghc82oyju.png)
Value of the circumference is 12pi inches