- #1

rocomath

- 1,755

- 1

[tex]r=radius[/tex]

[tex]W=2r[/tex]

[tex]P=2\pi r+2(L+2r)[/tex]

And isn't the total area just the sum of the areas? [tex]A=A_{circle}+A_{rectange}\rightarrow A=\pi r^2 + 2rL[/tex]