Diameter=d=2ft
We know
[tex]\boxed{\sf Lateral\:Surface\:Area=2πrh}[/tex]
[tex]\\ \sf\longmapsto Lateral\: Surface\:Area=2\times 3.14\times 2\times 1[/tex]
[tex]\\ \sf\longmapsto Lateral\;Surface\:Area=4(3.14)[/tex]
[tex]\\ \sf\longmapsto Lateral\:Surface\:Area=12.56ft^3[/tex]
[tex]\begin{gathered} {\underline{\boxed{ \rm { \purple{Surface \: \: area \: = \: 2 \: \pi \: r \: h \: + \: 2 \: \pi \: {r}^{2} }}}}}\end{gathered}[/tex]
[tex]\large{\bf{{{\color{navy}{h \: = \: 2 \: ft. }}}}}[/tex]
[tex]\bf \large \longrightarrow \: \: r \: = \: \frac{Diameter}{2} [/tex]
[tex]\bf \large \longrightarrow \: \: r \: = \: \frac{2}{2} \\ [/tex]
[tex]\bf \large \longrightarrow \: \: r \: = \: \cancel\frac{2}{2} \: ^{1} \\ [/tex]
[tex]\large{\bf{{{\color{navy}{r \: = \: 1 \: ft. \: }}}}}[/tex]
[tex]\bf \hookrightarrow \: \: \: 2 \: \times \: 3.14 \times \: 1 \: ft \: \times \: 2 \: ft \: + \: 2 \: \times \: 3.14 \: \times \: {(1 \: ft)}^{2}[/tex]
[tex]\bf \hookrightarrow \: \: \:6.28 \: ft \: \times \: 2 \: ft\: \: + \: 6.28 \: ft[/tex]
[tex]\bf \hookrightarrow \: \: \:12.56 \: {ft}^{2} \: + \: 6.28 \: ft[/tex]
[tex]\bf \hookrightarrow \: \: \:18.84 \: {ft} \: ^{2} [/tex]
Hence , the surface area of cylinder is 18.84 ft²
Round to the nearest 10 of 18.84 is 18.8
