We have the following:
[tex]A=2\cdot\pi\cdot r^2[/tex]
we have that the radius is half the diameter, therefore
[tex]\begin{gathered} r=\frac{d}{2}=\frac{10}{2} \\ r=5 \end{gathered}[/tex]
replacing:
[tex]\begin{gathered} A=2\cdot\pi\cdot5^2 \\ A=50\pi\cong157 \end{gathered}[/tex]
The surface area is 50*pi yd^2 or approximately 157 yd^2
