Recall the surface area for the following figures.
![\begin{gathered} \text{Cylinder}=2\pi rh+2\pi r^2 \\ \\ \text{The term }2\pi r^2\text{ includes a cover both the top and bottom of the cylinder} \\ \text{Since we will be using only the bottom modify the formula such that it only} \\ \text{includes the bottom part} \\ \\ \text{Pen Holder Surface Area}=2\pi rh+\pi r^2 \end{gathered}]()
Given that
height = h = 10.5 cm
radius = r = 3 cm
π = 22/7
Substitute the following given and we have the surface area for the pen holder
![\begin{gathered} \text{Pen Holder Surface Area}=2\pi rh+\pi r^2 \\ \text{Pen Holder Surface Area}=2((22)/(7))(3\operatorname{cm})(10.5\operatorname{cm})+((22)/(7))(3\operatorname{cm})^2 \\ \text{Pen Holder Surface Area}=198\operatorname{cm}+((22)/(7))(9\operatorname{cm}) \\ \text{Pen Holder Surface Area}=198\operatorname{cm}+(198)/(7)\operatorname{cm} \\ \text{Pen Holder Surface Area}=(1584)/(7)\operatorname{cm}^2 \end{gathered}]()
Now that we have the surface area, multiply it by 35 since there are 35 competitors in the competition
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