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(D) The penholder is a cylinder with a base,so its surface area is the sum of the curved surface area and the area of the circular base.Radius $(r) =

Vedclass · Accepted Answer

$7920$

Vedclass · Answer

(D) The penholder is a cylinder with a base,so its surface area is the sum of the curved surface area and the area of the circular base.Radius $(r) = 3 \,cm$Height $(h) = 10.5 \,cm$Surface area of $1$ penholder $= 2\pi rh + \pi r^2 = \pi r(2h + r)$$= \frac{22}{7} 	imes 3 	imes (2 	imes 10.5 + 3) \,cm^2$$= \frac{22}{7} 	imes 3 	imes (21 + 3) \,cm^2$$= \frac{22}{7} 	imes 3 	imes 24 \,cm^2 = \frac{1584}{7} \,cm^2$Total cardboard required for $35$ competitors $= 35 	imes \left(\frac{1584}{7}ight) \,cm^2$$= 5 	imes 1584 \,cm^2 = 7920 \,cm^2$Thus,$7920 \,cm^2$ of cardboard is required.

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