What is the surface area of the solid right circular

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Source: Veritas Prep

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What is the surface area of the solid right circular cylinder with height 10 above?
$$A.\ 18\pi$$
$$B.\ 32\pi$$
$$C.\ 60\pi$$
$$D.\ 78\pi$$
$$E.\ 90\pi$$
The OA is D

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by Brent@GMATPrepNow » Wed Jan 16, 2019 2:55 pm
BTGmoderatorLU wrote:Source: Veritas Prep

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What is the surface area of the solid right circular cylinder with height 10 above?
$$A.\ 18\pi$$
$$B.\ 32\pi$$
$$C.\ 60\pi$$
$$D.\ 78\pi$$
$$E.\ 90\pi$$
The OA is D
Given: radius = 3
Height = 10
Surface area of cylinder = 2πr² + 2πrh
= 2π(3²) + 2π(3)(10)
= 18Ï€ + 60Ï€
= 78Ï€

Answer: D

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by Scott@TargetTestPrep » Mon Jan 21, 2019 5:48 pm
BTGmoderatorLU wrote:Source: Veritas Prep

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What is the surface area of the solid right circular cylinder with height 10 above?
$$A.\ 18\pi$$
$$B.\ 32\pi$$
$$C.\ 60\pi$$
$$D.\ 78\pi$$
$$E.\ 90\pi$$
A right circular cylinder has a circular top and a circular bottom, so the surface area of the the two circular pieces is πr^2 + πr^2 = 2πr^2. The side of the cylinder is a rectangle with height h and length equal to the circumference of the circle 2πr, so its area is 2πrh.

Thus, the surface area of the cylinder is 2Ï€r^2 + 2Ï€rh, so we have:

2Ï€(3^2) + 2Ï€(3)(10) = 18Ï€ + 60Ï€ = 78Ï€

Answer: D

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