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30% of the surface area of a right circular cylinder is shad [#permalink]
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30% of the surface area of a right circular cylinder is shaded. If the diameter of the base of the cylinder is 10 and the height is 4, what is the surface area of the unshaded region?


A. \(27\pi\)

B. \(40\pi\)

C. \(63\pi\)

D. \(70\pi\)

E. \(90\pi\)
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Re: 30% of the surface area of a right circular cylinder is shad [#permalink]
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Carcass wrote:
30% of the surface area of a right circular cylinder is shaded. If the diameter of the base of the cylinder is 10 and the height is 4, what is the surface area of the unshaded region?


A. \(27\pi\)

B. \(40\pi\)

C. \(63\pi\)

D. \(70\pi\)

E. \(90\pi\)


Total surface area of cylinder = 2πr² + 2πrh

r = 5 (since the DIAMETER = 10)
h = 4

So, surface area = 2π(5²) + 2π(5)(4)
= 50π + 40π
= 90π

If 30% is shaded, then 70% is UNSHADED.
70% of 90π = 63π

Answer: C

Cheers,
Brent
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Re: 30% of the surface area of a right circular cylinder is shad [#permalink]
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Re: 30% of the surface area of a right circular cylinder is shad [#permalink]
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