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A circular tub with radius 1/2 meter and height 1/4 meter has a band

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A circular tub with radius 1/2 meter and height 1/4 meter has a band [#permalink]

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New post 08 Dec 2017, 02:51
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A circular tub with radius 1/2 meter and height 1/4 meter has a band painted around its circumference, as shown above. What is the surface area of the painted band in square meters?

(A) π/20
(B) 5π/12
(C) 5π
(D) 10π
(E) 15π

[Reveal] Spoiler:
Attachment:
2017-12-08_1441_003.png
2017-12-08_1441_003.png [ 8.68 KiB | Viewed 372 times ]
[Reveal] Spoiler: OA

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Re: A circular tub with radius 1/2 meter and height 1/4 meter has a band [#permalink]

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New post 08 Dec 2017, 05:04
Bunuel wrote:
Image
A circular tub with radius 1/2 meter and height 1/4 meter has a band painted around its circumference, as shown above. What is the surface area of the painted band in square meters?

(A) π/20
(B) 5π/12
(C) 5π
(D) 10π
(E) 15π

[Reveal] Spoiler:
Attachment:
2017-12-08_1441_003.png


Height of band is 5 cm = 1/20 meter

Cylinder Surface Area: \(2πr^2+2πrh\)

Band Surface Area: \(2πr*height.of.band.around = 2π(1/2)(1/20) = π/20\)

(A) is the answer.
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A circular tub with radius 1/2 meter and height 1/4 meter has a band [#permalink]

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New post 08 Dec 2017, 14:45
Bunuel wrote:
Image
A circular tub with radius 1/2 meter and height 1/4 meter has a band painted around its circumference, as shown above. What is the surface area of the painted band in square meters?

(A) π/20
(B) 5π/12
(C) 5π
(D) 10π
(E) 15π

[Reveal] Spoiler:
Attachment:
2017-12-08_1441_003.png

The surface area of the band equals the band's height * (circumference of cylinder's base): \(h * 2\pi r\)

Band's height:
\(5cm *
\frac{1meter}{100cm}=\frac{5}{100}m=\frac{1}{20}m\)

Band's surface area, in square meters:
\(h * 2\pi r\)
\(\frac{1}{20} * (2)\pi (\frac{1}{2})\)
\(\frac{\pi}{20}\)

Answer A

* This problem has two wrinkles: you use only part of a cylinder's surface area formula, and the given height of the cylinder does not matter.
Surface area of a cylinder:
\((2\pi r)(2) + (2\pi r)(h) =
(4\pi r + 2\pi r h)\)
\((2\pi r)(2)\) is the circular area of the cylinder's top and bottom, i.e., (area of circle * 2)
\((2\pi r)(h)\) is the area of the rectangle (if "unrolled). One length of the rectangle is the cylinder's circumference. The other length of the rectangle is the height of cylinder.

In this case, the band's surface area has nothing to do with the first expression (areas of top and bottom). And the band's area, though calculated the same as the rectangle's (circumference * height), has nothing to do with the given height of the rectangle, because the band's height in centimeters must be translated to a fraction of a meter anyway.

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A circular tub with radius 1/2 meter and height 1/4 meter has a band   [#permalink] 08 Dec 2017, 14:45
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