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The surface of a cylinder of radius r and height h...

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The surface of a cylinder of radius r and height h...

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The surface of a cylinder of radius r and height h is painted. The two bases of the cylinder are painted red, and the side of the cylinder is painted green. if the red area is equal to the green area, which of the following must be true?

A. h = r
B. h = π·r
C. h = 2r
D. h = r/2
E. h = r^2

The OA is A

Please, can any expert explain this PS question for me? I tried to solve it but I can't get the correct answer. I need your help. Thanks.

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Quote:
The surface of a cylinder of radius r and height h is painted. The two bases of the cylinder are painted red, and the side of the cylinder is painted green. if the red area is equal to the green area, which of the following must be true?

A. h = r
B. h = π·r
C. h = 2r
D. h = r/2
E. h = r^2

The OA is A

Please, can any expert explain this PS question for me? I tried to solve it but I can't get the correct answer. I need your help. Thanks.
Hi swerve,
Let's take a look at your question.

The red area is the area of two circular bases of the cylinder.
$$\text{Red Area}=2\pi r^2$$

The green area if equal to the curved surface of the cylinder, i.e.
$$\text{Green Area}=2\pi rh$$

If the red area is equal to the green area the,
$$2\pi rh=2\pi r^2$$
$$rh=r^2$$
$$h=r$$

Therefore, Option A is correct.

Hope it helps.
I am available if you'd like any follow up.

_________________
GMAT Prep From The Economist
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Hi swerve,

We're told that the surface of a cylinder of radius R and height H is painted: the two bases of the cylinder are painted red, and the side of the cylinder is painted green and the red area is equal to the green area. We're asked which of the following must be true. To answer this question, you first need the area formulas involved.

The 'top' and 'bottom' of the cylinder are circles, so each of those two areas is (pi)(R^2). The total of those 2 areas is: (2)(pi)(R^2).
The 'side' of the cylinder is the CIRCUMFERENCE of the circle multiplied by the height. That area is (2)(pi)(R)(H).

We're told that the two areas are EQUAL to one another, so....
(2)(pi)(R^2) = (2)(pi)(R)(H)

You can 'cancel out' the 2s and the pis....
(R^2) = (R)(H)
R = H

Final Answer: A

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

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