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Is the length of the longest line that can be drawn inside a cube grea

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Joined: 02 Sep 2009
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Is the length of the longest line that can be drawn inside a cube grea  [#permalink]

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23 Jul 2017, 23:27
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25% (medium)

Question Stats:

90% (01:54) correct 10% (02:44) wrong based on 16 sessions

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Is the length of the longest line that can be drawn inside a cube greater than the length of the longest line that can be drawn inside a right circular cylinder?

(1) Each side of the cube is half the height of the cylinder but twice the radius of the cylinder.
(2) The sum of the lengths of the radius of cylinder, the height of the cylinder and a side of the cube is 14

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Is the length of the longest line that can be drawn inside a cube grea  [#permalink]

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24 Jul 2017, 18:41
Bunuel wrote:
Is the length of the longest line that can be drawn inside a cube greater than the length of the longest line that can be drawn inside a right circular cylinder?

(1) Each side of the cube is half the height of the cylinder but twice the radius of the cylinder.
(2) The sum of the lengths of the radius of cylinder, the height of the cylinder and a side of the cube is 14

It should be A

The longest line we can draw in a cube is a diagonal to the farthest side. The longest line we can draw inside a cylinder is a line between a point on the top edge and a diagonal across to the opposite side at the base.

Statement 1: Sufficient

Let us call side of cube $$X$$, radius of cylinder $$R$$, diameter $$2R$$ or $$B$$, and height $$H$$.

Square:

From the attached figure, it is apparent, diagonal of cube is $$X\sqrt{3}$$. Just use Pythagoras theorem twice and you would reach this answer.

Cylinder:

Diagonal of Cylinder can be calculated by considering the triangle in the figure attached, with height $$H$$ and radius $$R$$. Statement 1 tells us

$$R = \frac{X}{2}$$
$$H = 2X$$

Using Pythagoras' theorem, we know.

Diagonal

$$D = \sqrt{(2R)^2 + H^2}$$ -- Base of triangle is $$B$$ which is $$2R$$

Replacing values of $$R$$ and $$H$$

$$D = \sqrt{X^2 + 4X^2} = X\sqrt{5}$$

Since $$X\sqrt{5}$$ is greater than $$X\sqrt{3}$$, we know the longest line inside the cylinder is larger.

Statement 2: Insufficient

This does not give us any proportions that could be used to compare diagonals.
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Is the length of the longest line that can be drawn inside a cube grea   [#permalink] 24 Jul 2017, 18:41
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