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# The figure above shows a cylinder with radius 2 and height 5. If point

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Math Expert
Joined: 02 Sep 2009
Posts: 50570
The figure above shows a cylinder with radius 2 and height 5. If point  [#permalink]

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08 Dec 2017, 02:47
00:00

Difficulty:

25% (medium)

Question Stats:

68% (01:09) correct 32% (01:29) wrong based on 38 sessions

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The figure above shows a cylinder with radius 2 and height 5. If points A and B lie on the circumference of the top and bottom of the cylinder respectively, what is the greatest possible straight line distance between A and B?

(A) 3
(B) 5
(C) 7
(D) √29
(E) √41

Attachment:

2017-12-08_1441.png [ 3.55 KiB | Viewed 640 times ]

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Joined: 24 Nov 2016
Posts: 152
Re: The figure above shows a cylinder with radius 2 and height 5. If point  [#permalink]

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08 Dec 2017, 05:23
Bunuel wrote:

The figure above shows a cylinder with radius 2 and height 5. If points A and B lie on the circumference of the top and bottom of the cylinder respectively, what is the greatest possible straight line distance between A and B?

(A) 3
(B) 5
(C) 7
(D) √29
(E) √41

Attachment:
2017-12-08_1441.png

Imagine a rectangle that has one length = diameter of the cylinder (4), and the same height as the cylinder (5).

Now, the largest possible straight line is the diagonal: $$a^2+b^2=c^2 ; 4^2+5^2=c^2; c=√41$$

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Joined: 22 May 2016
Posts: 2090
The figure above shows a cylinder with radius 2 and height 5. If point  [#permalink]

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08 Dec 2017, 09:38
Bunuel wrote:

The figure above shows a cylinder with radius 2 and height 5. If points A and B lie on the circumference of the top and bottom of the cylinder respectively, what is the greatest possible straight line distance between A and B?

(A) 3
(B) 5
(C) 7
(D) √29
(E) √41

Attachment:
2017-12-08_1441.png

To find the length of a diagonal of a cylinder, use the length of the diameter of the cylinder's base and its height as two legs of a right triangle.

One leg runs from B straight across the bottom. The other leg runs straight down from A to meet the line from B. The hypotenuse is the diagonal.

Length of diameter with radius of 2 = 4
Height = 5

Pythagorean theorem, where hypotenuse, $$h$$ = diagonal of cylinder, $$d$$:
$$leg^2 + leg^2 = d^2$$
$$(4^2 + 5^2) = (16 + 25) =41 =d^2$$
$$41 = d^2$$
$$d = \sqrt{41}$$

The figure above shows a cylinder with radius 2 and height 5. If point &nbs [#permalink] 08 Dec 2017, 09:38
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