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The figure above represents a right circular cylindrical tube made out

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The figure above represents a right circular cylindrical tube made out  [#permalink]

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New post 08 Dec 2017, 03:46
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Question Stats:

33% (01:54) correct 67% (02:03) wrong based on 69 sessions

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The figure above represents a right circular cylindrical tube made out of paper. The circumference of each circular base is 10 centimeters, the length of AB is 12 centimeters, and BD is a diameter of a base. If the tube is cut along AB, opened and flattened, what is the length of AD in centimeters?

(A) 13
(B) 17
(C) 22
(D) 2√61
(E) √146.5

Attachment:
2017-12-08_1440_002.png
2017-12-08_1440_002.png [ 6.2 KiB | Viewed 1294 times ]

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Re: The figure above represents a right circular cylindrical tube made out  [#permalink]

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New post 08 Dec 2017, 04:01
Bunuel wrote:
Image
The figure above represents a right circular cylindrical tube made out of paper. The circumference of each circular base is 10 centimeters, the length of AB is 12 centimeters, and BD is a diameter of a base. If the tube is cut along AB, opened and flattened, what is the length of AD in centimeters?

(A) 13
(B) 17
(C) 22
(D) 2√61
(E) √146.5

Attachment:
2017-12-08_1440_002.png


Radius = >
2*pie * r = 10
r = 5/pie
ABD will be right angle Triangle
AD is Hypotenuse and hence can be calculated
AD = \(\sqrt{(12^2 + (5/pie)^2)}\)
AD = \(\sqrt{(144 + 25/~9.5)} = \sqrt{146.5}\)
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Re: The figure above represents a right circular cylindrical tube made out  [#permalink]

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New post Updated on: 09 Dec 2017, 09:06
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Bunuel wrote:
Image
The figure above represents a right circular cylindrical tube made out of paper. The circumference of each circular base is 10 centimeters, the length of AB is 12 centimeters, and BD is a diameter of a base. If the tube is cut along AB, opened and flattened, what is the length of AD in centimeters?

(A) 13
(B) 17
(C) 22
(D) 2√61
(E) √146.5

Attachment:
2017-12-08_1440_002.png


If the tube is opened and flattened, then we have a rectangle with length 12 = AB, and width 10 = the circumference.

Now, lines BD and AC are equal to half the width = 10/2 = 5.

So to find the distance AD, we need to find the diagonal of the rectangle with length 12 = AB and width 5 = BD

\(a^2+b^2=c^2, 12^2+5^2=c^2, 169=c^2, c=13\)

(A) is the answer.

Originally posted by exc4libur on 08 Dec 2017, 06:44.
Last edited by exc4libur on 09 Dec 2017, 09:06, edited 1 time in total.
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Re: The figure above represents a right circular cylindrical tube made out  [#permalink]

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New post 08 Dec 2017, 07:15
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Bunuel wrote:
Image
The figure above represents a right circular cylindrical tube made out of paper. The circumference of each circular base is 10 centimeters, the length of AB is 12 centimeters, and BD is a diameter of a base. If the tube is cut along AB, opened and flattened, what is the length of AD in centimeters?

(A) 13
(B) 17
(C) 22
(D) 2√61
(E) √146.5

Attachment:
2017-12-08_1440_002.png


When you cut along AB and open it, you get a rectangle. Notice that in this rectangle, A is on the corner and D is in the centre of the opposite edge. So ABD will form a right triangle.

AB^2 + BD^2 = AD^2

AB is simply the length of the tube at 12 cm but notice what BD is. It is half the circumference of the circular base (since it is opened and flattened) which is 10 cm. So BD = 10/2 = 5 cm

This will be a 5-12-13 triangle then and AD will be 13 cm.

Answer (A)
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Re: The figure above represents a right circular cylindrical tube made out  [#permalink]

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New post 08 Dec 2017, 21:15
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VeritasPrepKarishma wrote:
Bunuel wrote:
Image
The figure above represents a right circular cylindrical tube made out of paper. The circumference of each circular base is 10 centimeters, the length of AB is 12 centimeters, and BD is a diameter of a base. If the tube is cut along AB, opened and flattened, what is the length of AD in centimeters?

(A) 13
(B) 17
(C) 22
(D) 2√61
(E) √146.5

Attachment:
The attachment 2017-12-08_1440_002.png is no longer available


When you cut along AB and open it, you get a rectangle. Notice that in this rectangle, A is on the corner and D is in the centre of the opposite edge. So ABD will form a right triangle.

AB^2 + BD^2 = AD^2

AB is simply the length of the tube at 12 cm but notice what BD is. It is half the circumference of the circular base (since it is opened and flattened) which is 10 cm. So BD = 10/2 = 5 cm

This will be a 5-12-13 triangle then and AD will be 13 cm.

Answer (A)


Responding to a pm:

Quote:
Can you help me understand how BD is half the circumference of the circular base. I am unable to visualize and hence not getting the point.


Here is the original roll.

Attachment:
Rectangular Roll Original.jpg
Rectangular Roll Original.jpg [ 11.96 KiB | Viewed 1003 times ]


When you open it, this is what it looks like.

Attachment:
Rectangular Roll.jpg
Rectangular Roll.jpg [ 14.09 KiB | Viewed 1004 times ]

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Re: The figure above represents a right circular cylindrical tube made out  [#permalink]

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New post 22 Dec 2017, 06:48
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Got the answer to be A :|

Can anyone explain how it is D??
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Re: The figure above represents a right circular cylindrical tube made out  [#permalink]

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New post 22 Dec 2017, 06:57
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Re: The figure above represents a right circular cylindrical tube made out  [#permalink]

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New post 22 Dec 2017, 07:06
Bunuel wrote:
MSarmah wrote:
Got the answer to be A :|

Can anyone explain how it is D??


The correct answer is A. Edited. Thank you.



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Re: The figure above represents a right circular cylindrical tube made out  [#permalink]

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Re: The figure above represents a right circular cylindrical tube made out   [#permalink] 20 Jun 2019, 06:26
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