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Retired Moderator
Joined: 16 Apr 2020
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The surface distance between two points on the surface of a cube is
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01 Aug 2021, 11:18
01 Aug 2021, 11:18
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Question Stats:
41% (01:16) correct
58% (01:13) wrong based on 67 sessions
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Attachment:
The surface distance between two points on the surface of a cube.png [ 3.23 KiB | Viewed 5882 times ]
Quantity A |
Quantity B |
surface distance (in cm) between vertices A and B |
\(6\sqrt{3}\) |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Retired Moderator
Joined: 19 Nov 2020
Posts: 326
Given Kudos: 64
GRE 1: Q160 V152
The surface distance between two points on the surface of a cube is
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01 Aug 2021, 13:07
01 Aug 2021, 13:07
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Aha, initially I thought the question is asking about a distance between two points, vertices 
, which is not asked in this question. Cube's diagonal would be \(d=side*{\sqrt{3}}\).
Here, surface distance is the distance on the surface of cube's sides. Like, we are drawing a line on the sides of a cube shaped box. Therefore, the required distance is a diagonal between two sides and the side itself \(side*{\sqrt{2}}+side\). Answer is \(6*{\sqrt{2}}+6\), and it's indeed A.
Thanks for posting a good question.
, which is not asked in this question. Cube's diagonal would be \(d=side*{\sqrt{3}}\).Here, surface distance is the distance on the surface of cube's sides. Like, we are drawing a line on the sides of a cube shaped box. Therefore, the required distance is a diagonal between two sides and the side itself \(side*{\sqrt{2}}+side\). Answer is \(6*{\sqrt{2}}+6\), and it's indeed A.
Thanks for posting a good question.
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Re: The surface distance between two points on the surface of a cube is
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02 Aug 2021, 03:01
02 Aug 2021, 03:01
2
Explanation:
The key thing to notice in this question is - "the shortest path on the surface of the cube"
Refer to the figure below,
Surface distance = \(\sqrt{12^2 + 6^2} = \sqrt{144 + 36} = \sqrt{180} = 6\sqrt{5}\)
Hence, option A
The key thing to notice in this question is - "the shortest path on the surface of the cube"
Refer to the figure below,
Surface distance = \(\sqrt{12^2 + 6^2} = \sqrt{144 + 36} = \sqrt{180} = 6\sqrt{5}\)
Hence, option A
Attachments
surface distance (in cm) between vertices A and B.png [ 3.17 KiB | Viewed 5611 times ]
