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A rectangular solid with dimensions X × Y × Z is inscribed in a sphere
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14 May 2021, 00:24
14 May 2021, 00:24
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Question Stats:
41% (02:23) correct
58% (02:48) wrong based on 141 sessions
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A rectangular solid with dimensions X × Y × Z is inscribed in a sphere with all of its eight vertices on the sphere. X, Y and Z are all positive integers, and the radius of the sphere is \(\frac{\sqrt{14}}{2}\)
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Volume of the rectangular solid |
2\(π\) |
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Re: A rectangular solid with dimensions X × Y × Z is inscribed in a sphere
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15 May 2021, 01:58
15 May 2021, 01:58
14
Explanation:
Since all the vertices lie on the sphere, the diameter of the sphere is the diagonal of the rectangular solid
i.e. \(\sqrt{X^2 + Y^2 + Z^2} = 2(\frac{\sqrt{14}}{2})\)
\(X^2 + Y^2 + Z^2 = 14\)
Also X, Y and Z are positive integers, we can clearly see that \(1^2 + 2^2 + 3^2 = 14\)
Therefore, \(XYZ = (1)(2)(3) = 6\)
Col. A: \(6\)
Col. B: \(2(3.14) = 6.28\)
Hence, option B
Since all the vertices lie on the sphere, the diameter of the sphere is the diagonal of the rectangular solid
i.e. \(\sqrt{X^2 + Y^2 + Z^2} = 2(\frac{\sqrt{14}}{2})\)
\(X^2 + Y^2 + Z^2 = 14\)
Also X, Y and Z are positive integers, we can clearly see that \(1^2 + 2^2 + 3^2 = 14\)
Therefore, \(XYZ = (1)(2)(3) = 6\)
Col. A: \(6\)
Col. B: \(2(3.14) = 6.28\)
Hence, option B
General Discussion
Re: A rectangular solid with dimensions X × Y × Z is inscribed in a sphere
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14 May 2021, 10:37
14 May 2021, 10:37
explanation plz
