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24 Dec 2018, 09:33
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Fresh GMAT Club Tests' Question:
A rectangular solid, x by y by z, is inscribed in a sphere, so that all eight of its vertices are on the sphere. If x, y and z are positive integers, and the radius of the sphere is \(\frac{\sqrt{14}}{2}\), what is the volume of the rectangular solid?
A. 6
B. 11
C. 12
D. 18
E. 22
M37-61
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24 Dec 2018, 09:54
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Given radius of the sphere = \(\sqrt{14}/2\)
Diameter = \(\sqrt{14}\)
Also, diameter of the sphere is the diagonal of the rectangle.
Diagonal of the rectangle = \(\sqrt{x^2+y^2+y^2}\) = \(\sqrt{14}\)
\(x^2+y^2+y^2\) = 14.
Since, x, y, and z are all positive integers.
Only possible value is 1,2 and 3.
As \(1^2+2^2+3^2 = 14\)
Hence the volume = 1*2*3 = 6.
A is the answer.
Diameter = \(\sqrt{14}\)
Also, diameter of the sphere is the diagonal of the rectangle.
Diagonal of the rectangle = \(\sqrt{x^2+y^2+y^2}\) = \(\sqrt{14}\)
\(x^2+y^2+y^2\) = 14.
Since, x, y, and z are all positive integers.
Only possible value is 1,2 and 3.
As \(1^2+2^2+3^2 = 14\)
Hence the volume = 1*2*3 = 6.
A is the answer.
29 Nov 2021, 08:47
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Official Solution:
A rectangular solid, \(x\) by \(y\) by \(z\), is inscribed in a sphere, so that all eight of its vertices are on the sphere. If \(x\), \(y\) and \(z\) are positive integers, and the radius of the sphere is \(\frac{\sqrt{14}}{2}\), what is the volume of the rectangular solid?
A. \(6\)
B. \(11\)
C. \(12\)
D. \(18\)
E. \(22\)
The radius of \(\frac{\sqrt{14}}{2}\) means that the diameter of the sphere is \(\sqrt{14}\).
Notice that the diameter of the sphere is equal to the long diagonal of the rectangular solid, which is equal to \(\sqrt{x^2+y^2+z^2}\). So, we have that \(\sqrt{x^2+y^2+z^2}=\sqrt{14}\).
Square: \(x^2+y^2+z^2=14\)
Since \(x\), \(y\) and \(z\) are positive integers, then \((x, y, z)=(1, 2, 3)\) (in any order).
The volume of the rectangular solid is therefore \(xyz=6\).
Answer: A
A rectangular solid, \(x\) by \(y\) by \(z\), is inscribed in a sphere, so that all eight of its vertices are on the sphere. If \(x\), \(y\) and \(z\) are positive integers, and the radius of the sphere is \(\frac{\sqrt{14}}{2}\), what is the volume of the rectangular solid?
A. \(6\)
B. \(11\)
C. \(12\)
D. \(18\)
E. \(22\)
The radius of \(\frac{\sqrt{14}}{2}\) means that the diameter of the sphere is \(\sqrt{14}\).
Notice that the diameter of the sphere is equal to the long diagonal of the rectangular solid, which is equal to \(\sqrt{x^2+y^2+z^2}\). So, we have that \(\sqrt{x^2+y^2+z^2}=\sqrt{14}\).
Square: \(x^2+y^2+z^2=14\)
Since \(x\), \(y\) and \(z\) are positive integers, then \((x, y, z)=(1, 2, 3)\) (in any order).
The volume of the rectangular solid is therefore \(xyz=6\).
Answer: A
