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29 Nov 2021, 08:46
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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. \(22\)
A. \(6\)
B. \(11\)
C. \(12\)
D. \(18\)
E. \(22\)
29 Nov 2021, 08:46
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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
