The diagonals of three faces of a rectangular parallelepiped that meet at a single corner measure \(\sqrt{5}\) cm, \(\sqrt{10}\) cm. and \(\sqrt{13}\) cm. What is the volume of this parallelepiped?
6 \(cm^{3}\)
Let the three edges of the parallelepiped that meet at the corner be \(a\), \(b\) and \(c\). Each face of the box is a rectangle, so a face diagonal is the hypotenuse built on two of these edges.
The three given face diagonals therefore satisfy \(\sqrt{a^2+b^2}=\sqrt{5}\), \(\sqrt{b^2+c^2}=\sqrt{10}\) and \(\sqrt{a^2+c^2}=\sqrt{13}\).
Squaring removes the roots: \(a^2+b^2=5\), \(b^2+c^2=10\) and \(a^2+c^2=13\).
Add all three equations. Every edge appears twice, so \(2(a^2+b^2+c^2)=5+10+13=28\), giving \(a^2+b^2+c^2=14\).
Subtract each pair-sum from this total to isolate the remaining square: \(c^2=14-5=9\), \(a^2=14-10=4\) and \(b^2=14-13=1\).
Taking positive roots (edges are lengths): \(a=2\), \(b=1\), \(c=3\).
Volume of a rectangular parallelepiped is the product of its edges: \(V=abc=2\times1\times3=6\ cm^{3}\). The key idea is that each face diagonal squared equals the sum of two edge-squares, so the system solves cleanly.
The volume of the parallelepiped is 6 cm3.
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