The diagonals of the three faces of a cuboid are $x, y, z$ respectively. What is the volume of the cuboid?

\frac{x y z}{2 \sqrt{2}}

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\frac{\sqrt{(y^{2} + z^{2}) (z^{2} + x^{2}) (x^{2} + y^{2})}}{2 \sqrt{2}}

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\frac{\sqrt{(y^{2} + z^{2} - x^{2}) (z^{2} + x^{2} - y^{2}) (x^{2} + y^{2} - z^{2})}}{2 \sqrt{2}}

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None of the above

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Solution

The correct option is D $\frac{\sqrt{(y^{2} + z^{2} - x^{2}) (z^{2} + x^{2} - y^{2}) (x^{2} + y^{2} - z^{2})}}{2 \sqrt{2}}$
If $l, b, h$ be the dimensions of the cuboid.
$x^{2} = l^{2} + b^{2}; y^{2} = b^{2} + h^{2}; z^{2} = h^{2} + l^{2}$
Therefore, $x^{2} + y^{2} - z^{2} = 2 b^{2}$ , so $b = \sqrt{\frac{1}{2} (x^{2} + y^{2} - z^{2})}$
Similarly, $l = \sqrt{\frac{1}{2} (x^{2} + z^{2} - y^{2})}$ and
$h = \sqrt{\frac{1}{2} (y^{2} + z^{2} - x^{2})}$
$Volume = l \times b \times h$ is the same as (c).

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