Faces diagonals

If a cuboid's diagonals are x, y, and z (wall diagonals or three faces), then find the cuboid volume.

Solve for x=1.3, y=1, z=1.2

Correct result:

V =  0.4996

Solution:

$$\begin{gathered} x=1.3 \\ y=1 \\ z=1.2 \\ {x}^2=a^2+b^2 \\ {y}^2=b^2+c^2 \\ {z}^2=a^2+c^2 \\ {a}^2=x^2-b^2=x^2-y^2+c^2=x^2-y^2+z^2-a^2 \\ 2a^2=x^2-y^2+z^2 \\ a=\sqrt{{\displaystyle \frac{x^2-y^2+z^2}{2}}}=\sqrt{{\displaystyle \frac{1.3^2-1^2+1.2^2}{2}}}\doteq 1.032 \\ \dots \\ b=\sqrt{{\displaystyle \frac{x^2+y^2-z^2}{2}}}=\sqrt{{\displaystyle \frac{1.3^2+1^2-1.2^2}{2}}}\doteq 0.7906 \\ \dots \\ c=\sqrt{{\displaystyle \frac{-x^2+y^2+z^2}{2}}}=\sqrt{{\displaystyle \frac{-1.3^2+1^2+1.2^2}{2}}}\doteq 0.6124 \\ V=a\cdot b\cdot c=1.032\cdot 0.7906\cdot 0.6124=0.4996 \end{gathered}$$

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Matematik

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https://www.hackmath.net/en/math-problem/16153

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