Faces diagonals

If the diagonals of a cuboid are x, y, and z (wall diagonals or three faces) respectively than find the volume of a cuboid.

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

Result

V =  0.5

Solution:

x=1.3 y=1 z=1.2  x2=a2+b2 y2=b2+c2 z2=a2+c2  a2=x2−b2=x2−y2+c2=x2−y2+z2−a2 2a2=x2−y2+z2  a=x2−y2+z22=1.32−12+1.222≐1.032  ... b=x2+y2−z22=1.32+12−1.222≐0.7906  ... c=−x2+y2+z22=−1.32+12+1.222≐0.6124  V=a⋅ b⋅ c=1.032⋅ 0.7906⋅ 0.6124≐0.4996≐0.5x=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{ \dfrac{ x^2-y^2+z^2 }{ 2 } }=\sqrt{ \dfrac{ 1.3^2-1^2+1.2^2 }{ 2 } } \doteq 1.032 \ \\ \ \\ ... \ \\ b=\sqrt{ \dfrac{ x^2+y^2-z^2 }{ 2 } }=\sqrt{ \dfrac{ 1.3^2+1^2-1.2^2 }{ 2 } } \doteq 0.7906 \ \\ \ \\ ... \ \\ c=\sqrt{ \dfrac{ -x^2+y^2+z^2 }{ 2 } }=\sqrt{ \dfrac{ -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 \doteq 0.4996 \doteq 0.5

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