Cuboid face diagonals

The lengths of the cuboid edges are in the ratio 1: 2: 3. Will the lengths of its diagonals be the same ratio?

The cuboid has dimensions of 5 cm, 10 cm, and 15 cm. Calculate the size of the wall diagonals of this cuboid.

Result

y =  0

Solution:

a=5 b=10 c=15  a:b:c=1:2:3  u1=a2+b2=52+102=5 511.1803 u2=a2+c2=52+152=5 1015.8114 u3=b2+c2=102+152=5 1318.0278  k1=u1:u2=11.1803:15.81140.7071 k2=u3:u2=18.0278:15.81141.1402 k3=u3:u1=18.0278:11.18031.6125  k1,k2,k31,2,3  y=0a = 5 \ \\ b = 10 \ \\ c = 15 \ \\ \ \\ a:b:c = 1:2:3 \ \\ \ \\ u_{ 1 } = \sqrt{ a^2+b^2 } = \sqrt{ 5^2+10^2 } = 5 \ \sqrt{ 5 } \doteq 11.1803 \ \\ u_{ 2 } = \sqrt{ a^2+c^2 } = \sqrt{ 5^2+15^2 } = 5 \ \sqrt{ 10 } \doteq 15.8114 \ \\ u_{ 3 } = \sqrt{ b^2+c^2 } = \sqrt{ 10^2+15^2 } = 5 \ \sqrt{ 13 } \doteq 18.0278 \ \\ \ \\ k_{ 1 } = u_{ 1 }:u_{ 2 } = 11.1803:15.8114 \doteq 0.7071 \ \\ k_{ 2 } = u_{ 3 }:u_{ 2 } = 18.0278:15.8114 \doteq 1.1402 \ \\ k_{ 3 } = u_{ 3 }:u_{ 1 } = 18.0278:11.1803 \doteq 1.6125 \ \\ \ \\ k_{ 1 },k_{ 2 },k_{ 3 } \ne 1,2,3 \ \\ \ \\ y = 0



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Pythagorean theorem is the base for the right triangle calculator.

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