The lengths of the diagonals of the three faces of the rectangular parallelepiped having a common

The lengths of the diagonals of the three faces of the rectangular parallelepiped having a common vertex are 5 cm, 2√13 cm, and 3√5. find the diagonal of the parallelepiped. I know the answer is 61.

Let us introduce the notation: the lengths of the edges of the parallelepiped – a, b, c, the diagonals of the faces of the parallelepiped – d1, d2, d3, the diagonal of the parallelepiped – D.

The square of any diagonal of a face is equal to the sum of the squares of the edges of that face. We can conditionally write a system of equations:

a ^ 2 + b ^ 2 = d1 ^ 2;

a ^ 2 + c ^ 2 = d2 ^ 2;

b ^ 2 + c ^ 2 = d3 ^ 2.

Adding the left and right sides of all three equations, we get:

a ^ 2 + b ^ 2 + a ^ 2 + c ^ 2 + b ^ 2 + c ^ 2 = d1 ^ 2 + d2 ^ 2 + d3 ^ 2;

2 * (a ^ 2 + b ^ 2 + c ^ 2) = d1 ^ 2 + d2 ^ 2 + d3 ^ 2.

It is known that the square of the diagonal of a parallelepiped is equal to the sum of the squares of its three edges:

a ^ 2 + b ^ 2 + c ^ 2 = D ^ 2.

Consequently:

2 * D ^ 2 = d1 ^ 2 + d2 ^ 2 + d3 ^ 2;

D ^ 2 = 0.5 * (d1 ^ 2 + d2 ^ 2 + d3 ^ 2) = 0.5 * (5 ^ 2 + (2√13) ^ 2 + (3√5) ^ 2) = 0, 5 * (25 + 52 + 45) = 61;

D = √61 cm – the required diagonal of the given parallelepiped.