The diagonal of the rectangular parallelepiped, at the base of which the square lies, is 8cm

The diagonal of the rectangular parallelepiped, at the base of which the square lies, is 8cm, and the diagonal of the side face is 7cm. Find the height of the box.

Consider a right-angled triangle B1AD, in which the angle A is straight, since the parallelepiped is rectangular, and the diagonal AB1 belongs to the lateral one, which is perpendicular to the ABP.

Then, by the Pythagorean Theorem, AD ^ 2 = DB1 ^ 2 – AB1 ^ 2.

AD ^ 2 = 8 ^ 2 – 7 ^ 2 = 64 – 49 = 15.

AD = AB = BC = CD = √15.

Consider a right-angled triangle AB1B, and by the Pythagorean theorem, define the leg BB1, which is the height of the parallelepiped.

BB1 ^ 2 = AB1 ^ 2 – AB ^ 2 = 7 ^ 2 – (√15) ^ 2 = 49 – 15 = 34.

BB1 = √34 cm.

Answer: BB1 = √34 cm.