# The base of the rectangle is a rhombus with a perimeter of 40 cm. One of the diagonals of the rhombus is 12 cm.

The base of the rectangle is a rhombus with a perimeter of 40 cm. One of the diagonals of the rhombus is 12 cm. Find the volume of the parallelepiped if its large diagonal is 20 cm.

ABCD is the rhombus that lies at the base of the parallelepiped. AC = 12 cm (see picture).
Since the perimeter is 40 cm, each side of the rhombus is 10 cm.
The diagonals of the rhombus bisect each other and intersect at right angles.
AO = 6, AB = 10. By the Pythagorean theorem, we find OB:
OB * OB = AB * AB – AO * AO = 100 – 36 = 64
OB = 8, then
BD = 2 * OB = 2 * 8 = 16 cm.
Find the area of ​​the rhombus:
S = (AC * BD) / 2 = 12 * 16/2 = 96 sq. Cm.

In the parallelepiped ABCDEFGH (see figure), the largest diagonal is HB, since BD> AC.
The HDB corner is right because the parallelepiped is rectangular. Then, by the Pythagorean theorem:
HD * HD = HB * HB – BD * BD = 400 – 256 = 144
HD = 12 cm.
Find the volume of the parallelogram:
V = S * HD = 96 * 12 = 1152 cc cm.