# 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.

Answer: 1152 cubic meters. cm.