# The base of a straight parallelepiped is a rhombus with a larger diagonal of 30 cm. The smaller diagonal

**The base of a straight parallelepiped is a rhombus with a larger diagonal of 30 cm. The smaller diagonal of the parallelepiped makes an angle of 45 degrees with the base plane and is equal to 24√2. Find the total surface area.**

The diagonal ВD of the rhombus ABCD is the projection of the smaller diagonal DВ1 of the parallelepiped onto the base plane, then the angle ВDВ1 = 45, and then the triangle ВDВ1 is rectangular and isosceles, BB1 = ВD.

By the Pythagorean theorem, 2 * BD ^ 2 = DB1 ^ 2 = 2 * 576.

BD = BB1 = 24 cm.

The diagonals of the rhombus, at the intersection point, are halved and intersect at right angles. AO = AC / 2 = 30/2 = 15 cm, OD = BD / 2 = 24/2 = 12 cm.

By the Pythagorean theorem, AD ^ 2 = AO ^ 2 + OD ^ 2 = 225 + 144 = 369.

AD = 3 * √41 cm.

Determine the area of the base of the parallelepiped.

Sb = АС * ВD / 2 = 30 * 24/2 = 360 cm2.

The lateral surface area is:

Sside = Ravsd * BB1 = 4 * 3 * √41 * 24 = 288 * √41 cm2.

Then Sпов = 2 * Sсн + S side = 2 * 360 + 288 * √41 = 720 + 288 * √41 cm2.

Answer: The total surface area is 720 + 288 * √41 cm2.