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.



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