# At the base of a straight parallelepiped lies a rhombus with an acute angle of 60 degrees.

**At the base of a straight parallelepiped lies a rhombus with an acute angle of 60 degrees. The diagonal of the side face is inclined to the base plane at an angle of 45 degrees, and the area of this face is 10 cm2. Find the total surface area.**

The area of the side face of the rhombus, by condition, is 10 cm2. The angle С1DC = 45, then the angle CC1D = 180 – 90 – 45 = 45, then the triangle CC1D is rectangular and isosceles, CC1 = CD.

Sd1s1sd = 10 = CC1 * CD = CD ^ 2.

СD = √10 cm.

Since there is a rhombus at the base of the prism, AB = AD = BC = CD = √10 cm.

Determine the area of the base of the prism.

Sbn = AD * AB * Sin60 = √10 * √10 * √3 / 2 = 5 * √3 cm2.

The area of the lateral surface of the prism is: Sbok = 4 * Sd1s1sd = 4 * 10 = 40 cm2.

Determine the total surface area of the rhombus.

S floor = 2 * S main + S side = 2 * 5 * √3 + 40 = 10 * √3 + 40 = 10 * (√3 + 4) cm2.

Answer: The total surface area of the prism is 10 * (√3 + 4) cm2.