At the base of a straight parallelepiped lies a parallelogram with sides a = 3√2 and b = √2 and an acute angle

At the base of a straight parallelepiped lies a parallelogram with sides a = 3√2 and b = √2 and an acute angle of 45 degrees. The area of the lateral surface is 4 times the area of its base. Find the height of the box.

Determine the area of the parallelogram lying at the base of the parallelepiped through the sides and the angle.

Sbad = AB * BC * SinBAD = 3 * √2 * √2 * (√2 / 2) = 3 * √2 cm2.

By convention, the lateral surface area is four times the base area.

Side = 4 * Sb = 4 * 3 * √2 = 12 * √2 cm2.

Also Sbok = Rosn * AA1.

12 * √2 = 2 * (√2 + 3 * √2) * AA1.

AA1 = H = (12 * √2) / (8 * √2) = 1.5.

Answer: The height of the parallelepiped is 1.5 cm.