# At the base of a rectangular parallelepiped there is a rhombus with a side of 12 cm and an angle of 60.

**At the base of a rectangular parallelepiped there is a rhombus with a side of 12 cm and an angle of 60. The smaller diagonal of the parallelepiped is 13 cm. Find the total surface area of the parallelepiped.**

The smaller diagonal of the rhombus lies opposite the acute angle of the rhombus, the diagonals of the rhombus are the bisectors of its corners.

In a right-angled triangle formed by the halves of the diagonals of the rhombus and its side, half of the smaller diagonal lies opposite the half of the acute angle of the rhombus. Half of the smaller diagonal can be found as the product of the side of the rhombus by the sine of half of the acute angle: d / 2 = 12 * sin60 / 2 = 12 * sin30 = 12 * 0.5 = 6 cm.

This means that the length of the smaller diagonal of the rhombus is 6 * 2 = 12 cm.

Consider a rectangular triangle, in which the legs are the height of the parallelepiped and the smaller diagonal of its base, the hypotenuse is the smaller diagonal of the parallelepiped.

The square of the height of a parallelepiped can be found as the difference between the squares of the smaller diagonal of the parallelepiped and the smaller diagonal of its base-rhombus:

h ^ 2 = 13 ^ 2-12 ^ 2 = 169-144 = 25;

h = √25 = 5 cm.

The total surface area of a parallelepiped is equal to the sum of the surfaces of the side and two bases:

Stot = 2Sn + S side.

The area of the base-rhombus is equal to the product of the square of its side by the sine of the angle between the sides:

Sb = 12 ^ 2 * sin60 = 144 * (√3 / 2) = 72√3 cm2.

The lateral surface area of a straight parallelepiped can be found as the product of the parallelepiped’s height by the perimeter of the base:

Sside = h * P = 5 * 12 * 4 = 240 cm2.

S total = 2Sn + S side = 2 * 72√3 + 240 = 144√3 + 240≈489.4 cm2.