At the base of the rectangular parallelepiped lies a rhombus with diagonals equal to 12cm and 16cm.

univerkov | June 15, 2021 | education

At the base of the rectangular parallelepiped lies a rhombus with diagonals equal to 12cm and 16cm. The height of the parallelepiped is 8 cm. Find its total surface area.

Since there is a rhombus at the base, the sides of the base are equal, AB = AD = BC = CD.

The diagonals of the rhombus intersect at right angles and at the point of intersection are divided in half, then the triangle AOD is rectangular, in which OA = AC / 2 = 16/2 = 8 cm, OD = BD / 2 = 12/2 = 6 cm.Then, by the theorem Pythagoras AD ^ 2 = ОА ^ 2 + ОD ^ 2 = 64 + 36 = 100.

AD = CD = BC = AB = 10 cm.

Let’s define the area of a rhombus through its diagonals.

Sb = (АС * ВD) / 2 = 16 * 12/2 = 96 cm2.

Determine the area of the side face of the parallelepiped.

Sside = АА1 * АD = 8 * 10 = 80 cm2.

Determine the total area of the parallelogram.

S floor = 2 * Sb + 4 * S side = 2 * 96 + 4 * 80 = 192 + 320 = 512 cm2.

Answer: The total area of the parallelogram is 512 cm2.

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