In a right parallelepiped, the sides of the base are equal to m and n, one of the corners of the base is 60 degrees.

| September 9, 2021 | education

In a right parallelepiped, the sides of the base are equal to m and n, one of the corners of the base is 60 degrees. Find the side surface of the parallelepiped if the larger dioganal of the base is equal to the smaller diogonal of the parallelepiped.

In triangle ABD, we define the length of the side BD by the cosine theorem.

BD ^ 2 = AB ^ 2 + AD ^ 2 – 2 * AB * AD * Cos120 = m ^ 2 + n ^ 2 – 2 * m * n * (1/2) = m2 + n2 – m * n.

Since there is a parallelogram at the base, the sum of its adjacent angles is 180, then the angle ABC = (180 – BAD) = (180 – 60) = 120.

In the triangle ABC, by the cosine theorem, we define the length of the side of the AC.

AC ^ 2 = AB ^ 2 + BC ^ 2 – 2 * AB * BC * Cos120 = m ^ 2 + n ^ 2 – 2 * m * n * (-1 / 2) = m ^ 2 + n ^ 2 + m * n.

By hypothesis, AC = DВ1, then in a right-angled triangle DВВ1, by the Pythagorean theorem,

BB1 ^ 2 = DB1 ^ 2 – BD ^ 2 = m ^ 2 + n ^ 2 + m * n – (m ^ 2 + n2 – m * n) = 2 * m * n.

BB1 = √ (2 * m * n).

The perimeter of the base of the parallelepiped is: Rosn = 2 * (m + n), then Sbok = Rosn * BB1 =

2 * (m + n) * √ (2 * m * n) cm2.

Answer: The lateral surface area is 2 * (m + n) * √ (2 * m * n) cm2.



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