At the base of the rectangular parallelepiped ABCDA1B1C1D1 there is a rhombus ABCD

| May 10, 2021 | education

At the base of the rectangular parallelepiped ABCDA1B1C1D1 there is a rhombus ABCD with a side equal to a and an angle BAD equal to 60 degrees. Plane BC1D makes an angle of 60 degrees with the base plane. Find the total surface area of the parallelepiped.

At the base of the parallelepiped lies a rhombus, the acute angle of which is 60, then the obtuse angle is 120. Diagonal BD divides obtuse angles by 60, therefore, triangles ABD and ACD are equilateral, so they have all angles equal to 60. Then BD = AB = a.

Consider a right-angled triangle COD, in which the hypotenuse CD = a, the angle OCD = 30.

Then the leg CO = CD * Cos30 = a * √3 / 2.

Consider a right-angled triangle С1СО, which has a leg СО = a * √3 / 2, an angle С1ОС = 60, then CC1 = СО * tg60 = (a * √3 / 2) * √3 = a * 3/2.

Then S side = 4 * a * a * 3/2 = 6 * a ^ 2 cm2.

The area of ​​the rhombus at the base is: Srombus = a ^ 2 * SinBAD = a ^ 2 * √3 / 2.

The area of ​​the base of the parallelepiped is equal to two areas of the rhombus.

Sbasn = 2 * a ^ 2 * √3 / 2 = a ^ 2 * √3.

Then the total area is: S = Sbok + Sbasn = 6 * a ^ 2 + a ^ 2 * √3 = a ^ 2 * (6 + √3).

Answer: S = a ^ 2 * (6 + √3).



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