In a right parallelepiped ABCDA1B1C1D1 BC = 7, CD = 15, angle BCD = 60. A plane is drawn through the diagonal BD

| February 17, 2021 | education

In a right parallelepiped ABCDA1B1C1D1 BC = 7, CD = 15, angle BCD = 60. A plane is drawn through the diagonal BD and the vertex C1 at an angle of 45 degrees to the base plane. Find the area of the side surface of the parallelepiped.

Determine the area of ​​the base of the parallelepiped.

Sbn = ВС * СD * Sin60 = 7 * 15 * √3 / 2 = 105 * √3 / 2 cm2.

In the triangle of the ВСD, we define the segment of the ВD by the cosine theorem.

ВD ^ 2 = BC ^ 2 + СD ^ 2 – 2 * BC * СD * Cos60 = 49 + 225 – 2 * 7 * 15 * 1/2 = 274 – 105 = 169.

ВD = 13 cm.

The diagonal of the parallelogram at the base of the parallelepiped divides the base into two equal triangles, then Svsd = Sbn / 2 = 105 * √3 / 4 cm2.

Svsd = ВD * CH / 2 = 13 * CH / 2 = 105 * √3 / 4.

CH = 210 * √3 / 52 = 105 * √3 / 26 cm.

The right-angled triangle CC1H is isosceles, since the angle CHC1 = 450, then CC1 = 105 * √3 / 26 cm.

Then Sbok = Rosn * CC1 = 44 * 105 * √3 / 26 = 2310 * √3 / 13 cm2.

Answer: The lateral surface area is 2310 * √3 / 13 cm2.



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