Side AD of rectangle ABCD lies in the alpha plane, making an angle of 60 degrees with the plane of the rectangle.

| January 11, 2021 | education

Side AD of rectangle ABCD lies in the alpha plane, making an angle of 60 degrees with the plane of the rectangle. Line BC is removed from the alpha plane by 4 roots of 3 dm. Find the area of a given rectangle if its orthogonal projection onto the plane is alpha square.

Let us denote the orthogonal projections of points B and C onto the plane α through B1 and C1.
ВСǁ AD, means ВСǁα. BB1 is perpendicular to α, so the distance from BC to α is 4√3 dm, and triangle ABB1 is rectangular (angle B1 is straight), ∟A = 60˚.
В ∆АВВ1: АВ1 = ВВ1 / tgА = 4√3 / √3 / 3 = 12 (dm)
S AB1C1D = 12 ^ 2 = 144 (dm ^ 2) – area of a square – orthogonal prection of rectangle ABCD.
The area of the figure is equal to the area of the orthogonal projection divided by the cosine of the angle between the planes.
S ABCD = 144 / cos 60 ˚ = 144 / 0.5 = 288 (dm ^ 2).



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