At the base of the straight prism there is a parallelogram with sides of 4 dm and 5 dm and an angle between them of 30 degrees.

| January 9, 2021 | education

At the base of the straight prism there is a parallelogram with sides of 4 dm and 5 dm and an angle between them of 30 degrees. Find the cross-sectional area of the prism by the plane if it is known that it intersects all lateral edges and forms an angle of 45 degrees with the base plane.

Let us determine the area of the parallelogram of the prism lying in the area.
Savsd = AВ * AD * SinВAD = 4 * 5 * Sin30 = 4 * 5 * (1/2) = 10 dm2.
The area of the parallelogram AVSD is the orthogonal projection of the section ADS1V1 onto the plane of the base of the prism.
The area of the orthogonal projection of the AВСD is equal to the area of the projected section ADS1V1 multiplied by the cosine of the angle between the planes.
Sads1 = Sads1v1 * Cos45, Then, Sads1v1 = Sads1v1 * Cos45.
Sads1in1 = 10 / (√2 / 2) = 20 / √2 = 20 * √2 / (√2 * √2) = 10 * √2 dm2.
Answer: The cross-sectional area is 10 * √2 dm2.



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