Section in a regular quadrangular prism. Section in a regular quadrangular prism In a regular quadrangular prism abcda1b1c1d1

In a regular quadrangular prism ABCDA 1 B 1 C 1 D 1, the base sides are equal to 2, and the side edges are equal to 5. Point E is marked on the edge AA 1 so that AE: EA 1 = 3: 2. Find the angle between the planes ABC and BED 1 .

Solution. Let line D 1 E intersect line AD at point K. Then planes ABC and BED 1 will intersect along line KB.

From the point E we drop the perpendicular EH to the line KB, then the segment AH (projection of EH) will be perpendicular to the line KB (three perpendiculars theorem).

Angle AHE is the linear angle of the dihedral angle formed by planes ABC and BED 1 .

Since AE: EA 1 = 3: 2, we get: .

From the similarity of triangles A 1 D 1 E and AKE we get: .

In a right triangle AKB with right angle A: AB = 2, AK = 3, ; where is the height
.

From right triangle AHE with right angle A we get: and ∠AHE = arctg(√13/2).

Answer: arctg(√13/2).

Tasks for independent solution

1. In cuboid ABCDA 1 B 1 C 1 D 1 AB 1 \u003d 2, AD \u003d AA 1 \u003d 1. Find the angle between the line AB and the plane ABC 1.

2. In a right hexagonal prism ABCDEFA 1 B 1 C 1 D 1 E 1 F 1 all angles are equal to 1. Find the distance from point B to the plane DEA 1 .

3. In a rectangular parallelepiped ABCDA 1 B 1 C 1 D 1 AB \u003d 1, AA 1 \u003d 2. Find the angle between the line AB 1 and the plane ABC 1.

Let's consider another two-point stereometric problem from the training KIMs.

Task.In the right quadrangular prism ABCDA 1 B 1 C 1 D 1 the side AB of the base is equal to 5, and the side edge AA 1 is equal to the square root of five. On the ribs of the sun and C 1 D 1 points K and L are marked respectively, and SC=2, and C 1 L =1. Plane gparallel to line B D and contains points K and L.

a) Prove that the line A 1 C is perpendicular to the planeg.

b) Find the volume of the pyramid, the top of which is the point A 1, and the base is the section of the given prism by the planeg.

Solution.a) Carefully execute the drawing and analyze the data. Because ABCDA 1 B 1 C 1 D 1 - a regular quadrangular prism, which means the base ABCD - a square with a side of 5. The side edges are perpendicular to the bases. Since the planegpasses through point K and is parallel to line B D , then the line of intersection of the planegand plane ABC is parallel to line B D (If another plane is drawn through a line parallel to a given plane, then the line of intersection of these planes will be parallel to the given line.).

Draw a line parallel to B through point K. D to the intersection with CD at the point M. Hence, KM is perpendicular to AC ( because diagonals of a square BD and AC are perpendicular ).

Triangles BCD and SKM are similar (both rectangular and isosceles), so CM=KS=2. By the Pythagorean theorem from the triangle SKM we find that KM=2√2, and from the triangle BCD BD=5 √2 . The diagonals of a square are equal, so AC = BD =5 √2 .

Now, through the dot L draw a straight line parallel to D to the intersection with B 1 C 1 at the point T. Along the segment T L plane KM L crosses the upper base ( If two parallel planes are intersected by a third plane, then the lines of intersection will be parallel). So T C 1 = C 1 L =1. From triangle T LC1 according to the Pythagorean theorem T L = √2.

In an isosceles trapezoid CT L M point H - the middle of the upper base, point N - the middle of the lower base, which means H N – trapezium height, N N perpendicular to KM. This means that the KM is perpendicular to the plane AA 1 C, including the straight line A 1 C.

Consider the diagonal section of a prism rectangle AA 1 C 1 C. From the point H we drop a perpendicular to AC. Then N E \u003d EU \u003d H C 1 \u003d 0.5 √2. NOT \u003d C C 1 \u003d √5.

In triangles AA 1 C and N PC angle PCA is common. The tangent of the angle AA 1 C is 5√2 : √5 = √10 Tangent of angle H N E from triangle H N E is equal to √5: 0.5 √2 = √10 . So the angles AA 1 C and H N E are equal. But then the remaining angles A 1 AC \u003d N PC=90 ⁰ . We have A 1 C perpendicular to straight lines H N and KM, so A 1 C is perpendicular to the plane of the trapezoid CT L M. What was required to be proved.

To find the volume of the pyramid A 1 ct L M, you need to find the area of ​​the trapezoid KT L M and height A 1 R. From triangle H N E by the Pythagorean theorem H N 2 =5.5. Trapezium area CT L M is equal to H N * (T L + KM) / 2 \u003d √5.5 * (√2 + 2 √2) / 2 \u003d 1.5 √11.

Exercise.

In a regular quadrangular prism ABCDA 1 B 1 C 1 D 1, the sides of the base are equal to 3, and the side edges are equal to 4. Point E is marked on the edge AA 1 so that AE: EA 1 = 1: 3.

a) Construct the line of intersection of the planes ABC and BED 1 .

b) Find the angle between planes ABC and BED 1 .

Solution:

a) Construct a line of intersection of the planesABC andBed 1.

Let's build a plane BED 1 . Points E and D 1 lie in the same plane, so we draw a straight line ED 1 .

Points E and B lie in the same plane, so we draw a line EB. Since the faces of a regular quadrangular prism are parallel, we draw a line BF in the face of BB 1 C 1 C parallel to the line ED 1. Points F and D 1 lie in the same plane, so let's draw a straight line FD 1 . We got the required plane BED 1 .

Since the line ED 1 and the line AD lie in the same plane ADD 1, they intersect at the point K, which lies in the plane ABC. Points K and B lie in the planes ABC and BED 1, therefore, the planes ABC and BED 1 intersect along the straight line KB. The desired line of intersection of the planes ABC and BED 1 is built.

b) Find the angle between the planesABC andBed 1

The segment AE is perpendicular to the plane ABC, from the point E we drop the perpendicular EH to the line KB. The point H lies in the plane ABC, then AH is the projection of EH onto the plane ABC. Through the point H there passes a line perpendicular to the inclined line EH, then by the theorem on three perpendiculars the segment AH is perpendicular to the line KB.

The angle ∠EHA is the linear angle of the dihedral angle formed by the planes ABC and BED 1 . Angle ∠EHA is the required angle between planes ABC and BED 1 . Let's find the value of this angle.

Consider right triangle EHA (∠A = 90˚):

By condition AE: EA 1 = 1: 3, then AE: AA 1 = 1: 4.

Triangles AKE and A 1 D 1 E are similar, then

A 1 D 1 = 3, AE = 1, A 1 E = AA 1 - AE = 3

Consider a right triangle AKB (∠А = 90˚).