The cross section in the correct quadrangular prism. Section in the right quadrangular prism in the correct quadrangular prism ABCDA1B1C1D1

In the correct four-trigger prism of ABCDA 1 B 1 C 1 D 1, the base sides are 2, and the side ribs are equal to 5. On the edge AA 1, point E is noted that AE: EA 1 \u003d 3: 2. Find the angle between the ABC and Bed planes 1 .

Decision. Let the direct D 1 E crosses the direct AD at the point K. Then the ABC and Bed 1 planes will be reset in a straight line KB.

From point E, we lower the perpendicular EH into direct Kb, then the segment AH (projection Eh) will be perpendicular to the direct Kb (three perpendicular theorem).

The angle of AHE is a linear angle of a couphed angle formed by the ABC and Bed 1 planes.

Since AE: EA 1 \u003d 3: 2, we get :.

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

In a rectangular triangle AKB with a direct angle A: AU \u003d 2, AK \u003d 3 ,; Where height is from
.

From a rectangular triangle AHE with a straight angle A we get: and ∠ ahe \u003d arctg (√13 / 2).

Answer: arctg (√13 / 2).

Tasks for self-decide

1. B. rectangular parallelepiped ABCDA 1 B 1 C 1 D 1 AB 1 \u003d 2, AD \u003d AA 1 \u003d 1. Find the angle between direct AV and the ABC 1 plane.

2. In the direct hexagonal prism of ABCDEFA 1 B 1 C 1 D 1 E 1 F 1 1 All angles are equal to 1. Find the distance from the point to the DEA 1 plane.

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

The task.

In the right one quadrangular prize ABCDA 1 B 1 C 1 D 1 The sides of the base are equal to 3, and the side ribs are equal to 4. On the edge Aa 1, point E is noted so that AE: Ea 1 \u003d 1: 3.

a) Build the direct intersection of the ABC and Bed 1 planes.

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

Decision:

a) Build a direct intersection of the planesABC I.Bed 1..

We construct the plane BED 1. The points E and D 1 lie in the same plane, so we will spend direct ED 1.

The points E and B lie in the same plane, so we will spend direct ev. So the verge of the correct quadrangular prism is parallel, we carry out in the boundaries of BB 1 C 1 with direct BF parallel to the direct ED 1. Points F and D 1 lie in the same plane, so we will spend direct FD 1. Received the desired plane Bed 1.

Since the straight line ED 1 and direct AD lie in the same plane of Add 1, then they intersect at the point K, lying in the ABC plane. Points to and B lie in the planes ABC and BED 1, therefore, the plane ABC and BED 1 intersect in a straight line. The desired direct intersection of the ABC and Bed 1 planes is built.

b) Find the angle between the planesABC I.Bed 1.

The segment AE is perpendicular to the ABC plane, out of the point E to lower the EH perpendicular to the direct quarter. Point H lies in the ABC plane, then AH is the projection of EH on the ABC plane. Through the point H passes direct, perpendicular to the inclined Eh, then by the theorem about the three perpendicular sections of AH perpendicular to the direct quarter.

The angle of ∠eha is a linear angle of a couphed angle formed by the ABC and Bed 1 planes. The angle of ∠eha is the desired angle between the planes ABC and BED 1. We find the magnitude of this corner.

Consider right triangle EHA (∠A \u003d 90˚):

By condition AE: Ea 1 \u003d 1: 3, then AE: AA 1 \u003d 1: 4.

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

A 1 D 1 \u003d 3, AE \u003d 1, A 1 E \u003d AA 1 - AE \u003d 3

Consider the rectangular triangle AKB (∠A \u003d 90˚).

Consider the next two-tier stereometric task of training kim.

A task. In the right quadrangular prismABCDA 1 B 1 C 1 D 1 Side AB foundation 5, and the side edge AA 1 is equal to the root square of five. On the edges of the sun andC 1 D 1 Answered points K and L accordingly, with SK \u003d 2, andC 1 L \u003d 1. Plane g. parallel to direct B.D. and contains points to andL.

a) Prove that straight a 1 with perpendicular planeg..

b) Find the volume of the pyramid, the vertex of which is point A 1, and the base is a cross section of this prism planeg..

Decision. a) carefully perform the drawing and analyze the data. AsABCDA 1 B 1 C 1 D 1 - the right quadrangular prism, it means the basisAbcd. - Square with side 5. Side edges perpendicular to the bases. Since the planeg. passes through the point to and parallel to the straightD. then the plane intersection lineg. and the plane of ABC parallel to the straightD. (If the other plane is directly parallel to this plane, then the intersection line of these planes will be parallel to this direct).

Through the point to spending a straight parallel inD. before the intersection of S.CD At the point M. means km perpendicular to AS ( as diagonal Square BD. and speakers perpendicular ).

Triangles bcd. and SCM are similar (both rectangular and equal), it means CM \u003d COP \u003d 2. According to the Pythagore Theorem of the SCM triangle, we find that KM \u003d 2√2, and from the triangleBCD BD \u003d 5 √2 . The diagonal of the square is equal, it means both \u003dBd \u003d 5 √2.

Now, through the pointL. we carry out a straight parallel inD. before the intersection of S.B 1 C 1 At the point of T. by section TL Plane km l crosses the top base ( If two parallel planes cross the third plane, then the intersection lines will be parallel). So T.C 1 \u003d C 1 L \u003d 1. From the triangle T.LC 1. According to Pythagore's theorem TL \u003d √2.

In an equalized trapezium KTL. M dot n - middle top base, pointN. - the middle of the lower base, which meansN. - Height of the trapezium, nN. perpendicular to km. So the CM is perpendicular to the AA 1 C plane, including direct A 1 S.

Consider the diagonal cross section of the prism rectangleAA 1 C 1 C. From the point n downward perpendicular on the AU. ThenN E \u003d EU \u003d H C 1 \u003d 0.5 √2. Not \u003d C 1 \u003d √5.

In triangles AA 1 C andN. RS Angle of Rs - General. Tangent angle AA 1 C is 5√2: √5 \u003d √10 Tangent Angle H n E from the triangle nN E is equal to √5: 0.5 √2 \u003d √10 . So angles AA 1 C and NN. E is equal. But then the remaining angles of 1 spell \u003dN RS \u003d 90 ⁰ . We have 1 with perpendicular directN. and km, it means a 1 with perpendicular to the plane of the KT trapezL. M. What was required to prove.

In order to find the volume of the pyramid A 1 CTL. M, you need to find the square of the KT trapezL. M and height A 1 r. From the triangle nN. E by the theorem of Pythagora nN 2. \u003d 5.5. KT trapezium squareL m is equal to n * (T l + km) / 2 \u003d √5.5 * (√2 + 2 √2) / 2 \u003d 1.5 √11.