I will consider the question of the formula for the projections of the faces of a rectangular tetrahedron. I will first consider the orthogonal projection of a segment lying in the plane α , highlighting two cases of the location of this segment relative to the line l=α∩π .
Case 1 AB∥l(Fig. 8). Segment A 1 B 1 , which is an orthogonal projection of segment AB, is equal and parallel to segment AB.

Rice. 8

Case 2 CD⊥l(Fig. 8). By the three perpendiculars theorem, the line C 1 D 1 , which is the orthogonal projection of the line CD, is also perpendicular to the line l. Therefore, ∠CEC 1 is the angle between the plane α and the plane of projections π , i.e. , where C 0 D=C 1 D 1. Therefore |C 1 D 1 |=|CD|∙cosφ
Now consider the issue of orthogonal projection of a triangle.
The area of the orthogonal projection of a triangle onto a plane is equal to the area of the projected triangle multiplied by the cosine of the angle between the plane of the triangle and the plane of projections.

Proof. Projection area of a triangle.
a) Let one of the sides, for example AC, of the projected triangle ABC be parallel to the line l=α∩π (Fig. 9) or lie on it.

Rice. 9

Then its height VN is perpendicular to the straight line l, and the area is equal to, i.e.

Based on the above properties of the orthogonal projection of the segment, I have:

According to the theorem on three perpendiculars, the line B 1 H 1 - the orthogonal projection of the line BH - is perpendicular to the line l, therefore, the segment B 1 H 1 is the height of the triangle A 1 B 1 C 1. That's why . Thus, .
b) None of the sides of the projected triangle ABC is parallel to the line l (Fig. 10). Draw a line through each vertex of the triangle parallel to line l. One of these lines lies between the other two (in the figure, it is line m), and, therefore, divides the triangle ABC into triangles ABD and ACD with heights BH and CE, respectively, drawn to their common side AD (or its continuation), which is parallel to l. The line m 1 - orthogonal projection of the line m - also divides the triangle A 1 B 1 C 1 - the orthogonal projection of the triangle ABC - into triangles A 1 B 1 D 1 and A 1 C 1 D 1 , where . Taking into account (9) and (10), we get

GEOMETRY
Lesson Plans for Grades 10

Lesson 56

Subject. Area of an orthogonal projection of a polygon

The purpose of the lesson: the study of the theorem on the area of \u200b\u200bthe orthogonal projection of a polygon, the formation of students' skills to apply the studied theorem to solving problems.

Equipment: stereometric set, cube model.

During the classes

I. Checking homework

1. Two students reproduce the solutions to problems No. 42, 45 on the board.

2. Frontal interrogation.

1) Define the angle between two planes that intersect.

2) What is the angle between:

a) parallel planes;

b) perpendicular planes?

3) To what extent can the angle between two planes change?

4) Is it true that a plane that intersects parallel planes intersects them at the same angles?

5) Is it true that a plane that intersects perpendicular planes intersects them at the same angles?

3. Checking the correctness of the solution of problems No. 42, 45, which the students recreated on the board.

II. Perception and awareness of new material

Assignment to students

1. Prove that the projection area of a triangle with one side in the projection plane is equal to the product of its area and the cosine of the angle between the plane of the polygon and the projection plane.

2. Prove the theorem for the case when the lattice triangle has one side parallel to the projection plane.

3. Prove the theorem for the case when the lattice triangle has none of its sides parallel to the projection plane.

4. Prove the theorem for any polygon.

Problem solving

1. Find the area of the orthogonal projection of a polygon whose area is 50 cm2 and the angle between the plane of the polygon and its projection is 60°.

2. Find the area of the polygon if the area of the orthogonal projection of this polygon is 50 cm2, and the angle between the plane of the polygon and its projection is 45°.

3. The area of the polygon is 64 cm2, and the area of the orthogonal projection is 32 cm2. Find the angle between the planes of the polygon and its projection.

4. Or maybe the area of the orthogonal projection of the polygon is equal to the area of this polygon?

5. The edge of the cube is a. Find the cross-sectional area of a cube by a plane passing through the top of the base at an angle of 30° to this base and intersecting all side edges. (Answer. )

6. Problem No. 48 (1, 3) from the textbook (p. 58).

7. Problem No. 49 (2) from the textbook (p. 58).

8. The sides of the rectangle are 20 and 25 cm. Its projection onto a plane is similar to it. Find the projection perimeter. (Answer. 72 cm or 90 cm.)

III. Homework

§4, n. 34; Security Question No. 17; tasks No. 48 (2), 49 (1) (p. 58).

IV. Summing up the lesson

Question for the class

1) Formulate a theorem on the area of the orthogonal projection of a polygon.

2) Can the area of the orthogonal projection of a polygon be greater than the area of the polygon?

3) Through the hypotenuse AB of a rectangular triangle ABC the plane α is drawn at an angle of 45° to the plane of the triangle and the perpendicular CO to the plane α. AC \u003d 3 cm, BC \u003d 4 cm. Indicate which of the following statements are correct and which are incorrect:

a) the angle between the planes ABC and α is equal to the angle CMO, where the point H is the base of the altitude CM of the triangle ABC;

b) SD = 2.4 cm;

c) triangle AOC is an orthogonal projection of triangle ABC onto the plane α;

d) the area of triangle AOB is 3 cm2.

(Answer. a) Correct; b) wrong; c) wrong; d) correct.)

Detailed proof of the polygon orthogonal projection theorem

If - projection of a flat n -gon to a plane, then, where is the angle between the planes of the polygons and. In other words, the projection area of a flat polygon is equal to the product of the area of the projected polygon and the cosine of the angle between the projection plane and the plane of the projected polygon.

Proof. I stage. Let's do the proof first for the triangle. Let's consider 5 cases.

1 case. lie in the projection plane .

Let be the projections of points onto the plane, respectively. In our case. Let's assume that. Let - height, then by the theorem of three perpendiculars, we can conclude that - height (- the projection of the inclined, - its base and the straight line passes through the base of the inclined, moreover).

Consider. It is rectangular. By definition of cosine:

On the other hand, since and, then by definition - linear angle dihedral angle formed by the half-planes of the planes and with the boundary line, and, therefore, its measure is also the measure of the angle between the projection planes of the triangle and the triangle itself, that is.

Find the ratio of the area to:

Note that the formula remains true even when . In this case

2nd case. Only lies in the projection plane and is parallel to the projection plane .

Let be the projections of points onto the plane, respectively. In our case.

Let's draw a straight line through the point. In our case, the straight line intersects the projection plane, which means, by the lemma, the straight line also intersects the projection plane. Let it be at a point Since, then the points lie in the same plane, and since it is parallel to the projection plane, it follows from the sign of parallelism of the straight line and the plane that. Therefore, is a parallelogram. Consider and. They are equal on three sides (- common, like opposite sides of a parallelogram). Note that the quadrilateral is a rectangle and is equal (along the leg and hypotenuse), therefore, it is equal on three sides. That's why.

For 1 case is applicable:, i.e..

3rd case. Only lies in the projection plane and is not parallel to the projection plane .

Let the point be the point of intersection of the line with the projection plane. Let us note that i. On 1 occasion: i. Thus we get that

4 case. Vertices do not lie in the projection plane . Consider perpendiculars. Take the smallest among these perpendiculars. Let it be perpendicular. It may turn out that either only, or only. Then we still take it.

Let us set aside a point from a point on a segment, so that and from a point on a segment, a point, so that. Such a construction is possible, since - the smallest of the perpendiculars. Note that is a projection and, by construction. Let us prove that and are equal.

Let's consider a quadrilateral. By condition - perpendiculars to one plane, therefore, according to the theorem, therefore. Since by construction, then, on the basis of a parallelogram (on parallel and equal opposite sides), we can conclude that - a parallelogram. Means, . It is proved similarly that, . Therefore, and are equal on three sides. That's why. Note that and, as opposite sides of parallelograms, therefore, on the basis of the parallelism of the planes, . Since these planes are parallel, they form the same angle with the projection plane.

For the previous cases apply:

5 case. The projection plane intersects the sides . Let's look at straight lines. They are perpendicular to the projection plane, so by the theorem they are parallel. On co-directed rays with origins at points, we set aside equal segments, respectively, so that the vertices lie outside the projection plane. Note that is a projection and, by construction. Let's show that it is equal.

Since and, by construction, then. Therefore, on the basis of a parallelogram (on two equal and parallel sides), - a parallelogram. It can be proved similarly that and are parallelograms. But then, and (as opposite sides), therefore, is equal in three sides. Means, .

In addition, and, therefore, on the basis of the parallelism of the planes. Since these planes are parallel, they form the same angle with the projection plane.

For applicable case 4:.

II stage. Let's split a flat polygon into triangles using diagonals drawn from the vertex: Then, according to the previous cases for triangles: .

Q.E.D.

IN Lately in task C2 there are problems in which it is necessary to construct a section of a polyhedron by a plane and find its area. Such a task is proposed in the demo version. It is often convenient to find the area of a section through the area of its orthogonal projection. The presentation contains a formula for such a solution and a detailed analysis of the problem, which is accompanied by a series of drawings.

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Preparation for the Unified State Examination - 2014 in mathematics. Finding the cross-sectional area through the area of its orthogonal projection. Task C2 Mathematics teacher MBOU secondary school No. 143 of Krasnoyarsk Knyazkina T.V.

Consider the solution of such a problem: cuboid, . The section of the parallelepiped passes through points B and D and forms an angle with the plane ABC. Find the sectional area. It is often convenient to find the area of a section through the area of its orthogonal projection. Finding the area of a triangle in terms of the area of its orthogonal projection is easily illustrated by the following figure:

CH is the height of triangle ABC, C ‘H is the height of triangle ABC ", which is an orthogonal projection of triangle ABC. From right triangle CHC ": The area of triangle ABC" is The area of triangle ABC is Therefore, the area of triangle ABC is equal to the area of triangle ABC' divided by the cosine of the angle between the planes of triangle ABC and triangle ABC ", which is the orthogonal projection of triangle ABC .

Since the area of any polygon can be represented as the sum of the areas of triangles, the area of a polygon is equal to the area of its orthogonal projection onto a plane divided by the cosine of the angle between the planes of the polygon and its projection. We use this fact to solve our problem (see slide 2) The solution plan is as follows: A) We build a section. B) Find its orthogonal projection onto the plane of the base. C) Find the area of the orthogonal projection. D) Find the cross-sectional area.

1. First we need to build this section. Obviously, the segment BD belongs to the section plane and the base plane, that is, it belongs to the line of intersection of the planes:

The angle between two planes is the angle between two perpendiculars that are drawn to the line of intersection of the planes and lie in these planes. Let the point O be the point of intersection of the diagonals of the base. OC - perpendicular to the line of intersection of the planes, which lies in the plane of the base:

2. Determine the position of the perpendicular, which lies in the section plane. (Remember that if a straight line is perpendicular to the projection of an oblique one, then it is also perpendicular to the most oblique one. We are looking for an oblique one by its projection (OC) and the angle between the projection and the oblique one). Find the tangent of the angle COC ₁ between OC ₁ and OC

Therefore, the angle between the section plane and the base plane is greater than between OC ₁ and OC. That is, the section is located somehow like this: K is the intersection point of OP and A ₁C₁, LM||B₁D₁ .

So, here is our section: 3. Find the projection of the BLMD section onto the base plane. To do this, we find the projections of the points L and M .

Quadrilateral BL ₁M₁D is the projection of the section onto the plane of the base. 4. Find the area of the quadrilateral BL ₁M₁D . To do this, subtract the area of the triangle L ₁CM₁ from the area of the triangle BCD Find the area of the triangle L ₁CM₁ . Triangle L ₁CM₁ is similar to triangle BCD . Let's find the similarity coefficient.

To do this, consider m triangles OPC and OKK₁ : Therefore, the area of the triangle L₁CM₁ is 4/25 of the area of the triangle BCD (the ratio of the areas of similar figures is equal to the square of the similarity coefficient). Then the area of quadrilateral BL₁M₁D is equal to 1-4/25=21/25 of the area of triangle BCD and is equal to

5. Now find 6 . And finally, we get: Answer: 112

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Chapter IV. Lines and planes in space. Polyhedra

§ 55. Projection area of a polygon.

Recall that the angle between a line and a plane is the angle between a given line and its projection onto the plane (Fig. 164).

Theorem. The area of the orthogonal projection of the polygon onto the plane is equal to the area of the projected polygon multiplied by the cosine of the angle formed by the plane of the polygon and the projection plane.

Each polygon can be divided into triangles, the sum of the areas of which is equal to the area of the polygon. Therefore, it suffices to prove the theorem for a triangle.

Let /\ ABC is projected onto a plane R. Consider two cases:
a) one of the parties /\ ABC is parallel to the plane R;
b) none of the parties /\ ABC is not parallel R.

Consider first case: let [AB] || R.

Draw through the (AB) plane R 1 || R and project orthogonally /\ ABC on R 1 and on R(Fig. 165); we get /\ ABC 1 and /\ A"B"S".
By the projection property, we have /\ ABC 1 /\ A"B"C", and therefore

S /\ ABC1=S /\ A"B"C"

Let's draw _|_ and the segment D 1 C 1 . Then _|_ , a = φ is the angle between the plane /\ ABC and plane R 1 . That's why

S /\ ABC1 = 1 / 2 | AB | | C 1 D 1 | = 1 / 2 | AB | | CD 1 | cos φ = S /\ ABC cos φ

and hence S /\ A"B"C" = S /\ ABC cos φ.

Let's move on to consideration second case. Draw a plane R 1 || R over that peak /\ ABC, the distance from which to the plane R the smallest (let it be vertex A).
We will design /\ ABC on a plane R 1 and R(Fig. 166); let its projections be respectively /\ AB 1 C 1 and /\ A"B"S".

Let (sun) p 1 = D. Then

S /\ A"B"C" = S /\ AB1 C1 = S /\ ADC1-S /\ ADB1 = (S /\ ADC-S /\ ADB) cos φ = S /\ ABC cos φ

Task. A plane is drawn through the side of the base of a regular triangular prism at an angle φ = 30° to the plane of its base. Find the area of the resulting section if the side of the base of the prism A= 6 cm.

Let's depict the section of this prism (Fig. 167). Since the prism is regular, its side edges are perpendicular to the plane of the base. Means, /\ ABC is a projection /\ ADC, so