I will consider the question of the formula for the projections of the faces of a rectangular tetrahedron. Preliminarily, I will consider the orthogonal projection of a segment lying in the plane α, highlighting two cases of the location of this segment relative to the straight line l = α∩π.
Case 1.
AB∥l(fig. 8). The segment A 1 B 1, which is the orthogonal projection of the segment AB, is equal and parallel to the segment AB.
Rice. eight
Case 2.
CD⊥l(fig. 8). By the three perpendicular 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 I will consider the issue of orthogonal triangle design.
The area of an 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 the projections.
Proof. The projected area of the triangle. According to the three perpendicular theorem, the line B 1 H 1 - the orthogonal projection of the line BN - 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, .
a) Let one of the sides, for example AC, of the projected triangle ABC be parallel to the straight line l = α∩π (Fig. 9) or lie on it.
Then its height VN is perpendicular to the straight line l, and the area is equal, i.e.
Rice. nine
b) None of the sides of the projected triangle ABC is parallel to the line l (Fig. 10). Draw a straight line through each vertex of the triangle parallel to the straight line l. One of these lines lies between two others (in the figure it is line m), and, therefore, splits triangle ABC into triangles ABD and ACD with heights, respectively, BH and CE, drawn to their common side AD (or its continuation), which is parallel l. Line m 1 - orthogonal projection of line m - also splits triangle A 1 B 1 C 1 - orthogonal projection of triangle ABC - into triangles A 1 B 1 D 1 and A 1 C 1 D 1, where. Taking into account (9) and (10), I get
Recall that the angle between a straight line and a plane is the angle between a given straight line and its projection onto a 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 plane of the projection.
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 \ (\ Delta \) ABC be projected onto the plane R... Consider two cases:
a) one of the sides \ (\ Delta \) ABC is parallel to the plane R;
b) none of the sides \ (\ Delta \) ABC is parallel R.
Consider first case: let [AB] || R.
Let's draw a plane through (AB) R 1 || R and project orthogonally \ (\ Delta \) ABC on R 1 and on R(fig. 165); we get \ (\ Delta \) ABC 1 and \ (\ Delta \) A'B'S '.
By the property of projection, we have \ (\ Delta \) ABC 1 \ (\ cong \) \ (\ Delta \) A'B'C ', and therefore
S \ (\ Delta \) ABC1 = S \ (\ Delta \) A'B'C '
Draw ⊥ and the segment D 1 C 1. Then ⊥, a \ (\ widehat (CD_ (1) C_ (1)) \) = φ is the value of the angle between the plane \ (\ Delta \) ABC and the plane R 1 . That's why
S \ (\ Delta \) ABC1 = 1/2 | AB | | C 1 D 1 | = 1/2 | AB | | CD 1 | cos φ = S \ (\ Delta \) ABC cos φ
and therefore S \ (\ Delta \) A'B'C '= S \ (\ Delta \) ABC cos φ.
Let's move on to considering second case... Let's draw a plane R 1 || R through that vertex \ (\ Delta \) ABC, the distance from which to the plane R the smallest (let it be vertex A).
Let's design \ (\ Delta \) ABC on the plane R 1 and R(fig. 166); let its projections be, respectively, \ (\ Delta \) AB 1 C 1 and \ (\ Delta \) A'B'S '.
Let (ВС) \ (\ cap \) p 1 = D. Then
S \ (\ Delta \) A'B'C '= S \ (\ Delta \) AB1 C1 = S \ (\ Delta \) ADC1 - S \ (\ Delta \) ADB1 = (S \ (\ Delta \) ADC - S \ (\ Delta \) ADB) cos φ = S \ (\ Delta \) 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 draw a cross-section of this prism (Fig. 167). Since the prism is correct, its lateral edges are perpendicular to the plane of the base. Hence, \ (\ Delta \) ABC is the projection of \ (\ Delta \) ADC, therefore
$$ S _ (\ Delta ADC) = \ frac (S _ (\ Delta ABC)) (cos \ phi) = \ frac (a \ cdot a \ sqrt3) (4cos \ phi) $$
or
$$ S _ (\ Delta ADC) = \ frac (6 \ cdot 6 \ cdot \ sqrt3) (4 \ cdot \ frac (\ sqrt3) (2)) = 18 (cm ^ 2) $$
Detailed proof of the theorem on the orthogonal projection of a polygon
If is the projection of the flat n -gon to the plane, then, where is the angle between the planes of the polygons and. In other words, the projected area of a plane polygon is equal to the product of the area of the projected polygon by the cosine of the angle between the plane of the projection and the plane of the projected polygon.
Proof. I stage. Let us carry out the proof first for a 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. Suppose that. Let be the height, then by the theorem of three perpendiculars we can conclude that - the height (- the projection is inclined, - its base and the straight line passes through the base of the inclined, moreover).
Let's consider. It is rectangular. By definition of cosine:
On the other hand, since and, then, by definition, is the linear angle of the 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.
Let's find the ratio of the area to:
Note that the formula remains valid even when. In this case
2 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 a straight line and a plane that. Therefore, is a parallelogram. Consider and. They are equal on three sides (- common, like the opposite sides of a parallelogram). Note that a quadrilateral is a rectangle and is equal (along the leg and hypotenuse), therefore, it is equal on three sides. That's why.
For case 1 is applicable:, i.e.
3 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 straight line with the projection plane. Note that and. On 1 occasion: and. Thus, we get that
4 case. Vertices do not lie in the projection plane ... Consider perpendiculars. Take the smallest of these perpendiculars. Let it be perpendicular. It may turn out that, either only, or only. Then we take it anyway.
Let us set aside a point from a point on a segment, so that a point from a point on a segment, so that. Such a construction is possible, since - is the smallest of the perpendiculars. Note that is a projection and, by construction. Let us prove that and are equal.
Consider a quadrilateral. By hypothesis - perpendiculars to one plane, therefore, by the theorem, therefore. Since by construction, then by the sign of a parallelogram (along parallel and equal opposite sides), we can conclude that is a parallelogram. Means, . It is proved similarly that,. Therefore, they are equal on three sides. That's why. Note that and, as opposite sides of parallelograms, therefore, by the sign of parallelism of the planes,. Since these planes are parallel, they form the same angle with the projection plane.
For the preceding cases apply:
5 case. The projection plane intersects the sides ... Consider straight lines. They are perpendicular to the projection plane, therefore, by the theorem, they are parallel. On co-directed rays with origins at points, respectively, set equal segments, so that the vertices lie outside the projection plane. Note that is a projection and, by construction. Let us show that is equal to.
Since and, by construction, then. Therefore, by the sign of a parallelogram (along two equal and parallel sides), it is a parallelogram. It is proved in a similar way that and are parallelograms. But then, and (as opposing sides), therefore, is equal on 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 case 4 applies:
II stage. Divide a flat polygon into triangles using the diagonals drawn from the vertex: Then, according to the previous cases for triangles:.
Q.E.D.
GEOMETRY
Lesson plans for 10 grades
Lesson 56
Theme. Polygon orthogonal projection area
The purpose of the lesson: studying the theorem on the area of the 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 solutions to problems No. 42, 45 on the blackboard.
2. Frontal polling.
1) Give a definition of 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 to 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 polygon plane and the projection plane.
2. Prove the theorem for the case when there is a lattice triangle with one side parallel to the projection plane.
3. Prove the theorem for the case when there is a lattice triangle with none of its sides parallel to the projection plane.
4. Prove the theorem for any polygon.
Solving problems
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 a 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 an orthogonal projection of a polygon is equal to the area of this polygon?
5. The edge of the cube is equal to a. Find the cross-sectional area of a cube by a plane passing through the apex of the base at an angle of 30 ° to this base and intersecting all the side edges. (Answer. )
6. Problem number 48 (1, 3) from the textbook (p. 58).
7. Problem number 49 (2) from the textbook (p. 58).
8. The sides of the rectangle are 20 and 25 cm. Its projection onto the plane is similar to it. Find the perimeter of the projection. (Answer. 72 cm or 90 cm.)
III. Homework
§4, p. 34; security question number 17; tasks No. 48 (2), 49 (1) (p. 58).
IV. Lesson summary
Question to the class
1) Formulate the theorem on the area of the orthogonal projection of a polygon.
2) Can the area of an orthogonal projection of a polygon be larger than the area of a polygon?
3) Through the hypotenuse AB of the right-angled 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 = 3 cm, BC = 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 point H is the base of the CM height of the triangle ABC;
b) CO = 2.4 cm;
c) triangle AOC is an orthogonal projection of triangle ABC onto the plane α;
d) the area of the AOB triangle is 3 cm2.
(Answer. A) Correct; b) wrong; c) wrong; d) correct.)
Recently, in the 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 in terms of 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 exam - 2014 in mathematics. Finding the cross-sectional area through the area of its orthogonal projection. Task C2 Mathematics teacher MBOU secondary school № 143 in Krasnoyarsk Knyazkina TV
Consider the solution to the following problem: In a rectangular parallelepiped,. The section of the parallelepiped passes through points B and D and forms an angle with the plane ABC. Find the cross-sectional area. It is often convenient to find the cross-sectional area in terms of the area of its orthogonal projection. Finding the area of a triangle through 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 the orthogonal projection of triangle ABC. From right-angled triangle CHC": Area of triangle ABC "is equal to 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 the 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) Build a section. B) Find its orthogonal projection onto the base plane. 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 point O be the point of intersection of the base diagonals. 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 inclined projection, then it is perpendicular to the most inclined one. We are looking for an inclined one in its projection (OC) and the angle between the projection and the inclined one). Find the tangent of the angle COC ₁ between OC ₁ and OC
Consequently, the angle between the section plane and the base plane is greater than between OC ₁ and OC. That is, the section is located something 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 plane of the base. To do this, we find the projections of the points L and M.
Quadrangle BL ₁M₁D - section projection onto the base plane. 4. Find the area of the quadrilateral BL ₁M₁D. To do this, subtract the area of the triangle L CM₁ from the area of the triangle BCD. Find the area of the triangle L CM₁. The triangle L CM₁ is similar to the triangle BCD. Let's find the coefficient of similarity.
To do this, consider the t reangles OPC and OKK₁: Consequently, 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 the quadrilateral BL₁M₁D is equal to 1-4/25 = 21/25 of the area of the triangle BCD and is equal to
5. Now we find 6. And finally, we get: Answer: 112
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