Recently, 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 provides a formula for such a solution and detailed analysis task, 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 " : Area of triangle ABC " is equal to 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. Straight 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 one . 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
A 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
Consider the plane p and the line intersecting it . Let BUT is an arbitrary point in space. Draw a line through this point , parallel to the line . Let . Dot is called point projection BUT to the plane p in parallel design along a given line . Plane p , on which the points of space are projected is called the projection plane.
p - projection plane;
- direct design; ;
; ; ;
Orthogonal design is a special case of parallel design. Orthogonal projection is a parallel projection in which the projection line is perpendicular to the projection plane. Orthogonal projection is widely used in technical drawing, where a figure is projected onto three planes - horizontal and two vertical.
Definition:
Orthographic projection of a point M to the plane p called the base M 1 perpendicular MM 1, lowered from the point M to the plane p.
Designation: , 
, .
Definition: Orthographic projection of the figure F to the plane p is the set of all points of the plane that are orthogonal projections of the set of points of the figure F to the plane p.
Orthogonal design, as a special case of parallel design, has the same properties:
p - projection plane;
- direct design; ;
1) ;
2) , .
- Projections of parallel lines are parallel.
PROJECTION AREA OF A FLAT FIGURE
Theorem: The area of the projection of a flat polygon onto a certain plane is equal to the area of the projected polygon multiplied by the cosine of the angle between the plane of the polygon and the projection plane.
Stage 1: The projected figure is a triangle ABC, the side of which AC lies in the projection plane a (parallel to the projection plane a).
Given:
Prove:
Proof:
1. ; ;
2. ; ; ; ;
3. ; 
;
4. According to the three perpendiculars theorem;
ВD - height; In 1 D - height;
5. - linear angle of the dihedral angle;
6. ; ; ; 
;
Stage 2: The projected figure is a triangle ABC, none of the sides of which lies in the projection plane a and is not parallel to it.
Given:
Prove:
Proof:
1. ; ;
2. ; ;
4. ; ; ;
(Stage 1);
5. ; ; ;
(Stage 1);
Stage: The designed figure is an arbitrary polygon.
Proof:
The polygon is divided by diagonals drawn from one vertex into a finite number of triangles, for each of which the theorem is true. Therefore, the theorem will also be true for the sum of the areas of all triangles whose planes form the same angle with the projection plane.
Comment: The proved theorem is valid for any flat figure, bounded by a closed curve.
Exercises:
1. Find the area of a triangle whose plane is inclined to the projection plane at an angle if its projection is a regular triangle with side a.
2. Find the area of a triangle whose plane is inclined to the projection plane at an angle if its projection is isosceles triangle with side 10 cm and base 12 cm.
3. Find the area of a triangle whose plane is inclined to the projection plane at an angle if its projection is a triangle with sides 9, 10 and 17 cm.
4. Calculate the area of the trapezoid, the plane of which is inclined to the projection plane at an angle if its projection is isosceles trapezium, whose larger base is 44 cm, the side is 17 cm and the diagonal is 39 cm.
5. Calculate the projection area of a regular hexagon with a side of 8 cm, the plane of which is inclined to the projection plane at an angle.
6. Rhombus with a side of 12 cm and acute angle forms an angle with the given plane. Calculate the area of the projection of the rhombus on this plane.
7. A rhombus with a side of 20 cm and a diagonal of 32 cm forms an angle with a given plane. Calculate the area of the projection of the rhombus on this plane.
8. The projection of the canopy on a horizontal plane is a rectangle with sides and . Find the area of the canopy if the side faces are equal rectangles inclined to the horizontal plane at an angle , and the middle part of the canopy is a square parallel to the projection plane.
11. Exercises on the topic "Lines and planes in space":
The sides of the triangle are 20 cm, 65 cm, 75 cm. A perpendicular equal to 60 cm is drawn from the vertex of the larger angle of the triangle to its plane. Find the distance from the ends of the perpendicular to the larger side of the triangle.
2. From a point separated from the plane at a distance of cm, two inclined ones are drawn, forming angles with the plane equal to , and between themselves - a right angle. Find the distance between the intersection points of the inclined plane.
3. Party right triangle is equal to 12 cm. The point M is chosen so that the segments connecting the point M with all the vertices of the triangle form angles with its plane. Find the distance from point M to the vertices and sides of the triangle.
4. A plane is drawn through the side of the square at an angle to the diagonal of the square. Find the angles at which two sides of the square are inclined to the plane.
5. The leg of an isosceles right triangle is inclined to the plane a passing through the hypotenuse at an angle. Prove that the angle between plane a and the plane of the triangle is .
6. The dihedral angle between the planes of triangles ABC and DBC is . Find AD if AB = AC = 5 cm, BC = 6 cm, BD = DC = cm.
Control questions on the topic "Lines and planes in space"
1. List the basic concepts of stereometry. Formulate the axioms of stereometry.
2. Prove the consequences of the axioms.
3. What is mutual arrangement two lines in space? Define intersecting, parallel, intersecting lines.
4. Prove the criterion for intersecting lines.
5. What is the relative position of the line and the plane? Give definitions of intersecting, parallel lines and planes.
6. Prove the sign of parallelism of a straight line and a plane.
7. What is the relative position of the two planes?
8. Define parallel planes. Prove a criterion for the parallelism of two planes. Formulate theorems about parallel planes.
9. Define the angle between lines.
10. Prove the sign of perpendicularity of a line and a plane.
11. Give definitions of the base of the perpendicular, the base of the oblique, the projection of the oblique onto a plane. Formulate the properties of the perpendicular and oblique, lowered to the plane from one point.
12. Define the angle between a straight line and a plane.
13. Prove the theorem on three perpendiculars.
14. Give definitions of a dihedral angle, a linear angle of a dihedral angle.
15. Prove the sign of perpendicularity of two planes.
16. Define the distance between two different points.
17. Define the distance from a point to a line.
18. Define the distance from a point to a plane.
19. Define the distance between a straight line and a plane parallel to it.
20. Define the distance between parallel planes.
21. Define the distance between skew lines.
22. Define the orthogonal projection of a point onto a plane.
23. Define the orthogonal projection of a figure onto a plane.
24. Formulate properties of projections onto a plane.
25. Formulate and prove a theorem on the projection area of a flat polygon.
GEOMETRY
Lesson Plans for Grades 10
Lesson 56
Topic. 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 the plane that intersects perpendicular planes intersects them at the same angle?
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 larger area polygon?
3) A plane α is drawn through the hypotenuse AB of a right triangle ABC at an angle of 45° to the plane of the triangle and a 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 equal to the angle CMO, where the point H is the base of the height 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.)
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 \(\Delta\)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.
Draw through the (AB) plane R 1 || R and project orthogonally \(\Delta\)ABC onto R 1 and on R(Fig. 165); we get \(\Delta\)ABC 1 and \(\Delta\)A'B'C'.
By the projection property, we have \(\Delta\)ABC 1 \(\cong\) \(\Delta\) A'B'C', and therefore
S \(\Delta\)ABC1 = S \(\Delta\)A'B'C'
Let's draw ⊥ and the segment D 1 C 1 . Then ⊥ , a \(\widehat(CD_(1)C_(1))\) = φ is the angle between the plane \(\Delta\) ABC and the plane R one . 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 hence S \(\Delta\)A'B'C' = S \(\Delta\)ABC cos φ.
Let's move on to consideration second case. 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 \(\Delta\)AB 1 C 1 and \(\Delta\)A'B'C', respectively.
Let (BC) \(\cap \) 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
A 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. Hence, \(\Delta\)ABC is the projection of \(\Delta\)ADC, so
$$ S_(\Delta ADC) = \frac(S_(\Delta ABC))(cos\phi) = \frac(a\cdot a\sqrt3)(4cos\phi) $$
or
$$ S_(\Delta ADC) = \frac(6\cdot 6\cdot \sqrt3)(4\cdot\frac(\sqrt3)(2)) = 18 (cm^2) $$