Perpendicular and oblique. Complete lessons - Knowledge Hypermarket. Perpendicular and oblique to a straight line

Lesson topic

Perpendicular and oblique.

Lesson Objectives

Get acquainted with new definitions and remember some already studied.
Learn to apply the properties of shapes when solving problems.
Understand some seemingly simple concepts and definitions.
Developmental – to develop students’ attention, perseverance, perseverance, logical thinking, mathematical speech.
Educational - through the lesson, cultivate an attentive attitude towards each other, instill the ability to listen to comrades, mutual assistance, and independence.

Lesson Objectives

Test students' problem-solving skills.
Learn to correctly perceive information.
Review the basics of perpendicular and oblique.

Lesson Plan

Introduction.
Repetition of previously studied material.
Perpendicular and oblique.
Examples of problem solving.

introduction

It's no secret that all elementary geometry came to us mainly from Egypt and Greece. In distant and ancient times, geometry was used as a science to measure the earth and also very closely in construction. All theorems, laws and axioms were derived and proven to facilitate measuring or construction work. Today's topic was very important for the people of that time since perpendicular and oblique are the main guidelines for this type of work.

There are many hypotheses regarding the construction technique of the Egyptian pyramids. It is obvious that this technique has changed over time, i.e. later pyramids were built differently than earlier ones. Most of hypotheses are based on the fact that the blocks were cut in quarries using punches, chisels, chisels, adzes, etc., the main material in the manufacture of which was copper. Accordingly, the extracted material had to be somehow transported to the construction site and installed. The discrepancies between the various hypotheses relate mainly to the methods of delivery and installation of blocks, as well as estimates of construction time and labor requirements.

Construction technique of the Great Pyramids according to Herodotus

Our the only written source, which describes the process of building the pyramids, serves as book II of the “History” of Herodotus, who visited Egypt ca. 450 BC uh. Without speaking the language of the Egyptians, Herodotus had to take notes from the words of the Greek settlers who lived in the country, as well as - through translators - from the words of representatives of the Egyptian priesthood. It was definitely difficult for him to find out how the Great Pyramids were built two thousand years ago before him, since it was unlikely to be known to the Egyptians themselves.

Some were obliged to drag huge blocks of stones from quarries in the Arabian Mountains to the Nile (the stones were transported across the river on ships), while others were ordered to drag them further to the so-called Libyan Mountains. One hundred thousand people performed this work continuously, changing every three months. It took ten years for the exhausted people to build the road along which these stone blocks were dragged - work, in my opinion, was almost as enormous as the construction of the pyramid itself. The construction of the pyramid itself lasted twenty years.

Other theories of block making and installation

There is also a theory that the very blocks that make up the pyramid were made using formwork. On the previous tier, rectangular formwork was installed, into which a mortar-like composition was then poured. The frozen block itself served as formwork for the next blocks of the growing tier. The components of the solution could be delivered relatively easily by numerous slaves without the use of complex equipment.

This theory explains well the ideal fit of the walls of individual blocks.

Alternative hypotheses

A number of authors put forward hypotheses that the pyramids were built by other developed civilizations, either terrestrial, which later disappeared, or alien. Also, one of the societies of amateur Egyptologists put forward a theory according to which huge stone blocks were moved with the help of kites. Egyptologists do not consider such hypotheses seriously.

Perpendicular and oblique

And so let's start with the simplest and let's repeat what perpendicular and oblique are.

Definition. Two lines are called perpendicular if they intersect at right angles.

Answer: 13.

Machines and Mechanisms.

Machines and Mechanisms, mechanical devices that facilitate work and increase its productivity. Machines can be of varying degrees of complexity - from a simple one-wheeled wheelbarrow to elevators, cars, printing, textile, and computing machines. Energy machines convert one type of energy into another. For example, hydroelectric generators convert the mechanical energy of falling water into electrical energy. The internal combustion engine converts the chemical energy of gasoline into thermal energy and then into the mechanical energy of vehicle movement.

A gear is a mechanism or part of a mechanism that includes gears.

Purpose:

broadcast rotational movement between shafts, which may have parallel, intersecting or crossing axes.
converting rotational motion into translational motion and vice versa.

In this case, the force is transmitted from one element to another using teeth. The gear wheel of the transmission with a smaller number of teeth is called a pinion, the second gear with a larger number of teeth is called a wheel. A pair of gears having the same number of teeth, in this case the driving gear is called a gear, and the driven gear is called a wheel.

Archimedes screw, Archimedes screw- a mechanism historically used to transfer water from low-lying bodies of water to irrigation canals. It was one of several inventions and discoveries traditionally attributed to Archimedes, who lived in the 3rd century BC. e. The Archimedes screw became the prototype of the auger.

The propeller is usually rotated by a wind wheel or manually. As the lower end of the pipe turns, it collects a certain amount of water. This amount of water will slide up the spiral pipe as the shaft rotates until finally the water flows out of the top of the pipe, supplying the irrigation system.

Questions

What is perpendicular?
Which line is called slanted?
Are the diagonals of a square divided in half by the point of intersection?
Are the diagonals of a square equal?
Where is an inclined plane used in practice?
What shape is called a rectangle?

List of sources used

"The Pyramid Builders" Notes by Dr. Z. Hawass
Perepelkin Yu. Ya. History of Ancient Egypt. - St. Petersburg: “Summer Garden”, 2000.
Kobycheva Marina Viktorovna, mathematics teacher
Mazur K. I. “Solving the main competition problems in mathematics of the collection edited by M. I. Skanavi”

We worked on the lesson

Poturnak S.A.

Kobycheva Marina Viktorovna

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GEOMETRY

Section II. STEREOMETRY

§8. PERPENDICULAR AND OBLIQUE. OBLIQUE PROJECTION ON A PLANE.

2. Properties of perpendicular and oblique.

Let's consider the properties of perpendicular and oblique.

1) The perpendicular drawn from a given point to the plane is less than any inclined one drawn from the same point to the plane.

In Figure 411: AN AK.

2) If two inclined slopes drawn from a given point to a plane are equal, then their projections are equal.

K 1 and perpendicular AN and AK = AK 1. Then by property: NK = NK 1.

3) If two inclined slopes drawn from a given point to a given plane have equal projections, then they are equal to each other.

In Figure 412, two inclined planes AK and A are drawn from point A to plane a K 1 and perpendicular to AN, with KH = K 1 N. Then by property: AK = AK 1 .

4) If two inclined planes are drawn from a given point to a plane, then the larger inclined one has a larger projection.

L and perpendicular to AN, A K > AL . Then by property: H K > HL.

5) If two inclined slopes are drawn from a given point to a plane, then the larger of them is the one that has a larger projection onto the given plane.

In Figure 413, two inclined planes AK and A are drawn from point A to plane a L and perpendicular to AN, NK> N L . Then by property: AK> A L .

Example 1. Two inclined slopes are drawn from a point to a plane, the lengths of which are 41 cm and 50 cm. Find the projections of the inclined ones, if they have a ratio of 3:10, and the distance from the point to the plane.

Solutions. 1) A L = 41 cm; AK = 50 cm (Fig. 413). By property we have H L NK. Let us denote H L = 3 x cm, NK = 10 x cm, AN = h see AN - distance from point A to the planeα .

4) Equating, we get 41 2 - 9x 2 = 50 2 - 100 x 2; x 2 = 9; x = 3 (considering x> 0). So, N L = 3 ∙ 3 = 9 (cm), NK = 10 ∙ 3 = 30 (cm).

Example 2. From this point to two inclined planes are drawn, eachin cm. The angle between the inclined ones is 60°, and the angle between their projections is straight. Find the distance from the point to the plane.

Theorem . If a perpendicular and inclined lines are drawn from one point outside the plane, then:

1) oblique ones having equal projections are equal;

2) of the two inclined ones, the one whose projection is larger is larger;

3) equal obliques have equal projections;

4) of the two projections, the one that corresponds to the larger oblique one is larger.

Three Perpendicular Theorem . In order for a straight line lying in a plane to be perpendicular to an inclined one, it is necessary and sufficient that this straight line be perpendicular to the projection of the inclined one (Fig. 12.3).

Theorem on the area of the orthogonal projection of a polygon onto a plane. The area of the orthogonal projection of a polygon onto a plane is equal to the product of the area of the polygon and the cosine of the angle between the plane of the polygon and the projection plane.

Example 1. Through a given point draw a straight line parallel to the given plane.

Solution. Analysis. Let's assume that the straight line is constructed (Fig. 12.4). A line is parallel to a plane if it is parallel to some line lying in the plane (based on the parallelism of the line and the plane). Two parallel lines lie in the same plane. This means that by constructing a plane passing through a given point and an arbitrary line in a given plane, it will be possible to construct a parallel line.

Construction.

1. On a plane  we conduct a direct A.

3. In plane  through the point A let's make a direct b, parallel to the line A.

4. A straight line has been built b, parallel to the plane  .

Proof. Based on the parallelism of a straight line and a plane, a straight line b parallel to the plane  , since it is parallel to the line A, belonging to the plane  .

Study. The problem has an infinite number of solutions, since the straight line A in the plane  is chosen randomly.

Example 2. Determine at what distance from the plane the point is located A, if straight AB intersects the plane at an angle of 45º, the distance from the point A to the point IN, belonging to the plane, is equal to
cm.

Solution. Let's make a drawing (Fig. 12.5):

AC– perpendicular to the plane  , AB– inclined, angle ABC– angle between straight line AB and plane  . Triangle ABC– rectangular,
because AC– perpendicular. The required distance from the point A to the plane - this is the leg AC right triangle. Knowing the angle
and hypotenuse
let's find the leg AC:

In response we get : AC = 3 cm.

Example 3. Determine at what distance from the plane of an isosceles triangle there is a point 13 cm distant from each of the vertices of the triangle if the base and height of the triangle are equal to 8 cm.

Solution. Let's make a drawing (Fig. 12.6). Dot S away from the points A, IN And WITH at the same distance. So, inclined S.A., S.B. And S.C. equal, SO– the common perpendicular of these inclined ones. By the theorem of obliques and projections AO = VO = CO.

Dot ABOUT– the center of a circle circumscribed about a triangle ABC. Let's find its radius:

Where Sun– base; AD– the height of a given isosceles triangle.

Finding the sides of a triangle ABC from a right triangle ABD according to the Pythagorean theorem:

Now we find OB:

Consider a triangle SOB:
S.B.= 13 cm, OB= 5 cm. Find the length of the perpendicular SO according to the Pythagorean theorem:

In response we get: SO= 12 cm.

Example 4. Given parallel planes  And  . Through the point M, which does not belong to any of them, straight lines are drawn A And b, which intersect the plane  at points A 1 and IN 1 and the plane  – at points A 2 and IN 2. Find A 1 IN 1 if it is known that MA 1 = 8 cm, A 1 A 2 = 12 cm, A 2 IN 2 = 25 cm.

Solution. Since the condition does not say how the point is located relative to both planes M, then two options are possible: (Fig. 12.7, a, b). Let's look at each of them. Two intersecting lines A And b define a plane. This plane intersects two parallel planes  And  along parallel lines A 1 IN 1 and A 2 IN 2 according to Theorem 5 about parallel lines and parallel planes.

Triangles MA 1 IN 1 and MA 2 IN 2 are similar (angles A 2 MV 2 and A 1 MV 1 – vertical, corners MA 1 IN 1 and MA 2 IN 2 – internal crosswise lying with parallel lines A 1 IN 1 and A 2 IN 2 and secant A 1 A 2). From the similarity of triangles follows the proportionality of the sides:

From here

Option a):

Option b):

We get the answer: 10 cm and 50 cm.

Example 5. Through the point A plane  a direct line was drawn AB, forming an angle with the plane  . Via direct AB a plane is drawn  , forming with the plane  corner  . Find the angle between the projection of a straight line AB to the plane  and plane  .

Solution. Let's make a drawing (Fig. 12.8). From point IN drop the perpendicular to the plane  .
Linear angle dihedral angle between planes  And  - this is the angle
Straight ADDBC, based on the perpendicularity of a straight line and a plane, since
And
Based on the perpendicularity of planes, the plane  perpendicular to the plane of the triangle DBC, since it passes through the line AD. We construct the desired angle by dropping a perpendicular from the point WITH to the plane  , let's denote it
Find the sine of this angle of a right triangle MYSELF. Let us introduce an auxiliary segment BC = a. From the triangle ABC:
From the triangle Navy (

) we'll find it.

A perpendicular dropped from a given point to a given plane is a segment connecting a given point with a point on the plane and lying on a straight line perpendicular to the plane. The end of this segment lying in the plane is called the base of the perpendicular. The distance from a point to a plane is the length of the perpendicular drawn from this point to the plane.

The slope drawn from a given point to a given plane is any segment that connects a given point to a point on the plane and is not perpendicular to this plane. The end of a segment lying in a plane is called the inclined base. A segment connecting the bases of a perpendicular and an oblique drawn from the same point is called an oblique projection.

In Figure 136, from point A, a perpendicular AB and an inclined AC are drawn to the plane. Point B is the base of the perpendicular, point C is the base of the inclined one, BC is the projection of the inclined AC onto plane a.

Since the distances from the points of a line to a plane parallel to it are the same, the distance from a line to a plane parallel to it is the distance from any point of it to this plane.

A straight line drawn on a plane through the base of an inclined plane perpendicular to its projection is also perpendicular to the inclined itself. And vice versa: if a straight line in a plane is perpendicular to an inclined one, then it is also perpendicular to the projection of the inclined one (the theorem of three perpendiculars).

In Figure 137, a perpendicular AB and an inclined AC are drawn to plane a. The straight line o, lying in the plane a, is perpendicular to BC - the projection of the inclined AC onto the plane a. According to T. 2.12, straight line a is perpendicular to inclined AC. If it were known that straight line a is perpendicular to inclined AC, then according to T. 2.12 it would be perpendicular to its projection - BC.

Example. The legs of the right triangle ABC are equal to 16 and from the vertex of the right angle C a perpendicular CD = 35 m is drawn to the plane of this triangle (Fig. 138). Find the distance from point D to the hypotenuse AB.

Solution. Let's do it. According to the condition, DC is perpendicular to the plane, i.e. DE is inclined, CE is its projection, therefore, by the theorem about three perpendiculars, it follows from the condition that

From we find To find the height CE in we find

On the other hand, where

From the Pythagorean theorem

46. Perpendicularity of planes.

Two intersecting planes are called perpendicular if any plane perpendicular to the line of intersection of these planes intersects them along perpendicular lines.

Figure 139 shows two planes that intersect along a straight line a. The plane y is perpendicular to the line a and intersects. In this case, the plane y intersects the plane a along the straight line c, and the plane intersects along the straight line d, and i.e., by definition

T. 2.13. If a plane passes through a line perpendicular to another plane, then these planes are perpendicular (a sign of perpendicularity of planes).

In Figure 140, the plane passes through a straight line, i.e., the plane is perpendicular.

If through some point taken outside a line we draw a line perpendicular to it, then for brevity the segment from this point to the line is called in one word perpendicular.

Segment CO is perpendicular to line AB. Point O is called base of the perpendicular CO (rice).

If a line drawn through a given point intersects another line, but is not perpendicular to it, then its segment from a given point to the point of intersection with another line is called inclined to this line.

Segment BC - inclined to straight line AO. Point C is called basis inclined (Fig.).

If we drop perpendiculars from the ends of some segment onto an arbitrary line, then the line segment enclosed between the bases of the perpendiculars is called projection of the segment to this straight line.

Segment АВ - projection of segment AB onto EC. The segment OM is also called the projection of the segment OM onto the EC.

The projection of the segment KP perpendicular to EC will be the point K (Fig.).

2. Properties of perpendicular and oblique.

Theorem 1. A perpendicular drawn from a point to a straight line is less than any oblique drawn from the same point to this straight line.

The segment AC (Fig.) is perpendicular to straight line OB, and AM is one of the inclined lines drawn from point A to straight line OB. It is required to prove that AM > AC.

In ΔMAC, the segment AM is the hypotenuse, and the hypotenuse is larger than each of the legs of this triangle. Therefore, AM > AC. Since we took the inclined AM arbitrarily, we can assert that any inclined line to a straight line is greater than a perpendicular to this line (and a perpendicular is shorter than any inclined line) if they are drawn to it from the same point.

The converse statement is also true, namely: if the segment AC (Fig.) is less than any other segment connecting the point AC with any point on the straight line OB, then it is perpendicular to OB. In fact, segment AC cannot be inclined to OB, since then it would not be the shortest of the segments connecting point A with points of straight line OB. This means that it can only be perpendicular to OB.

The length of the perpendicular dropped from a given point to a straight line is taken as the distance from a given point to this straight line.

Theorem 2. If two oblique lines drawn to a line from the same point are equal, then their projections are equal.

Let BA and BC be inclined lines drawn from point B to straight line AC (Fig.), and AB = BC. It is necessary to prove that their projections are also equal.

To prove this, let us lower the perpendicular BO from point B to AC. Then AO and OS will be projections of inclined AB and BC onto straight line AC. Triangle ABC is isosceles according to the theorem. VO is the height of this triangle. But the altitude in an isosceles triangle drawn to the base is at the same time the median of this triangle.

Therefore AO = OS.

Theorem 3 (converse). If two oblique lines drawn to a straight line from the same point have equal projections, then they are equal to each other.

Let AC and CB be inclined to straight line AB (Fig.). CO ⊥ AB and AO = OB.

It is required to prove that AC = BC.

In right triangles AOC and BOC, the legs AO and OB are equal. CO is the common leg of these triangles. Therefore, ΔAOC = ΔBOC. From the equality of triangles it follows that AC = BC.

Theorem 4. If two inclined lines are drawn from the same point to a straight line, then the one that has a larger projection onto this straight line is larger.

Let AB and BC be inclined to straight line AO; VO ⊥ AO and AO>CO. It is required to prove that AB > BC.

1) Inclined ones are located on one side of the perpendicular.

Angle ACE is external with respect to the right triangle COB (Fig.), and therefore ∠ACV > ∠COV, i.e. it is obtuse. It follows that AB > CB.

2) Inclined ones are located on both sides of the perpendicular. To prove this, let’s plot the segment OK = OS on AO from point O and connect point K to point B (Fig.). Then, by Theorem 3, we have: VC = BC, but AB > VC, therefore, AB > BC, i.e. the theorem is valid in this case as well.

Theorem 5 (converse). If two inclined lines are drawn from the same point to a straight line, then the larger inclined line also has a larger projection onto this straight line.

Let KS and BC be inclined to the straight line CV (Fig.), CO ⊥ CV and KS > BC. It is required to prove that KO > OB.

Between the segments KO and OB there can be only one of three relationships:

1) KO< ОВ,

2) KO = OV,

3) KO > OV.

KO cannot be less than OB, since then, according to Theorem 4, the inclined KS would be less than the inclined BC, and this contradicts the conditions of the theorem.

In the same way, KO cannot equal OB, since in this case, according to Theorem 3, KS = BC, which also contradicts the conditions of the theorem.

Consequently, only the last relation remains true, namely, that KO > OB.