Often it is necessary to draw ("build") an angle that would be equal to this corner, and the construction must be performed without the help of a protractor, but using only a compass and a ruler. Knowing how to build a triangle on three sides, we can solve this problem. Let on a straight line MN(dev. 60 and 61) is required to be built at the point K injection, equal to the angle B. This means that it is necessary from the point K draw a straight line constituting MN angle equal to B.
To do this, mark a point on each side of a given angle, for example A and WITH, and connect A and WITH straight line. Get a triangle ABC. Let's build now on a straight line MN this triangle so that its apex V was at the point TO: then this point will have an angle equal to the angle V. Build a triangle on three sides Sun, VA and AC we can: postpone (dev. 62) from the point TO section kl, equal sun; get a point L; around K, as near the center, we describe a circle with a radius VA, and around L- radius SA. Point R connect the intersections of the circles with TO and Z, - we get a triangle KPL, triangular ABC; it has a corner TO= ang. V.
This construction is faster and more convenient if from the top V set aside equal segments (with one dissolution of the compass) and, without moving its legs, describe with the same radius a circle around the point TO, like near the center.
How to cut a corner in half
Let it be required to divide the angle A(Fig. 63) into two equal parts using a compass and ruler, without using a protractor. We'll show you how to do it.
From the top A draw equal segments on the sides of the angle AB and AC(Fig. 64; this is done with one dissolution of the compass). Then we put the tip of the compass at the points V and WITH and describe with equal radii the arcs intersecting at the point D. straight line connecting A and D divides the angle A in half.
Let's explain why. If the point D connect with V and C (Fig. 65), then you get two triangles ADC and adb, u which have a common side AD; side AB equal to side AC, a BD is equal to CD. Triangles are equal on three sides, so the angles are equal. bad and DAC, lying opposite equal sides BD and CD. Therefore, a straight line AD divides the angle YOU in half.
Applications
12. Construct an angle of 45° without a protractor. At 22°30'. At 67°30'.
Solution. Dividing the right angle in half, we get an angle of 45 °. Dividing the angle of 45° in half, we get an angle of 22°30'. By constructing the sum of the angles 45° + 22°30', we get an angle of 67°30'.
How to draw a triangle given two sides and an angle between them
Let it be required on the ground to find out the distance between two milestones A and V(device 66), separated by an impenetrable swamp.
How to do it?
We can do this: aside from the swamp, we choose such a point WITH, from where both milestones are visible and it is possible to measure distances AC and Sun. Injection WITH we measure with the help of a special goniometric device (called an astrolabe). According to these data, i.e., according to the measured sides AC and sun and corner WITH between them, build a triangle ABC somewhere in a convenient location as follows. Having measured one known side in a straight line (Fig. 67), for example AC, build with it at the point WITH injection WITH; on the other side of this angle, a known side is measured Sun. Ends of known sides, i.e. points A and V connected by a straight line. It turns out a triangle in which two sides and the angle between them have pre-specified dimensions.
It is clear from the method of construction that only one triangle can be constructed given two sides and the angle between them. therefore, if two sides of one triangle are equal to two sides of another and the angles between these sides are the same, then such triangles can be superimposed on each other by all points, that is, they must also have third sides and other angles equal. This means that the equality of the two sides of the triangles and the angle between them can serve as a sign of the complete equality of these triangles. Shortly speaking:
Triangles are equal under two sides and angles between them.
In construction tasks, we will consider the construction of a geometric figure, which can be performed using a ruler and a compass.
With a ruler, you can:
arbitrary line;
an arbitrary line passing through a given point;
a straight line passing through two given points.
Using a compass, you can describe a circle of a given radius from a given center.
A compass can be used to draw a segment on a given line from a given point.
Consider the main tasks for the construction.
Task 1. Construct a triangle with given sides a, b, c (Fig. 1).
Solution. With the help of a ruler, draw an arbitrary straight line and take an arbitrary point B on it. With a compass opening equal to a, we describe a circle with center B and radius a. Let C be the point of its intersection with the line. With a compass opening equal to c, we describe a circle from the center B, and with a compass opening equal to b - a circle from the center C. Let A be the intersection point of these circles. Triangle ABC has sides equal to a, b, c.
Comment. In order for three line segments to serve as sides of a triangle, it is necessary that the larger of them be less than the sum of the other two (and< b + с).
Task 2.
Solution. This angle with vertex A and beam OM are shown in Figure 2.
Draw an arbitrary circle centered at the vertex A of the given angle. Let B and C be the points of intersection of the circle with the sides of the angle (Fig. 3, a). Let's draw a circle with radius AB with the center at the point O - the starting point of this ray (Fig. 3, b). The point of intersection of this circle with the given ray will be denoted as С 1 . Let us describe a circle with center C 1 and radius BC. Point B 1 of the intersection of two circles lies on the side of the desired angle. This follows from the equality Δ ABC \u003d Δ OB 1 C 1 (the third criterion for the equality of triangles).
Task 3. Construct the bisector of the given angle (Fig. 4).
Solution. From vertex A of a given angle, as from the center, we draw a circle of arbitrary radius. Let B and C be the points of its intersection with the sides of the angle. From points B and C with the same radius we describe circles. Let D be their intersection point, different from A. Ray AD divides angle A in half. This follows from the equality ΔABD = ΔACD (the third criterion for the equality of triangles).
Task 4. Draw a median perpendicular to this segment (Fig. 5).
Solution. With an arbitrary but identical compass opening (large 1/2 AB), we describe two arcs with centers at points A and B, which will intersect each other at some points C and D. The straight line CD will be the required perpendicular. Indeed, as can be seen from the construction, each of the points C and D is equally distant from A and B; therefore, these points must lie on the perpendicular bisector to segment AB.
Task 5. Divide this section in half. It is solved in the same way as problem 4 (see Fig. 5).
Task 6. Through a given point, draw a line perpendicular to the given line.
Solution. Two cases are possible:
1) given point O lies on the given line a (Fig. 6).
From point O we draw a circle with an arbitrary radius that intersects the line a at points A and B. From points A and B we draw circles with the same radius. Let О 1 be their intersection point different from О. We get ОО 1 ⊥ AB. Indeed, the points O and O 1 are equidistant from the ends of the segment AB and, therefore, lie on the perpendicular bisector to this segment.
The purpose of the lesson: Formation of the ability to build an angle equal to a given one. Task: Create conditions for mastering the construction algorithm using a compass and a ruler of an angle equal to a given one; create conditions for mastering the sequence of actions when solving a construction problem (analysis, construction, proof); improve the skill of using the properties of a circle, signs of equality of triangles to solve the problem of proof; provide the opportunity to apply new skills in solving problems
In geometry, construction tasks are distinguished that can be solved only with the help of two tools: a compass and a ruler without scale divisions. The ruler allows you to draw an arbitrary straight line, as well as build a straight line passing through two given points; using a compass, you can draw a circle of arbitrary radius, as well as a circle with a center at a given point and a radius equal to a given segment. I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I
Given: angle A. A Constructed: angle O. B C O D E Prove: A = O Proof: consider triangles ABC and ODE. 1.AC=OE, as radii of one circle. 2.AB=OD, as the radii of one circle. 3.BC=DE, as radii of one circle. ABC \u003d ODE (3 prizes) A \u003d O Task 2. Set aside an angle equal to this one from a given beam
Let us prove that the ray AB is the bisector of A 3. Proof: Additional construction (let's connect the point B with the points D and C). Consider ASV and ADB: A B C D 1.AC=AD as radii of one circle. 2.CB=DB, as radii of one circle. 3. AB - common side. ASV \u003d ADB, according to the III sign of equality of triangles Beam AB is a bisector 4. Research: The problem always has a unique solution.
Scheme for solving construction problems: Analysis (drawing the desired figure, establishing links between the given and desired elements, construction plan). Building according to plan. Proof that the figure satisfies the conditions of the problem. Research (when and how many solutions does the problem have?).
mathematics geometry skill lesson
Lesson summary “Constructing an angle equal to a given one. Construction of an angle bisector»
educational: to acquaint students with construction tasks, in the solution of which only compasses and a ruler are used; teach how to build an angle equal to a given one, build an angle bisector;
developing: development of spatial thinking, attention;
educational: education of diligence and accuracy.
Equipment: tables with the order of solving construction tasks; compass and ruler.
During the classes:
1. Actualization of the main theoretical concepts (5 min).
First, you can conduct a frontal survey on the following questions:
- 1. What figure is called a triangle?
- 2. What triangles are called equal?
- 3. Formulate signs of equality of triangles.
- 4. Which segment is called the bisector of a triangle? How many bisectors does a triangle have?
- 5. Define a circle. What is the center, radius, chord and diameter of a circle?
To repeat the signs of equality of triangles, you can suggest.
Exercise: indicate on which of the figures (Fig. 1) there are equal triangles.
Rice. 1
The repetition of the concept of a circle and its elements can be organized by offering the class the following exercise, with its execution by one student on the board: given a line a and a point A lying on the line and a point B not lying on the line. Draw a circle centered at point A passing through point B. Mark the intersection points of the circle with line a. Name the radii of the circle.
2. Learning new material ( practical work) (20 minutes)
Constructing an angle equal to a given one
To consider new material, it is useful for the teacher to have a table (table No. 1 of Appendix 4). The work with the table can be organized in different ways: it can illustrate the teacher's story or a sample solution record; you can invite students, using the table, to tell about the solution of the problem, and then independently complete it in notebooks. The table can be used when interviewing students and when repeating the material.
Task. Set aside from the given ray an angle equal to the given one.
Solution. This angle with vertex A and beam OM are shown in Figure 2.
Rice. 2
It is required to construct an angle equal to angle A, so that one of the sides coincides with the ray OM. Draw a circle of arbitrary radius centered at the vertex A of the given angle. This circle intersects the sides of the corner at points B and C (Fig. 3, a). Then we draw a circle of the same radius centered at the beginning of this ray OM. It intersects the beam at point D (Fig. 3, b). After that, we construct a circle with center D, the radius of which is equal to BC. Circles with centers O and D intersect at two points. Let us denote one of these points by the letter E. Let us prove that the angle MOE is the required one.
Consider triangles ABC and ODE. The segments AB and AC are the radii of the circle with the center A, and OD and OE are the radii of the circle with the center O. Since by construction these circles have equal radii, then AB=OD, AC=OE. Also, according to the construction, BC \u003d DE. Therefore, ABC = ODE on three sides. Therefore, DOE = YOU, i.e. the constructed angle MOE is equal to the given angle A.
Rice. 3
Constructing a bisector of a given angle
Task. Construct the bisector of the given angle.
Solution. Draw a circle of arbitrary radius centered at the vertex A of the given angle. It will intersect the sides of the corner at points B and C. Then we draw two circles of the same radius BC with centers at points B and C (only parts of these circles are shown in Figure 4). They intersect at two points. The one of these points that lies inside the angle BAC will be denoted by the letter E. Let us prove that the ray AE is the bisector of this angle.
Consider triangles ACE and ABE. They are equal on three sides. Indeed, AE is the common side; AC and AB are equal, as are the radii of the same circle; CE=BE by construction. From the equality of triangles ACE and ABE it follows that CAE \u003d BAE, i.e. the ray AE is the bisector of the given angle.
Rice. 4
The teacher can invite students to use this table (table No. 2 of Appendix 4) to build the bisector of the angle.
The student at the blackboard performs the construction, justifying each step of the actions performed.
The proof is shown by the teacher, it is necessary to dwell in detail on the proof of the fact that as a result of the construction, equal angles will indeed be obtained.
3. Fixing (10 min)
It is useful to offer students the following task to consolidate the material covered:
Task. Obtuse angle AOB is given. Construct the ray OX so that the angles XOA and XOB are equal obtuse angles.
Task. Use a compass and straightedge to construct angles of 30º and 60º.
Task. Construct a triangle given a side, an angle adjacent to its side, and a bisector of the triangle emanating from the vertex of the given angle.
- 4. Summing up (3 min)
- 1. During the lesson, we solved two building problems. Studied:
- a) build an angle equal to the given one;
- b) construct the bisector of the angle.
- 2. In the course of solving these problems:
- a) remembered the signs of equality of triangles;
- b) used the construction of circles, segments, rays.
- 5. To the house (2 min): No. 150-152 (see Appendix 1).
Lesson Objectives:
- Formation of skills to analyze the studied material and skills to apply it to solve problems;
- Show the significance of the concepts being studied;
- Development of cognitive activity and independence in obtaining knowledge;
- Raising interest in the subject, a sense of beauty.
Lesson objectives:
- To form skills in constructing an angle equal to a given one using a scale ruler, compass, protractor and drawing triangle.
- Check students' ability to solve problems.
Lesson plan:
- Repetition.
- Constructing an angle equal to a given one.
- Analysis.
- Construction of the first example.
- Construction of the second example.
Repetition.
Injection.
flat corner- unlimited geometric figure, formed by two rays (sides of the corner) coming out of one point (the vertex of the corner).
An angle is also called a figure formed by all points of the plane enclosed between these rays (Generally speaking, two such rays correspond to two angles, since they divide the plane into two parts. One of these angles is conditionally called internal, and the other external.
Sometimes, for brevity, an angle is called an angular measure.
To designate an angle, there is a generally accepted symbol: , proposed in 1634 by the French mathematician Pierre Erigon.
Injection- this is a geometric figure (Fig. 1), formed by two rays OA and OB (corner sides), emanating from one point O (corner apex).
An angle is denoted by a symbol and three letters indicating the ends of the rays and the vertex of the angle: AOB (moreover, the letter of the vertex is the middle one). The angles are measured by the amount of rotation of the ray OA around the vertex O until the ray OA passes into position OB. There are two commonly used units for measuring angles: radians and degrees. For radian measurement of angles, see below under "Arc length" and also in the chapter "Trigonometry".
Degree system for measuring angles.
Here, the unit of measure is the degree (its designation is °) - this is the rotation of the beam by 1/360 of a full turn. Thus, a full rotation of the beam is 360 o. One degree is divided into 60 minutes (notation ‘); one minute - respectively for 60 seconds (designation “). An angle of 90 ° (Fig. 2) is called right; an angle less than 90° (Fig. 3) is called acute; an angle greater than 90 ° (Fig. 4) is called obtuse.
Straight lines forming a right angle are called mutually perpendicular. If the lines AB and MK are perpendicular, then this is denoted: AB MK.
Constructing an angle equal to a given one.
Before starting construction or solving any problem, regardless of the subject, it is necessary to carry out analysis. Understand what the task is about, read it thoughtfully and slowly. If after the first time there are doubts or something was not clear or clear but not completely, it is recommended to read it again. If you are doing an assignment in class, you can ask the teacher. Otherwise, your task, which you misunderstood, may not be solved correctly, or you may find something that is not what was required of you and it will be considered incorrect and you will have to redo it. As for me - it is better to spend a little more time studying the task than to redo the task again.
Analysis.
Let a be a given ray with vertex A, and let (ab) be the desired angle. We choose points B and C on the rays a and b, respectively. Connecting points B and C, we get triangle ABC. V equal triangles the corresponding angles are equal, and hence the method of construction follows. If points C and B are chosen in some convenient way on the sides of a given angle, a triangle AB 1 C 1 equal to ABC is constructed from a given ray to a given half-plane (and this can be done if all sides of the triangle are known), then the problem will be solved.
When carrying out any constructions Be extremely careful and try to carry out all the constructions carefully. Since any inconsistencies can result in some kind of errors, deviations, which can lead to an incorrect answer. And if a task of this type is performed for the first time, then the error will be very difficult to find and fix.
Construction of the first example.
Draw a circle centered at the vertex of the given angle. Let B and C be the points of intersection of the circle with the sides of the angle. Draw a circle with radius AB centered at point A 1 - the starting point of this ray. The point of intersection of this circle with the given ray will be denoted by B 1 . Let's describe a circle with center B 1 and radius BC. The intersection point C 1 of the constructed circles in the specified half-plane lies on the side of the required angle.
Triangles ABC and A 1 B 1 C 1 are equal on three sides. Angles A and A 1 are the corresponding angles of these triangles. Therefore, ∠CAB = ∠C 1 A 1 B 1
For greater clarity, we can consider the same constructions in more detail.
Construction of the second example.
The task also remains to postpone from the given half-line to the given half-plane an angle equal to the given angle.
Construction.
Step 1. Let's draw a circle with an arbitrary radius and centers at the vertex A of the given angle. Let B and C be the intersection points of the circle with the sides of the angle. And draw the segment BC.
Step 2 Draw a circle with radius AB centered at point O, the starting point of this half-line. Denote the point of intersection of the circle with the ray B 1 .
Step 3 Now let's describe a circle with center B 1 and radius BC. Let the point C 1 be the intersection of the constructed circles in the specified half-plane. 
Step 4 Let's draw a ray from point O through point C 1 . Angle C 1 OB 1 will be the desired one.
Proof.
Triangles ABC and OB 1 C 1 are congruent as triangles with corresponding sides. And therefore the angles CAB and C 1 OB 1 are equal.
Interesting fact:
In numbers.
In the objects of the world around you, first of all, you notice their individual properties that distinguish one object from another.
The abundance of particular, individual properties overshadows the general properties inherent in absolutely all objects, and therefore it is always more difficult to discover such properties.
One of the most important common properties of objects is that all objects can be counted and measured. We reflect it common property objects in the concept of number.
People mastered the process of counting, that is, the concept of number, very slowly, for centuries, in a stubborn struggle for their existence.
In order to count, it is necessary to have not only objects to be counted, but already to have the ability to be distracted when considering these objects from all their other properties, except for number, and this ability is the result of a long historical development based on experience.
Every person now learns to count with the help of numbers imperceptibly in childhood, almost simultaneously with how he begins to speak, but this counting that we are accustomed to has gone a long way of development and has taken different forms.
There was a time when only two numbers were used to count objects: one and two. In the process of further expansion of the number system, parts of the human body were involved, and first of all, fingers, and if there were not enough such “numbers”, then sticks, pebbles and other things.
N. N. Miklukho-Maclay in his book "Travels" talks about a funny way of counting used by the natives of New Guinea:
Questions:
- What is the definition of an angle?
- What are the types of corners?
- What is the difference between diameter and radius?
List of sources used:
- Mazur K. I. "Solving the main competitive problems in mathematics of the collection edited by M. I. Scanavi"
- Mathematical ingenuity. B.A. Kordemsky. Moscow.
- L. S. Atanasyan, V. F. Butuzov, S. B. Kadomtsev, E. G. Poznyak, I. I. Yudina "Geometry, 7 - 9: a textbook for educational institutions"
Worked on the lesson:
Levchenko V.S.
Poturnak S.A.
Ask a question about modern education, express an idea or solve an urgent problem, you can Education Forum, where on international level an educational council of fresh thought and action gathers. Having created blog, You will not only improve your status as a competent teacher, but also make a significant contribution to the development of the school of the future. Education Leaders Guild opens the door to top-ranking specialists and invites you to cooperate in the direction of creating the best schools in the world.
Subjects > Mathematics > Mathematics Grade 7