Construct a circumscribed circle using a compass. Compass and ruler construction

Objectives:

to consolidate the concepts of "circle", "circle" in students; deduce the concept of "radius of a circle"; learn how to build circles of a given radius; develop the ability to reason, analyze.

Personal UUD:
to form a positive attitude towards math lessons;
interest in subject research activities;

Metasubject tasks

Regulatory UUD:
accept and save a learning task;
in cooperation with the teacher and the class, find several options for solutions;

Cognitive UUD:
problem statement and solution:
independently identify and formulate a problem;
general educational:
find the necessary information in the textbook;
build a circle of a given radius using a compass;
brain teaser:
form the concept of "radius";
carry out classification, comparison;
independently formulate conclusions;

Communicative UUD:
actively participate in collective work, while using speech means;
argue your point of view;

Item skills:
to identify essential features of the concepts of "radius of a circle";
build circles with different radii;
recognize radii in a drawing.

During the classes

Motivation for learning activities

- Let's check if everyone is ready for the lesson?

"Emotional entry into the lesson":

Smile like suns.

Frown like clouds

Cry like rains

Surprise as if you saw a rainbow

Now repeat after me

The game "Friendly Echo"

2.Knowledge update

Verbal counting

a) 60-40 36 + 12 10 + 20 58-12 90-50 31 + 13

Unravel the pattern. Continue the row.

Answer: 20, 48,30,46,40,44 50,42

b) Solve the problem:

1. On the first day, the store sold 42 kg of fruit, and on the second, 2 kg more. How many kilograms were sold on the second day?

What needs to be changed so that the task is solved in 2 steps.

Balls - 16 pcs.

Rope - 28 pcs.

Find a solution to this problem.

28-16 28+16

Modify the question so that the problem is solved by subtraction.

3. Statement of the educational problem

1. Name the geometric shapes

Circle circle oval ball

Which figure is superfluous?

What do the figures have in common? (Circle, circle, ball have the same shape)

What is the difference?

2.In

What points belong to the circle? What are the points outside the circle?

What does point O stand for? (center of circle)

What is the name of the OB segment?

How many radii can you draw in a circle?

Which line is not a radius? Why?

What conclusion can be drawn?

Conclusion: all radii have the same length .

3. How many circles are there in the picture?

How are circles different? (size)

What determines the size of a circle?

What conclusion can be drawn?

Conclusion: the larger the circle, the larger its radius.

Define the topic of the lesson.

Theme: Creates a circle of a given radius using a compass.

What tasks can we set ourselves for this lesson?

4. Work on the topic

a) Constructing a circle.

What do you need to know to draw a circle of a given size?

Draw a circle with a radius of 3 cm.

b) Preparation for project activities

1) Consider the drawing

What are the shapes of a butterfly? Circles with the same radius?

2) Work in pairs.

Restore the order of the stages to the project.

Project presentation or demo

Concept (make a sketch)

Build figures to implement the plan

Consider what radius the shapes should have

c) Work on the project.

Work in groups according to the compiled algorithm

This lesson focuses on the study of the circle and circle. Also, the teacher will teach you to distinguish between closed and open lines. You will get acquainted with the basic properties of a circle: center, radius and diameter. Learn their definitions. Learn to determine the radius if the diameter is known, and vice versa.

If we fill in the space inside the circle, for example, draw a circle with a compass on paper or cardboard and cut it out, we get a circle (Fig. 10).

Figure: 10. Circle

A circle is the part of the plane bounded by a circle.

Condition:Vitya Verhoglyadkin drew 11 diameters in his circle (Fig. 11). And when he counted the radii, he got 21. Did he count correctly?

Figure: 11. Illustration for the problem

Decision:the radii must be twice as large as the diameters, therefore:

Vitya counted incorrectly.

List of references

1. Mathematics. Grade 3. Textbook. for general education. institutions with adj. to the electron. carrier. At 2 pm Part 1 / [M.I. Moreau, M.A. Bantova, G.V. Beltyukova and others] - 2nd ed. - M .: Education, 2012 .-- 112 p .: ill. - (School of Russia).
2. Rudnitskaya V.N., Yudacheva T.V. Mathematics, grade 3. - M .: VENTANA-GRAF.
3. Peterson L.G. Mathematics, grade 3. - M .: Juventa.

1. Mypresentation.ru ().
2. Sernam.ru ().
3. School-assistant.ru ().

Homework

1. Mathematics. Grade 3. Textbook. for general education. institutions with adj. to the electron. carrier. At 2 pm Part 1 / [M.I. Moreau, M.A. Bantova, G.V. Beltyukova and others] - 2nd ed. - M .: Education, 2012., Art. 94 No. 1, art. 95 No. 3.

2. Solve the riddle.

My brother and I live together

We have so much fun together

We will put a mug on the sheet (Fig. 12),

Outline with a pencil.

It turned out what you need -

Called ...

3. It is necessary to determine the diameter of the circle if it is known that the radius is 5 m.

4. * Using a compass, draw two circles with radii: a) 2 cm and 5 cm; b) 10 mm and 15 mm.

In construction problems, compasses and a ruler are considered ideal tools, in particular, the ruler has no divisions and has only one side of infinite length, and the compass can have an arbitrarily large or arbitrarily small opening.

Valid constructions. The following operations are allowed in building tasks:

1. Mark a point:

• an arbitrary point of the plane;
• an arbitrary point on a given straight line;
• an arbitrary point on a given circle;
• the point of intersection of two given lines;
• points of intersection / tangency of a given line and a given circle;
• points of intersection / tangency of two specified circles.

2.Using a ruler, you can build a straight line:

• an arbitrary straight line on a plane;
• an arbitrary straight line passing through a given point;
• a straight line passing through two given points.

3.Using a compass, you can build a circle:

• an arbitrary circle on a plane;
• an arbitrary circle centered at a given point;
• an arbitrary circle with a radius equal to the distance between two specified points;
• a circle centered at the specified point and with a radius equal to the distance between the two specified points.

Solving building problems. The solution to the construction problem contains three essential parts:

1. Description of the method for constructing the desired object.
2. Proof that the object constructed in the described way is indeed the desired one.
3. Analysis of the described construction method for its applicability to different variants of initial conditions, as well as for the uniqueness or non-uniqueness of the solution obtained by the described method.

Draw a line segment equal to the given one. Let there be a ray with origin at the point $ O $ and a segment $ AB $. To construct the segment $ OP \u003d AB $ on the ray, you need to construct a circle with the center at the point $ O $ of radius $ AB $. The point of intersection of the ray with the circle will be the desired point of $ P $.

Constructs an angle equal to the given one. Let there be given a ray with origin at point $ O $ and angle $ ABC $. With the center at the point $ B $ we construct a circle with an arbitrary radius $ r $. Let us denote the points of intersection of the circle with the rays $ BA $ and $ BC $, respectively, $ A "$ and $ C" $.

Construct a circle centered at the point $ O $ of radius $ r $. The point of intersection of the circle with the ray will be denoted by $ P $. Construct a circle with center at point $ P $ of radius $ A "B" $. The intersection point of the circles will be denoted by $ Q $. Draw ray $ OQ $.

We get the angle $ POQ $ equal to the angle $ ABC $, since the triangles $ POQ $ and $ ABC $ are equal on three sides.

Constructs a midpoint perpendicular to a line segment. Let's construct two intersecting circles of arbitrary radius with centers at the ends of the segment. By connecting the two points of their intersection, we get the middle perpendicular.

Constructing the bisector of an angle. Let's draw a circle of arbitrary radius centered at the vertex of the corner. Let's construct two intersecting circles of arbitrary radius with centers at the points of intersection of the first circle with the sides of the corner. Connecting the vertex of the corner with any of the intersection points of these two circles, we get the bisector of the angle.

Construction of the sum of two segments.To construct a segment equal to the sum of two given segments on a given ray, you need to apply the method of constructing a segment equal to this one twice.

Plotting the sum of two angles. In order to postpone from a given ray an angle equal to the sum of two given angles, you need to apply the method of constructing an angle equal to this one twice.

Finding the midpoint of a line segment. In order to mark the middle of a given segment, you need to build a mid-perpendicular to the segment and mark the point of intersection of the perpendicular with the segment itself.

Creates a perpendicular line through a given point. Let it be required to construct a straight line perpendicular to a given point and passing through a given point. Draw a circle of arbitrary radius centered at a given point (regardless of whether it lies on a straight line or not), intersecting the straight line at two points. We build a midpoint perpendicular to a segment with ends at the points of intersection of a circle with a straight line. This will be the desired perpendicular line.

Draws a parallel straight line through a given point. Let it be required to construct a straight line parallel to the given one and passing through a given point outside the straight line. Build a straight line passing through a given point, perpendicular to this straight line. Then we build a straight line passing through this point, perpendicular to the constructed perpendicular. The straight line obtained in this case will be the desired one.

When manufacturing or processing wood parts, in some cases it is required to determine where their geometric center is. If the part has a square or rectangular shape, then this is not difficult to do. It is enough to connect opposite corners with diagonals, which will intersect exactly in the center of our figure.
For products that have the shape of a circle, this solution will not work, since they have no corners, and therefore no diagonals. In this case, you need some other approach based on different principles.

And they exist, and in many variations. Some of them are quite complex and require several tools, others are easy to implement and do not need a whole set of tools to implement them.
Now we will look at one of the easiest ways to find the center of a circle using just a regular ruler and pencil.

The sequence for finding the center of the circle:

A sentence that explains the meaning of a particular expression or name is called defining... We have already met with definitions, for example, with the definition of an angle, adjacent angles, an isosceles triangle, etc. Let us give the definition of another geometric figure - a circle.

Definition

This point is called center of circle, and the segment connecting the center with any point of the circle is circle radius (fig. 77). It follows from the definition of a circle that all radii have the same length.

Figure: 77

A segment connecting two points of a circle is called its chord. The chord passing through the center of the circle is called it diameter.

In Figure 78, the segments AB and EF are the chords of the circle, the segment CD is the diameter of the circle. Obviously, the diameter of the circle is twice its radius. The center of a circle is the midpoint of any diameter.

Figure: 78

Any two points of the circle divide it into two parts. Each of these parts is called a circular arc. In Figure 79, ALB and AMB are arcs bounded by points A and B.

Figure: 79

To depict a circle in the drawing, use compass (fig. 80).

Figure: 80

To draw a circle on the ground, you can use a rope (fig. 81).

Figure: 81

The part of the plane bounded by a circle is called a circle (Fig. 82).

Figure: 82

Compass and ruler construction

We have already dealt with geometric constructions: we drew straight lines, laid out segments equal to the data, drew angles, triangles and other shapes. In doing so, we used a scale ruler, compasses, protractor, drawing square.

It turns out that many constructions can be performed using only a compass and a ruler without scale divisions. Therefore, in geometry, those construction tasks are specially distinguished, which are solved using only these two tools.

What can you do with them? It is clear that 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. By performing these simple operations, we will be able to solve many interesting construction problems:

build an angle equal to the given one;
draw a straight line through this point, perpendicular to this straight line;
split this segment in half and other tasks.

Let's start with a simple task.

A task

On a given ray from its beginning to postpone a segment equal to the given one.

Decision

Let's depict the figures given in the condition of the problem: ray OS and segment AB (Fig. 83, a). Then, with a compass, we construct a circle of radius AB with center O (Fig. 83, b). This circle will intersect the OS ray at some point D. The segment OD is the required one.

Figure: 83

Examples of building tasks

Plotting an angle equal to a given one

A task

Set aside the given angle from the given ray.

Decision

This angle with the vertex A and the ray OM are shown in Figure 84. It is required to construct an angle equal to the angle A, so that one of its sides coincides with the ray OM.

Figure: 84

Let's 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. 85, a). Then we draw a circle of the same radius centered on the given ray OM. It crosses the ray at point D (Fig. 85, b). After that, construct a circle with center D, whose radius is equal to BC. The circles with centers O and D intersect at two points. We denote one of these points by the letter E. Let us prove that the angle MOE is the required one.

Figure: 85

Consider triangles ABC and ODE. The segments AB and AC are the radii of the circle with the center A, and the segments OD and OE are the radii of the circle with the center O (see Fig. 85, b). Since, by construction, these circles have equal radii, then AB \u003d OD, AC \u003d OE. Also, according to the construction, ВС \u003d DE.

Therefore, Δ ABC \u003d Δ ODE on three sides. Therefore, ∠DOE \u003d ∠BAC, i.e. the constructed angle MOE is equal to the given angle A.

The same construction can be performed on the ground if you use a rope instead of a compass.

Plotting the bisector of an angle

A task

Construct the bisector of the given angle.

Decision

This angle BAC is shown in Figure 86. Draw a circle of arbitrary radius centered at vertex A. It will intersect the sides of the angle at points B and C.

Figure: 86

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 the figure). They will intersect at two points, of which at least one lies inside the corner. Let us denote it by the letter E. Let us prove that the ray AE is the bisector of the given angle BAC.

Consider triangles ACE and ABE. They are equal on three sides. Indeed, AE is a common side; AC and AB are equal as the radii of the same circle; CE \u003d BE by construction.

From the equality of triangles ACE and ABE it follows that ∠CAE \u003d ∠BAE, that is, ray AE is the bisector of a given angle BAC.

Comment

Is it possible to divide a given angle into two equal angles using a compass and a ruler? It is clear that it is possible - for this you need to draw the bisector of this angle.

This angle can also be divided into four equal angles. To do this, you need to divide it in half, and then divide each half in half again.

Is it possible to divide this angle into three equal angles using a compass and a ruler? This task, dubbed angle trisection problems, has attracted the attention of mathematicians for centuries. Only in the 19th century it was proved that such a construction is impossible for an arbitrary angle.

Drawing perpendicular lines

A task

A straight line and a point on it are given. Construct a line passing through a given point and perpendicular to this line.

Decision

This line a and a given point M belonging to this line are shown in Figure 87.

Figure: 87

On the rays of the straight line a, outgoing from the point M, we postpone equal segments MA and MB. Then we construct two circles with centers A and B of radius AB. They intersect at two points: P and Q.

Let us draw a straight line through the point M and one of these points, for example the straight line MP (see Fig. 87), and prove that this line is the sought one, that is, that it is perpendicular to the given line a.

Indeed, since the median PM of the isosceles triangle PAB is also the height, PM ⊥ a.

Draw the midpoint of a line segment

A task

Construct the middle of this segment.

Decision

Let AB be a given segment. Let's construct two circles with centers A and B of radius AB. They intersect at points P and Q. Draw line PQ. The point O of the intersection of this line with the segment AB is the desired midpoint of the segment AB.

Indeed, the triangles APQ and BPQ are equal on three sides, so ∠1 \u003d ∠2 (Fig. 89).

Figure: 89

Consequently, the segment PO is the bisector of the isosceles triangle APB, and hence the median, that is, point O is the midpoint of the segment AB.

Tasks

143. Which of the segments shown in Figure 90 are: a) chords of the circle; b) the diameters of the circle; c) the radii of the circle?

Figure: 90

144. Segments AB and CD - diameters of a circle. Prove that: a) chords BD and AC are equal; b) chords AD and BC are equal; c) ∠BAD \u003d ∠BCD.

145. Segment MK - diameter of a circle with center O, and MP and PK - equal chords of this circle. Find ∠POM.

146. Segments AB and CD - diameters of a circle with center O. Find the perimeter of triangle AOD, if it is known that CB \u003d 13 cm, AB \u003d 16 cm.

147. Points A and B are marked on the circle with center O so that the angle AOB is a straight line. Segment BC - diameter of the circle. Prove that the chords AB and AC are equal.

148. Two points A and B are given on the straight line. On the extension of ray B A, set aside the segment BC so that BC \u003d 2AB.

149. Given a straight line a, a point B, not lying on it, and a segment PQ. Construct a point M on the line a so that BM \u003d PQ. Does a problem always have a solution?

150. Given a circle, point A, not lying on it, and a segment PQ. Construct a point M on the circle so that AM \u003d PQ. Does a problem always have a solution?

151. An acute angle BAC and an XY ray are given. Construct the YXZ corner so that ∠YXZ \u003d 2∠BAC.

152. An obtuse angle AOB is given. Construct the OX beam so that the XOA and XOB angles are equal obtuse angles.

153. Given a straight line a and a point M that does not lie on it. Construct a line passing through point M and perpendicular to line a.

Decision

Construct a circle centered at a given point M, intersecting a given line a at two points, which we denote by the letters A and B (Fig. 91). Then we construct two circles with centers A and B passing through point M. These circles intersect at point M and at one more point, which we denote by the letter N. We draw a straight line MN and prove that this line is the required one, that is, it is perpendicular to straight a.

Figure: 91

Indeed, the triangles AMN and BMN are equal on three sides, so ∠1 \u003d ∠2. It follows that the segment MC (C is the point of intersection of lines a and MN) is the bisector of the isosceles triangle AMB, and hence the height. Thus, MN ⊥ AB, that is, MN ⊥ a.

154. Given triangle ABC. Build: a) bisector AK; b) the median of the VM; c) the height of the CH triangle. 155. Using a compass and a ruler, build an angle equal to: a) 45 °; b) 22 ° 30 ".

Answers to problems

152. Indication. First, construct the bisector of the angle AOB.