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Name the shapes K E T S V A X

How many parts is the plane of the figure divided into?

Circle and circle Circle is a closed line Circle is a plane that lies inside the circle, together with the circle

Circle A circle divides a plane into two parts!

Construction O 1) Mark point O - the center of the circle. 2) Set the radius of the circle using a compass and ruler. 3) Place the leg of the compass at point O 4) Draw a circle.

All points of a circle are distant from its center. O – center of a circle and circle OA = OC = OE – radius – r AB – diameter - d AB = OA+OB d = 2r, r = d:2 O C A E B Radius – a segment connecting the center of the circle with a point lying on her. All radii of a circle are equal! Diameter is a segment connecting two points of a circle and passing through its center.

The diameter divides the circle into two semicircles, O C A B O C A B the circle into two semicircles.

Circular arc NE - arc NE, ends of the arc - points C and B. AC - arc AC, ends of the arc - points A and C. AB, BE O C A E B

Examples of a circle and a circle in life

Numbers for work: For consolidation of material: No. 850 (oral) No. 851 No. 853 No. 855 For repetition: No. 871(1) Independent work: No. 872(1)

Homework: paragraph 22, No. 874, No. 876, No. 878 (a, d, f)

No. 853 O A B r =3 cm OA= , OA r

No. 855 C D AC = 3cm, CB = 3cm D A = 4cm, B D =4cm B A

On the topic: methodological developments, presentations and notes

The image of a circle and its role in V. Nabokov’s story “The Circle”

"9 circles of hell according to Dante" A guide to the circles of hell from the Divine Comedy by Dante Alighieri.

“The Divine Comedy” (Italian: La Commedia, later La Divina Commedia) is a poem written by Dante Alighieri between 1307 and 1321 and provides the broadest synthesis of medieval culture...

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Circle The presentation was prepared by: Kislova Svetlana Igorevna Mathematics teacher MBOU Secondary School No. 2 G. Lyskovo

Goals and objectives: Systematize theoretical material on the topic “Circle”. Improve problem solving skills. Prepare students for the test. Prepare students to successfully solve the Geometry module when passing the OGE.

properties of tangent C-tangent A-point of tangency C OA O A C a b M A B O

Theorem about tangent and secant C M A V The square of the length of a tangent is equal to the product of the secant and its external part. D C A B O The product of one secant and its external part is equal to the product of another secant and its external part M O

Central and inscribed angles Central Inscribed B A O D A C B O

An inscribed angle is either equal to half of its corresponding central angle, or (2) complements half of this angle to 180 degrees. 12

Properties of inscribed angles O A B D C B K A C

Property of intersecting chords C B K A D

Incircle Each point of the bisector of an undeveloped angle is equidistant from its sides. Conversely: each point lying inside the angle and equidistant from the sides of the angle lies on its bisector O O - intersection of bisectors Property of a bisector A B C D Property of a circumscribed quadrilateral AB+CD=BC+AD The sums of opposite sides are equal.

Circumscribed circle Each point of the perpendicular bisector to a segment is equidistant from the ends of this segment Conversely: each point equidistant from the ends of the segment lies on the perpendicular bisector to it O - intersection of perpendicular bisectors Property of the perpendicular bisector A D C B Property of the cyclic quadrilateral Sum of opposite angles is 180* O

Oral problems on finished drawings 160 Answer: 80 ? Answer: 45 B A C B C A D A B C M K R 5 6 3 Answer: 28 ?

A C B D 7 8 P=? Answer: 30 M K T O 70°? Answer: 20° O

Must be able to: Apply definitions, properties of figures, and various theorems when solving problems. Be able to build a logical chain of reasoning. Apply theory to a new situation.

120° 60° 120° 240° 115° 65° 230° 40° 140° 140° AC CB AB R KTP PK PT KPT - - 4 3 5 2 , 5 30° 4 8 60° - - Answers:

Group 2 1 2 3 4 B A B A Group 1 1 2 3 4 A B B D Group 3 1 2 3 4 B A ABC B

On the topic: methodological developments, presentations and notes

A mathematics lesson in the 6th grade on the topic "Circumference. Circle. Circumference" is best conducted in the form of practical work....

Purpose of the lesson: to repeat the concept of a circle and a circle; calculating the value of Pi; introduce the concept of circumference and formulas for calculating the circumference....

First lesson on the topic Circumference in 6th grade. Practical work is carried out during which the children calculate the value of pi. Getting acquainted with the number Pi....

Rodionova G. M. Number circle on the coordinate plane // Algebra and beginnings of analysis, grade 10 //. Presentation contains material: number circle on the coordinate plane, basic...

Math lesson in 5th grade

on the topic "Circle and Circle".

• ©GBOU boarding school No. 1
• Mathematics teacher: Makarova N.A.
• St. Petersburg, 2015.

Goals and objectives of the lesson:

Educational:

• Ensure the understanding of the concepts of circle, circle and their elements (radius, diameter, chord, arc).
• Consider the relationship between the diameter and radius of a circle.
• Introduce the compass tool, teach how to draw a circle using a compass.
• Learn to find similarities and differences between a circle and a circle; broaden the horizons of students.

Educational:

• Development of logical thinking, attention, creative and cognitive abilities, imagination, ability to analyze, draw conclusions.
• Formation of accuracy and precision when making drawings.
• Application of information technologies in the study of mathematics.

Educational:

• Development of hard work, discipline, respect for classmates.
• Formation of interest in mathematics.

Equipment: interactive whiteboard, computer, drawing tools.

A compass is a drawing tool. It has a needle at one end and a pencil at the other.

You need to work with the compass carefully!!!

1. Mark a point in your notebook and name it the letter O.

2. Take a compass and spread the “legs” of the compass to a distance of 3 cm.

3. Place the needle of the compass at point O, and draw a closed line with the other “leg” of the compass.

A circle is a closed line consisting of points that are equally distant from the center.

Point O is called the center of the circle

Mark two points A and M on the circle.

Segments OA and OM are called radii of the circle.

The radius of a circle is the segment that connects the center of the circle and a point on the circle.

Let's connect the points O and M, O and A.

The radius is designated

Latin letter r.

Construct two circles with a radius of 2 cm in your notebook. Paint the inner area of ​​one circle.

CIRCLE is a geometric figure consisting of all points located at the same distance from the center of the circle.

CIRCLE is a geometric figure consisting of all points of the plane located inside the circle (including the circle itself).

Circle

Which objects are shaped like a circle and which ones are shaped like a circle?

Extend line segment AO until it intersects the circle.

Label the point of intersection with the letter K.

The segment AK is called the diameter of the circle.

The diameter of a circle is a line segment connecting two points on the circle and passing through its center.

The diameter is designated by the Latin letter d.

Connect the dots

M and K, A and M.

The segments MK and AM are called chords of the circle.

A chord is a line segment connecting two points on a circle.

Name all the radii, diameters and chords of a circle.

Draw a circle with center at point O.

Mark two points A and B on the circle.

Points A and B divided the circle into two parts, which are called arcs of the circle.

An arc of a circle is a part of a circle

between points A and B.

Name all the arcs on the circle:

Name the points

lying on a circle.

Name the points

not lying on the circle.

Name the points

lying on a circle.

Option 1

A1. What is the name of segment AB in drawing No. 1?

1) circle diameter

2) circle radius

3) chord of a circle

A2. Choose the correct continuation of the statement:

The radius of a circle is the segment that...

A3. Can a circle have two diameters of different lengths?

2) can't

3) make it difficult to answer

Option 2

A1. What is the name of segment AB in drawing No. 2?

1) chord of a circle

2) circle diameter

3) circle radius

A2. Choose the correct sentence of the statement:

The diameter of a circle is the segment that...

1) connects any two points on a circle

2) connects the center of the circle with any point on the circle

3) connects two points on a circle and passes through the center of the circle

A3. Can a circle have two radii of different lengths?

2) can't

3) find it difficult to answer

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The first lesson in the topic “Ordinary fractions”.

Textbook by N.Ya. Vilenkin “Mathematics 5”.

Lesson objectives: to familiarize students with the concept of circle and circumference; developing the ability to construct a circle using a compass along a given radius and diameter.

Learning objectives aimed at achieving:

Personal development:

• continue to develop the ability to clearly, accurately and competently express your thoughts in oral and written speech,
• develop creative thinking, initiative, resourcefulness, and activity in solving mathematical problems.

Meta-subject development:

• broaden your horizons, instill the ability to work together (a sense of camaraderie and responsibility for the results of your work);
• continue to develop the ability to understand and use mathematical visual aids.

Subject development:

• to form a theoretical and practical understanding of the circle and the circle as geometric figures and their elements;
• continue the development of visual skills (learn to use a compass to construct a circle of any radius);
• develop the ability to apply learned concepts to solve practical problems.

Lesson type: lesson in acquiring new knowledge, skills and abilities.

Forms of student work:

• individual;
• frontal;
• independent work;
• work in pairs;
• test control.

Necessary equipment:

• Projector and screen.
• Presentation “Circle and Circle”.
• Individual sheet for each student ( Annex 1).

Lesson structure and flow

TEST Find: sector, arc, radius, diameter, chord, segment

Through three points A, B and C that do not lie on the same line (through the vertices ABC), a circle can be drawn if such a fourth point exists. O, which is equally distant from points A, B and C. Let us prove that such a point exists and, moreover, only one. Every point equally distant from points A and B must lie on the perpendicular bisector MN to the segment AB, and in the same way every point equally distant from points B and C must lie on the perpendicular bisector PQ drawn to side BC. This means that if there is a point equally distant from three points A, B and C, then it must lie on both MN and PQ, which is possible only when it coincides with the intersection point of these two lines. Lines MN and PQ always intersect, since they are perpendicular to the intersecting lines AB and BC. Point O of their intersection will be a point equally distant from A, from B and from C, which means that if we take this point as the center, and take the distance OA (or OB, or OC) as the radius, then the circle will pass through points A, B and C. Since the lines MN and PQ can intersect only at one point, there can only be one center of the circle and only one length of its radius; This means that the circle we are looking for is unique.

Let's bend the drawing along the diameter AB so that its left side falls onto the right. Then the left semicircle will align with the right semicircle and the perpendicular KS will go along KD. It follows from this that point C, which is the intersection of the semicircle with the KS, will fall on D; therefore CK= KD; BC= BD, AC= AD. BC= BD AC= AD

Properties of the diameter of a circle 1. The diameter drawn through the middle of a chord is perpendicular to this chord and divides the arc subtended by it in half. 2. The diameter drawn through the middle of the arc is perpendicular to the chord subtending this arc and divides it in half.

1. Consider a circle with center O. AB = CD, P is the midpoint of the chord AB, Q is the midpoint of CD. 2. Consider ΔOAR and ΔOCQ (rectangular): OA = OS - radii, PA = CQ - halves of equal chords 3. ΔOAR = ΔOCQ (on the hypotenuse and leg). From the equality of triangles OP = OQ (equal legs), i.e. chords are equally distant from the center

Cases of relative position of a line and a circle d rd > r rd > r"> rd > r"> rd > r" title="Cases of relative position of a line and a circle d rd > r"> title="Cases of relative position of a line and a circle d rd > r"> !}

D

D>r If the distance from the center of the circle to the straight line is greater than the radius of the circle, then the straight line and the circle do not have common points. O d>r r r If the distance from the center of the circle to the straight line is greater than the radius of the circle, then the straight line and the circle do not have common points. O d>r r"> r If the distance from the center of the circle to the straight line is greater than the radius of the circle, then the straight line and the circle do not have common points. O d>r r"> r If the distance from the center of the circle to the straight line is greater than the radius of the circle, then the straight line and the circle do not have common points. O d>r r" title="d>r If the distance from the center of the circle to the straight line is greater than the radius of the circle, then the straight line and the circle do not have common points. O d>r r"> title="d>r If the distance from the center of the circle to the straight line is greater than the radius of the circle, then the straight line and the circle do not have common points. O d>r r"> !}

Tangent property. Let straight line p touch the circle at point A, i.e. A is their only common point. Proof by contradiction: 1. Let us assume that p is not perpendicular to the radius OA. Let's draw a perpendicular to OB on the river. 2. Let us plot the segment BC = BA on p. 3. OVA = OVS (on two legs). Therefore OS = OA. 4. C lies on the circle. Therefore, p and the circle have two common points, which is impossible. So, p OA, which is what was required

Take any point A of circle F and draw the radius OA. Then we draw a straight line p perpendicular to the radius OA. Any point B of straight line p, different from point A, is removed from O by more than a radius, since the inclined OB is longer than the perpendicular OA. Therefore, point B does not lie on F. This means that point A is the only common point of p and F, that is, p touches F at point A.

Various cases of relative position of two circles. d>R+R 1d>R+R 1 d=R+R 1d=R+R 1 d R+R 1d>R+R 1 d=R+R 1d=R+R 1 d"> R+R 1d>R+R 1 d=R+R 1d=R+R 1 d"> R+R 1d >R+R 1 d=R+R 1d=R+R 1 d" title="Different cases of relative position of two circles. d>R+R 1d>R+R 1 d=R+R 1d= R+R 1 d"> title="Various cases of relative position of two circles. d>R+R 1d>R+R 1 d=R+R 1d=R+R 1 d"> !}

1. The circles lie one outside the other, without touching in this case, obviously, d > R + R 1 R and R 1 are the radii of the circles d is the distance between the centers of the circles R + R 1 R and R 1 - radii of circles d - distance between centers of circles"> R + R 1 R and R 1 - radii of circles d - distance between centers of circles"> R + R 1 R and R 1 - radii of circles d - the distance between the centers of the circles" title="1. The circles lie one outside the other, without touching in this case, obviously, d > R + R 1 R and R 1 - the radii of the circles d - the distance between the centers of the circles"> title="1. The circles lie one outside the other, without touching in this case, obviously, d > R + R 1 R and R 1 are the radii of the circles d is the distance between the centers of the circles"> !}

3. The circles intersect then d

5. One circle lies inside the other without touching, then obviously d

R + R 1, then the circles are located one outside the other without touching. 2. If d = R + R 1, then the circles touch from the outside. 3. If d R – R 1, then the circles intersect. 4. If d = R – R 1, then the circles touch from the inside. 5." title="Converse sentences 1. If d > R + R 1, then the circles are located one outside the other, without touching. 2. If d = R + R 1, then the circles touch from the outside. 3. If d R – R 1, then the circles intersect. 4. If d = R – R 1, then the circles touch from the inside. 5." class="link_thumb"> 59 !} Converse propositions 1. If d > R + R 1, then the circles are located one outside the other, without touching. 2. If d = R + R 1, then the circles touch from the outside. 3. If d R – R 1, then the circles intersect. 4. If d = R – R 1, then the circles touch from the inside. 5. If d R + R 1, then the circles are located one outside the other, without touching. 2. If d = R + R 1, then the circles touch from the outside. 3. If d R – R 1, then the circles intersect. 4. If d = R – R 1, then the circles touch from the inside. 5."> R + R 1, then the circles are located one outside the other, without touching. 2. If d = R + R 1, then the circles touch from the outside. 3. If d R – R 1, then the circles intersect. 4. If d = R – R 1, then the circles touch from the inside. 5. If d R + R 1, then the circles are located one outside the other, without touching. 2. If d = R + R 1, then the circles touch from the outside. 3. If d R – R 1, then the circles intersect. 4. If d = R – R 1, then the circles touch from the inside. 5." title="Converse sentences 1. If d > R + R 1, then the circles are located one outside the other, without touching. 2. If d = R + R 1, then the circles touch from the outside. 3. If d R – R 1, then the circles intersect. 4. If d = R – R 1, then the circles touch from the inside. 5."> title="Converse propositions 1. If d > R + R 1, then the circles are located one outside the other, without touching. 2. If d = R + R 1, then the circles touch from the outside. 3. If d R – R 1, then the circles intersect. 4. If d = R – R 1, then the circles touch from the inside. 5.">!}

Given: circle with center O, ABC - inscribed Prove: ABC = ½ AC Proof: Consider the case when side BC passes through center O 1. Arc AC is less than a semicircle, AOC = AC (central) 2. Consider ΔABO, AO = OB ( radii). ΔABO isosceles 1 = 2, AOC – external angle ΔABO, AOC = = 2 1, therefore ABC = ½ AC 1 2

Given: circle with center O, ABC is inscribed Prove: ABC = ½ AC Proof: Consider the case when center O lies inside the inscribed angle. 1. Additional construction: diameter BD 2. Ray BO divides ABC into two angles 3. Ray BO intersects arc AC at point D 4. AC = AD + DC, therefore ABD = ½ AD and DBC = ½ DC or ABD + DBC = ½ AD + ½ DC or ABC = ½ AC

Given: circle with center O, ABC is inscribed Prove: ABC = ½ AC Proof: Consider the case when center O lies outside the inscribed angle. 1. Additional construction: diameter BD 2. Ray BO does not divide ABC into two angles 3. Ray BO does not intersect arc AC at point D 4. AC = AD - CD, therefore ABD = ½ AD and DBC = ½ DC or ABD - DBC = ½ AD - ½ DC or ABC = ½ AC

72

Proof. 1. Consider an arbitrary triangle ABC. Let us denote by the letter O the point of intersection of the bisectoral perpendiculars to its sides and draw the segments O A, O B and OS. 2. Since point O is equidistant from the vertices of triangle ABC, then OA = OB = OS. Therefore, a circle with center O of radius OA passes through all three vertices of the triangle and, therefore, is circumscribed about triangle ABC. Proof. 1. Consider an arbitrary triangle ABC and denote by the letter O the point of intersection of its bisectors. 2. Let's draw perpendiculars OK from point O. OL and OM respectively to sides AB, BC and CA. 3. Since point O is equidistant from the sides of triangle ABC, then OK = OL = OM. Therefore, a circle with center O of radius OK passes through points K, L and M. 4. The sides of triangle ABC touch this circle at points K, L, M, since they are perpendicular to the radii OK, OL and OM. This means that a circle with center O of radius OK is inscribed in triangle ABC.