A polyhedron inscribed in a sphere. Mathematics. The full course is repeatable. Open lesson on geometry

Description of the presentation by individual slides:

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municipal autonomous educational institution average comprehensive school № 45 Toolkit for 11th grade students Compiled by a mathematics teacher of the highest category, Elena Vyacheslavovna Gavinskaya. Kaliningrad 2016-2017 academic year

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Polyhedra inscribed in a sphere. The topic is similar to that of the planimetry course, where it was said that circles can be described around triangles and regular n-gons. The analogue of a circle in space is a sphere, and a polygon is a polyhedron. In this case, the analogue of a triangle is a triangular prism, and the analogue of regular polygons is regular polyhedra. Definition. A polyhedron is said to be inscribed in a sphere if all its vertices belong to this sphere. The sphere itself is said to be circumscribed about the polyhedron.

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“A sphere can be described around a straight prism if and only if a circle can be described around the base of this prism.” Proof: If a sphere is circumscribed around a straight prism, then all the vertices of the base of the prism belong to the sphere and, therefore, to the circle, which is the line of intersection of the sphere and the plane of the base. Conversely, let a circle with a center at point O1 and radius r be described near the base of a straight prism. Then, around the second base of the prism, a circle with the center at point O2 and the same radius can be described. Let O1O2=d, O – the middle of O1O2. Then the sphere with center O and radius R= will be the desired circumscribed sphere. Theorem 1.

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“A sphere can be described around any triangular pyramid, and only one.” Proof. Let us turn to a proof similar to that from the planimetry course. First of all, we need to find the locus of points equidistant from the two vertices of the triangle. For example, A and B. Such a geometric location is the perpendicular bisector drawn to the segment AB. Then we find the locus of points equidistant from A and C. This is the perpendicular bisector to the segment AC. The intersection point of these bisectoral perpendiculars will be the desired center O of the circle circumscribed about triangle ABC. Theorem 2.

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Now let's consider the spatial situation and make similar constructions. Let a triangular pyramid DABC be given, and points A, B and C define the plane α. The geometric locus of points equidistant from points A, B and C is a straight line a, perpendicular to the plane α and passing through the center O1 of the circle circumscribed about triangle ABC. The geometric locus of points equidistant from points A and D is the plane β, perpendicular to the segment AD and passing through its vertex - point E. The plane β and straight line a intersect at point O, which will be the desired center of the sphere circumscribed about the triangular pyramid DABC. Indeed, by virtue of the construction, point O is equally distant from all vertices of the DABC pyramid. Moreover, such a point will be unique, since the intersecting straight line and plane have a single common point.

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The ball described about regular pyramid. The ball can be described around any regular pyramid. The center of the ball lies on a straight line passing through the height of the pyramid and coincides with the center of a circle circumscribed about an isosceles triangle, the side of which is the lateral edge of the pyramid, and the height is the height of the pyramid. The radius of the ball is equal to the radius of this circle. The radius of the ball R, the height of the pyramid H and the radius of the circle r described near the base of the pyramid are related by the relation: R2=(H-R)2+r2 This relation is also valid in the case when H< R.

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The problem is about a ball circumscribed about a regular pyramid. “A sphere with a center at point O and a radius of 9√3 m is described near the regular pyramid PABC. The straight line PO, containing the height of the pyramid, intersects the base of the pyramid at point H so that PH:OH = 2:1. Find the volume of the pyramid if each of its side edges forms an angle of 45 degrees with the plane of the base.”

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Given: PABC – regular pyramid; the ball (O;R=9√3 m) is described near the pyramid; RO∩(ABC)=N; PH:OH=2:1; ∟RAN=∟ RVN=∟ RSN=45o. Find: Vpir. Solution: Since RN:OH=2:1 (by condition), then RN:OR=2:3 RN:9√3 =2:3 RN=6√3 (m) 2. RN _ (ABC) (as height of the pyramid) => => RN _ AN (by definition) => RAS - rectangular. 3. AT RAS:

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4. Since by condition RABC is a regular pyramid and PH is its height, then by definition ABC is correct; H is the center of a circle circumscribed about ABC, which means 5. Answer: 486 m3.

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A sphere circumscribed around a prism. A sphere can be described around a prism if it is straight and its bases are polygons inscribed in a circle. The center of the ball lies at the midpoint of the height of the prism connecting the centers of the circles described around the bases of the prism. The radius of the ball R, the height of the prism H and the radius of the circle r described around the base of the prism are related by the relation:

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The problem is about a sphere circumscribed around a prism. “A regular prism ABCDA1B1C1D1 with a height of 6 cm is inscribed in a sphere (so; R = 5 cm). Find the cross-sectional area of ​​the prism by a plane parallel to the planes of the base and passing through point O - the center of the ball.”

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Given: ABCDA1B1C1D1 – regular prism; a ball (O;R=5 cm) is described around a prism; the height of the prism h is 6 cm; α║(ABC); O with α. Find: Ssec α, Solution: Since, by condition, the prism is inscribed in a ball, then (r is the radius of the circle circumscribed around the base of the prism) But by condition, a regular prism is given, which means

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a) (АВВ1) ║(СС1D1) (by the property of a straight prism) α ∩ (АВВ1)=КМ α ∩ (СС1D1)=РН => KM ║ HP (by the property of parallel planes) Ho (BCC1) ║(ADD1) (by property of a straight prism) => KM=NR (by the property of parallel planes). This means that KMNR is a parallelogram (by attribute) => MN=KR and MN ║ KR b) α ║ (ABC) (by construction) α ∩ (ABB1)=KM (ABC) ∩ (ABB1)=AB => KM ║ AB (according to the property of parallel planes) 2. 3. Since according to the condition ABCDA1B1C1D1 is a regular prism, and the section by plane α is parallel to the bases, then the figure formed by the section is a square. Let's prove it: => => =>

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KMH= ABC=90o (as angles with correspondingly aligned sides) This means that the rhombus KMNR is a square (by definition), which is what needed to be proven. Moreover, the squares KMNR and ABCD are equal. Therefore, by property their areas are equal, and, therefore, Ssection α.=SABCD=32 (cm2) Answer: 32 cm2. c) KM ║ AB (proved) (BCC1) ║(ADD1) (by the property of a straight prism) => KM=AB=4√2 cm (by the property of parallel planes). d) Similarly, it is proven that MN ║ BC and MN = BC = 4√2 cm. This means that MN = KM => parallelogram MNRK is a rhombus (by definition). e) MN ║ BC (proved) KM ║ AB (proved) => =>

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A cylinder circumscribed around a prism. A cylinder can be described around a straight prism if its base is a polygon inscribed in a circle. The radius of the cylinder R is equal to the radius of this circle. The axis of the cylinder lies on the same straight line with the height H of the prism, connecting the centers of the circles described near the bases of the prism. In the case of a quadrangular prism (if the base is a rectangle), the axis of the cylinder passes through the intersection point of the diagonals of the bases of the prism.

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The problem is about a cylinder circumscribed around a prism. Straight prism ABCDA1B1C1D1, the base of which is a rectangle, is inscribed in a cylinder, the generatrix of which is 7 cm and the radius is 3 cm. Find the area of ​​the lateral surface of the prism if the angle between the diagonals ABCD is 60 degrees. ОО1 – cylinder axis.

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Given: ABCDA1B1C1D1 – straight prism; the cylinder is described near the prism; generatrix of the cylinder AA1=7 cm; the radius of the base of the cylinder is 3 cm; the angle between the diagonals ABCD is 60°; ОО1 – cylinder axis. Find: Sside prism. Solution: Since, according to the condition, a quadrangular prism, at the base of which is a rectangle, is inscribed in a ball, then according to the property AC∩ВD=O. This means AOB=60o and AO=OB=3cm. 2. In AOB using the cosine theorem.

Polyhedra inscribed in a sphere A convex polyhedron is called inscribed if all its vertices lie on some sphere. This sphere is called described for a given polyhedron. The center of this sphere is a point equidistant from the vertices of the polyhedron. It is the point of intersection of planes, each of which passes through the middle of the edge of the polyhedron perpendicular to it.

Formula for finding the radius of a circumscribed sphere Let SABC be a pyramid with equal lateral edges, h is its height, R is the radius of the circle circumscribed around the base. Let's find the radius of the circumscribed sphere. Note the similarity of right triangles SKO1 and SAO. Then SO 1 /SA = KS/SO; R 1 = KS · SA/SO But KS = SA/2. Then R 1 = SA 2 /(2SO); R 1 = (h 2 + R 2)/(2h); R 1 = b 2 /(2h), where b is a side edge.

A parallelepiped inscribed in a sphere Theorem: A sphere can be described around a parallelepiped if and only if the parallelepiped is rectangular, since in in this case it is straight and around its base - a parallelogram - a circle can be described (since the base is a rectangle).

Problem 1 Find the radius of a sphere circumscribed about a regular tetrahedron with edge a. Solution: SO 1 = SA 2 /(2SO); SO = = = a SO 1 = a 2 /(2 a) = a /4. Answer: SO 1 = a /4. Let us first construct an image of the center of a circumscribed ball using the image of a regular tetrahedron SABC. Let's draw the apothems SD and AD (SD = AD). In the isosceles triangle ASD, each point of the median DN is equidistant from the ends of the segment AS. Therefore, point O 1 is the intersection of the height SO and the segment DN. Using the formula from R 1 = b 2 /(2h), we get:

Problem 2 Solution: Using the formula R 1 =b 2 /(2h) to find the radius of the circumscribed ball, we find SC and SO. SC = a/(2sin(α /2)); SO 2 = (a/(2sin(α /2)) 2 – (a /2)2 = = a 2 /(4sin 2 (α /2)) – 2a 2 /4 = = a 2 /(4sin 2 ( α /2)) · (1 – 2sin 2 (α /2)) = = a 2 /(4sin 2 (α /2)) · cos α In a regular quadrangular pyramid, the side of the base is equal to a, and the plane angle at the apex is equal to α . Find the radius of the circumscribed ball. R 1 = a 2 /(4sin 2 (α /2)) · 1/(2a/(2sin(α /2))) =a/(4sin(α /2) ·). Answer : R 1 = a/(4sin(α /2) ·).

Polyhedra circumscribed about a sphere A convex polyhedron is called circumscribed if all its faces touch some sphere. This sphere is called inscribed for a given polyhedron. The center of an inscribed sphere is a point equidistant from all faces of the polyhedron.

Position of the center of an inscribed sphere Concept of a bisector plane of a dihedral angle. A bisector plane is a plane that divides a dihedral angle into two equal dihedral angles. Each point of this plane is equidistant from the faces of the dihedral angle. In the general case, the center of a sphere inscribed in a polyhedron is the intersection point of the bisector planes of all dihedral angles of the polyhedron. It always lies inside the polyhedron.

A pyramid circumscribed around a ball A ball is said to be inscribed in an (arbitrary) pyramid if it touches all faces of the pyramid (both lateral and base). Theorem: If the side faces are equally inclined to the base, then a ball can be inscribed in such a pyramid. Since the dihedral angles at the base are equal, their halves are also equal and the bisectors intersect at one point at the height of the pyramid. This point belongs to all bisector planes at the base of the pyramid and is equidistant from all faces of the pyramid - the center of the inscribed ball.

Formula for finding the radius of an inscribed sphere Let SABC be a pyramid with equal lateral edges, h is its height, r is the radius of the inscribed circle. Let's find the radius of the circumscribed sphere. Let SO = h, OH = r, O 1 O = r 1. Then, by the property of the bisector of the internal angle of a triangle, O 1 O/OH = O 1 S/SH; r 1 /r = (h – r 1)/ ; r 1 · = rh – rr 1 ; r 1 · (+ r) = rh; r 1 = rh/(+ r). Answer: r 1 = rh/(+ r).

A parallelepiped and a cube described around a sphere Theorem: A sphere can be inscribed into a parallelepiped if and only if the parallelepiped is straight and its base is a rhombus, and the height of this rhombus is the diameter of the inscribed sphere, which, in turn, is equal to the height of the parallelepiped. (Of all the parallelograms, only a circle can be inscribed in a rhombus) Theorem: A sphere can always be inscribed in a cube. The center of this sphere is the point of intersection of the diagonals of the cube, and the radius is equal to half the length of the edge of the cube.

Combinations of figures Inscribed and circumscribed prisms A prism circumscribed about a cylinder is a prism whose base planes are the planes of the bases of the cylinder, and the side faces touch the cylinder. A prism inscribed in a cylinder is a prism whose base planes are the planes of the bases of the cylinder, and the side edges are the generators of the cylinder. A tangent plane to a cylinder is a plane passing through the generatrix of the cylinder and perpendicular to the plane of the axial section containing this generatrix.

Inscribed and circumscribed pyramids A pyramid inscribed in a cone is a pyramid whose base is a polygon inscribed in the circle of the base of the cone, and the apex is the vertex of the cone. The lateral edges of a pyramid inscribed in a cone form the cone. A pyramid circumscribed around a cone is a pyramid whose base is a polygon circumscribed around the base of the cone, and the apex coincides with the apex of the cone. The planes of the side faces of the described pyramid are tangent to the plane of the cone. A tangent plane to a cone is a plane passing through the generatrix and perpendicular to the plane of the axial section containing this generatrix.

Other types of configurations A cylinder is inscribed in a pyramid if the circle of one of its bases touches all the lateral faces of the pyramid, and its other base lies on the base of the pyramid. A cone is inscribed in a prism if its vertex lies on the upper base of the prism, and its base is a circle inscribed in a polygon - the lower base of the prism. A prism is inscribed in a cone if all the vertices of the upper base of the prism lie on the lateral surface of the cone, and the lower base of the prism lies on the base of the cone.

Problem 1 In a regular quadrangular pyramid, the side of the base is equal to a, and the plane angle at the apex is equal to α. Find the radius of the ball inscribed in the pyramid. Solution: Let's express the sides of SOK in terms of a and α. OK = a/2. SK = KC cot(α /2); SK = (a · ctg(α /2))/2. SO = = (a/2) Using the formula r 1 = rh/(+ r), we find the radius of the inscribed ball: r 1 = OK · SO/(SK + OK); r 1 = (a/2) · (a/2) /((a/2) · ctg(α /2) + (a/2)) = = (a/2) /(ctg(α /2) + 1) = (a/2)= = (a/2) Answer: r 1 = (a/2)

Conclusion The topic “Polyhedra” is studied by students in grades 10 and 11, but in curriculum there is very little material on the topic “Inscribed and circumscribed polyhedra”, although it is very big interest students, since studying the properties of polyhedra contributes to the development of abstract and logical thinking, which will later be useful to us in study, work, life. While working on this essay, we studied all the theoretical material on the topic “Inscribed and circumscribed polyhedra,” examined possible combinations of figures and learned to apply all the studied material in practice. Problems involving the combination of bodies are the most difficult question in the 11th grade stereometry course. But now we can say with confidence that we will not have problems solving such problems, since during our research work we established and proved the properties of inscribed and circumscribed polyhedra. Very often, students have difficulties when constructing a drawing for a problem in this topic. But, having learned that to solve problems involving the combination of a ball with a polyhedron, the image of the ball is sometimes unnecessary and it is enough to indicate its center and radius, we can be sure that we will not have these difficulties. Thanks to this essay, we were able to understand this difficult but very fascinating topic. We hope that now we will not have any difficulties in applying the studied material in practice.

Polyhedra inscribed in a sphere A polyhedron is said to be inscribed in a sphere if all its vertices belong to this sphere. The sphere itself is said to be circumscribed about the polyhedron. Theorem. A sphere can be described around a pyramid if and only if a circle can be described around the base of this pyramid.

Polyhedra inscribed in a sphere Theorem. A sphere can be described near a straight prism if and only if a circle can be described near the base of this prism. Its center will be point O, which is the midpoint of the segment connecting the centers of the circles described near the bases of the prism. The radius of the sphere R is calculated by the formula where h is the height of the prism, r is the radius of the circle circumscribed around the base of the prism.

Exercise 3 The base of the pyramid is a regular triangle, the side of which is equal to 3. One of the side edges is equal to 2 and is perpendicular to the plane of the base. Find the radius of the circumscribed sphere. Solution. Let O be the center of the circumscribed sphere, Q the center of the circumscribed circle around the base, E the midpoint of SC. Quadrilateral CEOQ is a rectangle in which CE = 1, CQ = Therefore, R=OC=2. Answer: R = 2.

Exercise 4 The figure shows the pyramid SABC, for which the edge SC is equal to 2 and is perpendicular to the plane of the base ABC, the angle ACB is equal to 90 o, AC = BC = 1. Construct the center of the sphere circumscribed about this pyramid and find its radius. Solution. Through the middle D of edge AB we draw a line parallel to SC. Through the middle E of edge SC we draw a straight line parallel to CD. Their intersection point O will be the desired center of the circumscribed sphere. In the right triangle OCD we have: OD = CD = By the Pythagorean theorem, we find

Exercise 5 Find the radius of a sphere circumscribed about a regular triangular pyramid, the side edges of which are equal to 1, and the plane angles at the apex are equal to 90 degrees. Solution. In the tetrahedron SABC we have: AB = AE = SE = In the right triangle OAE we have: Solving this equation for R, we find

Exercise 4 Find the radius of a sphere circumscribed about a right triangular prism, at the base of which right triangle with legs equal to 1 and the height of the prism equal to 2. Answer: Solution. The radius of the sphere is equal to half the diagonal A 1 C of the rectangle ACC 1 A 1. We have: AA 1 = 2, AC = Therefore, R =

Exercise Find the radius of a sphere circumscribed about a regular 6-gonal pyramid, the edges of which are equal to 1, and the side edges are equal to 2. Solution. Triangle SAD is equilateral with side 2. The radius R of the circumscribed sphere is equal to the radius of the circle circumscribed about triangle SAD. Hence,

Exercise Find the radius of the sphere circumscribed about the unit icosahedron. Solution. In rectangle ABCD, AB = CD = 1, BC and AD are the diagonals of regular pentagons with sides 1. Therefore, BC = AD = By the Pythagorean theorem, AC = The required radius is equal to half of this diagonal, i.e.

Exercise Find the radius of a sphere circumscribed about a unit dodecahedron. Solution. ABCDE is a regular pentagon with side In the rectangle ACGF AF = CG = 1, AC and FG are the diagonals of the pentagon ABCDE and, therefore, AC = FG = By the Pythagorean theorem FC = The required radius is equal to half of this diagonal, i.e.

Exercise The figure shows a truncated tetrahedron obtained by cutting off the corners of a regular tetrahedron of triangular pyramids, the faces of which are regular hexagons and triangles. Find the radius of a sphere circumscribed about a truncated tetrahedron whose edges are equal to 1.

Exercise The figure shows a truncated octahedron obtained by cutting off triangular pyramids from the corners of the octahedron, the faces of which are regular hexagons and triangles. Find the radius of the sphere circumscribed about a truncated octahedron whose edges are equal to 1. Exercise The figure shows a truncated icosahedron obtained by cutting off the corners of the icosahedron of pentagonal pyramids, the faces of which are regular hexagons and pentagons. Find the radius of a sphere circumscribed about a truncated icosahedron whose edges are equal to 1.
Exercise The figure shows a truncated dodecahedron obtained by cutting off triangular pyramids from the corners of the dodecahedron, the faces of which are regular decagons and triangles. Find the radius of a sphere circumscribed about a truncated dodecahedron whose edges are equal to 1.
Exercise Find the radius of a sphere circumscribed about a unit cuboctahedron. Solution. Recall that a cuboctahedron is obtained from a cube by cutting off regular triangular pyramids with vertices at the vertices of the cube and side edges equal to half the edge of the cube. If the edge of the octahedron is equal to 1, then the edge of the corresponding cube is equal to The radius of the circumscribed sphere is equal to the distance from the center of the cube to the middle of its edge, i.e. is equal to 1. Answer: R = 1.

Mathematic teacher high school №2,

city ​​of Taldykorgan N.Yu.Lozovich

Public lesson in geometry

Lesson topic: “Ball. InscribedAnddescribed polyhedra"

Lesson objectives:

- educational - ensure during the lesson repetition, consolidation and testing of students’ mastery of definitions ball And spheres, and related concepts ( center, radii, diameters,diametrically opposed points, tangent planes And straight); concepts of inscribed and circumscribed polyhedra, knowledge of theorems on the section of a ball by a plane (20.3), on the symmetry of a ball (20.4), on the tangent plane to a ball (20.5), on the intersection of two spheres (20.6), on the construction of the center of a circumscribed (inscribed) sphere a pyramid and the construction of the center of a sphere described around a regular prism;

continue to develop the skills to independently apply the entire body of this knowledge in variable situations based on the model and non-standard ones that require creative activity;

educational - to instill in students responsibility for the results of their studies, perseverance in achieving goals, self-confidence, the desire to achieve great results, a sense of beauty (the beauty of geometric shapes, an elegant, beautiful solution to a problem).

developing - develop in students: the ability for specific and generalized thinking, creative and spatial imagination; associativity (the ability to rely on different connections: by similarity, analogy, contrast, cause-and-effect), the ability to logically and consistently express one’s thoughts, the need for learning and development, to create conditions in the lesson for the manifestation of students’ cognitive activity.

Lesson type

lesson of testing and correction of knowledge and skills.

Teaching methods

Introductory conversation (setting the purpose of the lesson, motivating students' learning activities, creating the necessary emotional and moral atmosphere, instructing students on organizing work in the lesson).

Frontal survey (oral testing of students’ knowledge of basic concepts, theorems, abilities to explain their essence, and to justify their reasoning).

Leveled independent work, based on the principle of a gradual increase in the level of knowledge and skills, i.e. from the reproductive level to the productive and creative level. The essence of the method is the individual independent work of students, constantly controlled and encouraged by the teacher.

Educational visual aids

Stereometric models geometric bodies, posters, drawings, flashcards for individual independent work.

Update

a) Basic knowledge.

It is necessary to activate the concepts: tangent to a circle, convex polygons inscribed in a circle and circumscribed about a circle, calculation of the radii of inscribed and circumscribed circles for regular polygons from planimetry; from the 10th grade course, the definition of symmetry relative to a plane, the concept of figures that are symmetrical with respect to a point, an axis (straight line), and a plane.

b) Ways to form motives and arouse interest.

In introductory conversation ensure that students understand the goal, recognize their personal interest in achieving it, reveal the meaning of the goal for the students themselves, emphasize the significance of this topic not only in itself, but also its propaedeutic nature for studying the next topic, saturate the lesson with material of an emotional nature (the beauty of geometric shapes , soap bubbles, Earth and Moon); emphasize the level nature of independent work: on the one hand, this will ensure a high scientific level of the material being studied, and on the other hand, accessibility, the students’ point is that each of them has the right to pedagogical support (“insurance”) for identifying, analyzing real or potential problems of the child, joint design of a possible way out of them; rating system assessment of knowledge is an additional incentive for the children.

c) Forms of monitoring the progress of work, mutual control. Mutual control (exchange of notebooks) is carried out after students have completed the first part of the 1st (student) level of independent work - students’ written answers to the teacher’s oral questions (mathematical dictation).

After exchanging notebooks, all correct answers are spoken out loud (if possible, visual aids are used: models of stereometric bodies, drawings, posters). Then the guys proceed to the rating assessment of the first part of the independent work: the correct complete answer is scored 1 point, if there are minor comments, then - 0.5 points, otherwise - 0 points. The number of points scored by each student is recorded on the board by the teacher. After which the guys begin to work on individual cards. Those who have completed the tasks of the 1st level and received the go-ahead from the teacher move on to completing the task of the next level. The success of solving the Problem should not be left without attention, encouragement, and praise. At the same time, the teacher carries out correctional work: understanding the student’s strengths and weaknesses, he helps him rely on his own strengths and complements him where the student, no matter how hard he tries, is still objectively unable to cope with something.

When checking operation, the following notation system is used:

The problem is not solved;

The problem is not solved, but there are some reasonable considerations in the work;

Only the answer is given to a problem where one answer is clearly not enough;

± - the problem is solved, but the solution contains minor omissions and inaccuracies;

The problem is completely solved;

+! – the solution to the problem contains unexpected bright ideas.

Great importance attached to a sheet of open accounting of the children’s activities, which is filled out as independent work is completed.

This ensures the indispensable conditions for assessing students' knowledge in the classroom - objectivity, efficiency, goodwill and transparency.

I level

Mathematical dictation.

1) I option. What property do all the vertices of a polyhedron inscribed in a sphere have?

II option. What property does each face of a polyhedron inscribed in a sphere have?

2)I option. If a sphere can be described around some polyhedron, then how can one construct its center?

II option. ABOUT How many parallelepipeds can be used to describe a sphere? Explain your answer.

3) I option. Where lies the center of the sphere described about the correct P- carbon prism?

II option. Where is the center of the sphere described around a regular pyramid?

4)I option. How to construct the center of a sphere inscribed in a regular n-gonal pyramid?

// option. Is it possible to fit a sphere into any regular prism?

Option I

I level

The radius of the ball is 6 cm; a plane is drawn through the end of the radius at an angle of 60° to it. Find the cross-sectional area.

Level II

A regular quadrangular prism is inscribed in a sphere of radius 5 cm. The edge of the base of the prism is 4 cm. Find the height of the prism.

Level III

Calculate the radius of a sphere inscribed in a regular tetrahedron with an edge of 4 cm.

IV level

A ball of radius R is inscribed in a truncated cone. The angle of inclination of the generatrix to the plane of the lower base of the cone is equal to A. Find the radii of the bases and the generatrix of the truncated cone.

Option II

I level

A sphere whose radius is 10 cm is intersected by a plane at a distance of 6 cm from the center. Find the cross-sectional area.

Level II

Find the radius of a sphere circumscribed about a cube with a side of 4 cm.

Level III.

A. Find the radius of the circumscribed sphere.

IV level

A ball of radius R is inscribed in a truncated cone. The angle of inclination of the generatrix to the plane of the lower base of the cone is equal to a. Find the radii of the bases and the generatrix of the truncated cone.

Ш option

I level

A plane perpendicular to it is drawn through the middle of the radius of the ball. How does the area of ​​the great circle relate to the area of ​​the resulting cross-section?

Level II

A regular triangular prism is inscribed in a sphere of radius 4 cm. The edge of the base of the prism is 3 cm. Find the height of the prism.

Level III

In a regular quadrangular pyramid, the side of the base is 4 cm, and the plane angle at the apex is A. Find the radius of the inscribed sphere.

IV level

A regular triangular pyramid with plane corners is inscribed in a ball of radius R A at its top. Find the height of the pyramid.

IV option

I level

Three points are given on the surface of the ball. The straight-line distances between them are 6 cm, 8 cm, 10 cm. The radius of the ball is 11 cm. Find the distance from the center of the ball to the plane passing through these points.

II level

A regular hexagonal prism is inscribed in a sphere of radius 5 cm. The edge of the base of the prism is 3 cm. Find the height of the technique.

Ш level

Find the radius of a sphere circumscribed about a regular n-gonal pyramid if the side of the base is 4 cm and the side edge is inclined to the plane of the base at an angle A.

IV level

A regular triangular pyramid with flat angles a at its vertex is inscribed into a ball of radius R. Find the height of the pyramid.

Lesson summary

The results of independent work are announced and analyzed. Students who need correctional work, are invited to correction lessons.

Set homework(with the necessary comments), consisting of mandatory and variable parts.

Mandatory part: paragraphs 187 - 193 - repeat; No. 44,45,39

Variable part No. 35