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Sphere and ball. Municipal educational institution secondary school No. 256, Fokino. A sphere is a surface that consists of all points in space located at a given distance from a given point. This point is called the center, and the given distance is the radius of the sphere, or ball - a body bounded by a sphere. A ball consists of all points in space located at a distance no more than a given point from a given point. A sphere is a surface that consists of all points in space located at a given distance from a given point. This point is called the center, and the given distance is the radius of the sphere, or ball - a body bounded by a sphere. A ball consists of all points in space located at a distance no more than a given point from a given point. The segment connecting the center of the ball with a point on its surface is called the radius of the ball. A segment connecting two points on the surface of a ball and passing through the center is called the diameter of the ball, and the ends of this segment are called diametrically opposite points of the ball. The segment connecting the center of the ball with a point on its surface is called the radius of the ball. A segment connecting two points on the surface of a ball and passing through the center is called the diameter of the ball, and the ends of this segment are called diametrically opposite points of the ball. What is the distance between diametrically opposite points of the ball if the distance of the point lying on the surface of the ball from the center is known? What is the distance between diametrically opposite points of the ball if the distance of the point lying on the surface of the ball from the center is known?

A ball can be considered as a body obtained by rotating a semicircle around a diameter as an axis. A ball can be considered as a body obtained by rotating a semicircle around a diameter as an axis. Let the area of ​​the semicircle be known. Find the radius of the ball, which is obtained by rotating this semicircle around the diameter. Let the area of ​​the semicircle be known. Find the radius of the ball, which is obtained by rotating this semicircle around the diameter.

Theorem. Any section of a ball by a plane is a circle. A perpendicular dropped from the center of the ball onto a cutting plane ends up in the center of this circle. Given: Prove: Proof: Consider a right triangle whose vertices are the center of the ball, the base of a perpendicular dropped from the center onto the plane, and an arbitrary section point. Consequence. If the radius of the ball and the distance from the center of the ball to the section plane are known, then the radius of the section is calculated using the Pythagorean theorem. Let the diameter of the ball and the distance from the center of the ball to the cutting plane be known. Find the radius of the circle of the resulting section. Let the diameter of the ball and the distance from the center of the ball to the cutting plane be known. Find the radius of the circle of the resulting section.

The smaller the distance from the center of the ball to the plane, the larger the radius of the section. A ball of radius five has a diameter and two sections perpendicular to this diameter. One of the sections is located at a distance of three from the center of the ball, and the second is at the same distance from the nearest end of the diameter. Mark the section whose radius is larger. A ball of radius five has a diameter and two sections perpendicular to this diameter. One of the sections is located at a distance of three from the center of the ball, and the second is at the same distance from the nearest end of the diameter. Mark the section whose radius is larger.

Task. On a sphere of radius R three points are taken that are the vertices of a regular triangle with side A. At what distance from the center of the sphere is the plane passing through these three points?

Given:

Find:

The largest radius of the section is obtained when the plane passes through the center of the ball. The circle obtained in this case is called a great circle. A large circle divides the ball into two hemispheres. The largest radius of the section is obtained when the plane passes through the center of the ball. The circle obtained in this case is called a great circle. A large circle divides the ball into two hemispheres. In a ball whose radius is known, two large circles are drawn. What is the length of their common segment? In a ball whose radius is known, two large circles are drawn. What is the length of their common segment?

A plane and a line, tangent to a sphere. A plane that has only one common point with a sphere is called a tangent plane. The tangent plane is perpendicular to the radius drawn to the point of tangency. Let a ball whose radius is known lie on a horizontal plane. In this plane through the point of contact and the point IN a segment is drawn, the length of which is known. What is the distance from the center of the ball to the opposite end of the segment? Let a ball whose radius is known lie on a horizontal plane. In this plane through the point of contact and the point IN a segment is drawn, the length of which is known. What is the distance from the center of the ball to the opposite end of the segment?

A straight line is called tangent if it has exactly one common point with the sphere. Such a straight line is perpendicular to the radius drawn to the point of contact. An infinite number of tangent lines can be drawn through any point on the sphere. A straight line is called tangent if it has exactly one common point with the sphere. Such a straight line is perpendicular to the radius drawn to the point of contact. An infinite number of tangent lines can be drawn through any point on the sphere. Given a ball whose radius is known. A point is taken outside the ball and a tangent to the ball is drawn through it. The length of the tangent segment from a point outside the ball to the point of contact is also known. How far from the center of the ball is the outer point? Given a ball whose radius is known. A point is taken outside the ball and a tangent to the ball is drawn through it. The length of the tangent segment from a point outside the ball to the point of contact is also known. How far from the center of the ball is the outer point?

The sides of the triangle are 13cm, 14cm and 15cm. Find the distance from the plane of the triangle to the center of the ball touching the sides of the triangle. The radius of the ball is 5 cm. The sides of the triangle are 13 cm, 14 cm and 15 cm. Find the distance from the plane of the triangle to the center of the ball touching the sides of the triangle. The radius of the sphere is 5 cm.

Given:

Find:

Through a point on a sphere whose radius is given, a great circle and a section are drawn intersecting the plane of the great circle at an angle of sixty degrees. Find the cross-sectional area. Through a point on a sphere whose radius is given, a great circle and a section are drawn intersecting the plane of the great circle at an angle of sixty degrees. Find the cross-sectional area.

The relative position of two balls. If two balls or spheres have only one common point, then they are said to touch. Their common tangent plane is perpendicular to the line of centers (the straight line connecting the centers of both balls). The contact of the balls can be internal or external. The contact of the balls can be internal or external. The distance between the centers of two touching balls is five, and the radius of one of the balls is three. Find the values ​​that the radius of the second ball can take. The distance between the centers of two touching balls is five, and the radius of one of the balls is three. Find the values ​​that the radius of the second ball can take.

Two spheres intersect in a circle. The line of centers is perpendicular to the plane of this circle and passes through its center. Two spheres intersect in a circle. The line of centers is perpendicular to the plane of this circle and passes through its center. Two spheres of the same radius, equal to five, intersect, and their centers are at a distance of eight. Find the radius of the circle along which the spheres intersect. To do this, it is necessary to consider the section passing through the centers of the spheres. Two spheres of the same radius, equal to five, intersect, and their centers are at a distance of eight. Find the radius of the circle along which the spheres intersect. To do this, it is necessary to consider the section passing through the centers of the spheres.

Inscribed and circumscribed spheres. A sphere (ball) is said to be circumscribed about a polyhedron if all the vertices of the polyhedron lie on the sphere. What quadrilateral can lie at the base of a pyramid inscribed in a sphere? What quadrilateral can lie at the base of a pyramid inscribed in a sphere?

A sphere is said to be inscribed in a polyhedron, in particular, in a pyramid, if it touches all the faces of this polyhedron (pyramid). A sphere is said to be inscribed in a polyhedron, in particular, in a pyramid, if it touches all the faces of this polyhedron (pyramid). At the base of a triangular pyramid lies an isosceles triangle, the base and sides are known. All lateral edges of the pyramid are equal to 13. Find the radii of the circumscribed and inscribed spheres. At the base of a triangular pyramid lies an isosceles triangle, the base and sides are known. All lateral edges of the pyramid are equal to 13. Find the radii of the circumscribed and inscribed spheres.

Given:

Find:

Stage II. Finding the radius of an inscribed sphere.

The second way to calculate the radius of an inscribed sphere is based on the fact that the center of a ball inscribed in a dihedral angle is equidistant from its sides, and therefore lies on the bisector plane. The second way to calculate the radius of an inscribed sphere is based on the fact that the center of a ball inscribed in a dihedral angle is equidistant from its sides, and therefore lies on the bisector plane. The side of the base of a regular quadrangular pyramid is 6, and the angle between the base and the side face is 600. Determine the radius of the inscribed sphere.

Given:

Find:

• The segment connecting the center of the sphere with the middle of the side of the base bisects the dihedral angle at the base.

Sphere
Lesson-lecture on the topic:
Geometry – 11th grade
5klass.net

Presentation plan
Definition of sphere, ball. Equation of a sphere. . Area of ​​a sphere. Lesson summary.
Def.environment

Circle and Circle
The part of the plane bounded by a circle is called a circle.
A circle is a geometric figure consisting of all points of the plane located at a given distance r from a given point.
r – radius;
d – diameter
Def. spheres

Definition of sphere
A sphere is a surface consisting of all points in space located at a given distance (R) from a given point (the center of the point.O).
A sphere is a body obtained as a result of the rotation of a semicircle around its diameter.
t. O – center of the sphere
ABOUT
D – sphere diameter – a segment connecting any 2 points of the sphere and passing through the center.
D=2R
ball
R – radius of the sphere – a segment connecting any point of the sphere with the center.

Ball
A body bounded by a sphere is called a ball. The center, radius and diameter of a sphere are also the center, radius and diameter of a sphere. A ball of radius R and center O contains all points in space that are located from point O at a distance not exceeding R.

Historical information about the sphere and ball
Both the words "ball" and "sphere" come from the Greek word "sphaira" - ball. In ancient times, the sphere and ball were held in high esteem. Astronomical observations over the firmament evoked the image of a sphere. The Pythagoreans, in their semi-mystical reasoning, argued that spherical celestial bodies are located from each other at a distance proportional to the intervals of the musical scale. This was seen as elements of world harmony. This is where the expression “music of the sphere” comes from. Aristotle believed that the spherical shape, as the most perfect, is characteristic of the Sun, Earth, Moon and all world bodies. He also believed that the Earth was surrounded by a number of concentric spheres. The sphere and ball have always been widely used in various fields of science and technology.

d/z approx.

How to draw a sphere?
R
1. Mark the center of the sphere (TO)
2. Draw a circle with center at t.O
3. Draw a visible vertical arc (meridian)
4. Draw an invisible vertical arc
5. Draw a visible horizontal arc (parallel)
6. Draw an invisible horizontal arc
7. Draw the radius of the sphere R
ABOUT
ur. env.

Equation of a circle
C(x0;y0)
M(x;y)
X
at
ABOUT
therefore the equation of a circle has the form: (x – x0)2 + (y – y0)2 = r2
Let's define a rectangular coordinate system Oxy
Let's construct a circle with center at point C and radius r
The distance from an arbitrary point M (x;y) to point C is calculated by the formula:
MC = (x – x0)2 + (y – y0)2
MC = r, or MC2 = r2

Task 1. Knowing the coordinates of the center C(2;-3;0), and the radius of the sphere R=5, write down the equation of the sphere.
The solution is as follows: the equation of a sphere with radius R and center at point C(x0;y0;z0) has the form (x-x0)2 + (y-y0)2 + (z-z0)2=R2, and the coordinates of the center of this sphere C(2;-3;0) and radius R=5, then the equation of this sphere is (x-2)2 + (y+3)2 + z2=25 Answer: (x-2)2 + (y+3 )2 + z2=25
ur. spheres

Sphere equation
(x – x0)2 + (y – y0)2 + (z – z0)2 = R2
X
at
z
M(x;y;z)
R
Let's define a rectangular coordinate system Оxyz
Let's construct a sphere with center at point C and radius R
MC = (x – x0)2 + (y – y0)2 + (z – z0)2
MC = R, or MC2 = R2
C(x0;y0;z0)
Therefore, the equation of the sphere has the form:

The relative position of a circle and a straight line
r
d
If d d= r
d>r
If d = r, then the straight line and the circle have 1 common point.
If d > r, then the line and the circle have no common points.
There are 3 possible cases
Sphere and plane

The relative position of the sphere and the plane
Depending on the ratio of d and R, 3 cases are possible...
Let us introduce the rectangular coordinate system Oxyz
Let us construct a plane α coinciding with the Oxy plane
Let us depict a sphere with a center at t.C, lying on the positive semi-axis Oz and having coordinates (0;0;d), where d is the distance (perpendicular) from the center of the sphere to the plane α.

The section of a sphere by a plane is a circle.
r
The relative position of the sphere and the plane
Let's consider 1 case
d r = R2 - d2
M
As the cutting plane approaches the center of the ball, the radius of the circle increases. The plane passing through the diameter of the ball is called diametrical. The circle obtained as a result of the section is called a great circle.

d = R, i.e. if the distance from the center of the sphere to the plane is equal to the radius of the sphere, then the sphere and plane have one common point
The relative position of the sphere and the plane
Let's consider case 2

d > R, i.e. if the distance from the center of the sphere to the plane is greater than the radius of the sphere, then the sphere and the plane do not have common points.
The relative position of the sphere and the plane
Let's consider case 3

Problem 2. A ball with a radius of 41 dm is intersected by a plane located at a distance of 9 dm from the center. Find the radius of the section.
Given: Sphere with center at point O R=41 dm α - cutting plane d = 9 dm
Find: rsec = ?
Solution: Consider ∆OMK – rectangular OM = 41 dm; OK = 9 dm; MK = r, r = R2 - d2 according to the Pythagorean theorem: MK2 = r2 = 412- 92 = 1681 - 81 = 1600 hence rsec = 40 dm
Answer: rsec = 40 dm
r

Area of ​​a sphere
Area of ​​a sphere of radius R: Sсф=4πR2
The sphere cannot be turned onto a plane.
Let us describe a polyhedron around the sphere, so that the sphere touches all its faces.
The area of ​​a sphere is taken to be the limit of the sequence of surface areas of polyhedra described around the sphere as the largest size of each face tends to zero
i.e.: The surface area of ​​the ball is equal to four times the area of ​​the larger circle
Sballs=4 Scircles

Problem 3. Find the surface area of ​​a sphere whose radius = 6 cm.
Given: sphere R = 6 cm Find: Ssf =?
Solution: Sсф = 4πR2 Sсф = 4π 62 = 144π cm2 Answer: Sсф = 144π cm2

Lesson summary
definition of sphere, ball; equation of the sphere; the relative position of the sphere and the plane; surface area of ​​the sphere.
Today you met:

In the course of studying stereometry - one of the main sections in geometry, studying figures in space, attention is paid to the consideration of such bodies as a sphere and a ball. Definitions and main characteristics of these stereometric bodies are given in the presentation. With its help you can create a structured lesson for 10th grade students.

Before moving on to studying the sphere and ball itself, it is proposed to remember what a circle and a circle are. A large number of lessons were previously devoted to the study of these figures on a plane, during which the basic formulas, concepts and properties of these figures were discussed. A large number of problems and interesting examples were solved.

If we take a certain point in space, then the set of all equidistant points will form a figure called a sphere. On the second slide of the presentation, after demonstrating the definitions of a circle and a circle, an image of a sphere is shown. A theoretical definition is given that must be understood and remembered, and also be able to reproduce.

Each sphere has parameters such as radius, diameter, center, etc. The diameter, as in the case of a circle and a circle, is twice the product of the radius.

They are designated by analogy with the designations for a circle, that is, through the Latin letters r and d. To imagine a sphere, you can look at a ball - it represents a geometric figure.

What is a ball? The definition of this body is given a separate slide, which provides the definition and some justifications.

The center, radius and diameter of the ball coincide with the center, radius and diameter of the sphere to which it is bounded.

Is it possible to get a sphere as a result of rotation? Of course yes. This question can be asked to schoolchildren so that they have the opportunity to develop spatial thinking.

In order to obtain a sphere as a result of motion, it is necessary to take a semicircle and bring it into rotation around its diameter. This is demonstrated on slide 5.

The last slides of the presentation “Ball and Sphere” are devoted to the consideration of practical problems. Using these problems as an example, you can solve similar examples that can occur both in homework and in tests at school.

This presentation will be useful for both novice tutors or teachers and experienced professionals. By using a presentation, you can achieve a more effective result.