Cone. Frustum

Tapered surface called the surface formed by all straight lines passing through each point of the given curve and a point outside the curve (Fig. 32).

This curve is called guide , direct - generating , dot - summit conical surface.

Straight circular tapered surface called the surface formed by all lines passing through each point of the given circle and a point on the line that is perpendicular to the plane of the circle and passes through its center. In what follows, this surface will be briefly referred to as conical surface (fig.33).

cone (straight circular cone ) is called a geometric body bounded by a conical surface and a plane that is parallel to the plane of the guide circle (Fig. 34).

Rice. 32 Fig. 33 Fig. 34

A cone can be considered as a body obtained by rotating a right triangle around an axis containing one of the legs of the triangle.

The circle that bounds the cone is called basis . The vertex of a conical surface is called summit cone. The line segment connecting the top of a cone with the center of its base is called tall cone. The segments that form a conical surface are called generating cone. axis of a cone is a straight line passing through the vertex of the cone and the center of its base. Axial section called the section passing through the axis of the cone. Lateral surface development A cone is a sector whose radius is equal to the length of the generatrix of the cone, and the length of the arc of the sector is equal to the circumference of the base of the cone.

For a cone, the following formulas are true:

where R is the radius of the base;

H- height;

l- the length of the generatrix;

S main- base area;

S side

S full

V is the volume of the cone.

truncated cone called the part of the cone enclosed between the base and the cutting plane parallel to the base of the cone (Fig. 35).

A truncated cone can be considered as a body obtained by rotating a rectangular trapezoid around an axis containing the lateral side of the trapezoid, perpendicular to the bases.

The two circles that bound the cone are called its grounds . Height of a truncated cone is the distance between its bases. The segments that form the conical surface of a truncated cone are called generating . The straight line passing through the centers of the bases is called axis truncated cone. Axial section called the section passing through the axis of the truncated cone.

For a truncated cone, the following formulas are true:

(8)

where R is the radius of the lower base;

r is the radius of the upper base;

H is the height, l is the length of the generatrix;

S side is the lateral surface area;

S full is the total surface area;

V is the volume of the truncated cone.

Example 1 The section of the cone parallel to the base divides the height in a ratio of 1:3, counting from the top. Find the area of ​​the lateral surface of a truncated cone if the radius of the base and the height of the cone are 9 cm and 12 cm.

Decision. Let's make a drawing (Fig. 36).

To calculate the area of ​​the lateral surface of a truncated cone, we use formula (8). Find the radii of the bases About 1 A and About 1 V and generating AB.

Consider similar triangles SO 2 B and SO 1 A, coefficient of similarity , then

From here

Since then

The area of ​​the lateral surface of a truncated cone is equal to:

Answer: .

Example2. A quarter circle of radius is folded into a conical surface. Find the radius of the base and the height of the cone.

Decision. The quadruple of a circle is a development of the lateral surface of the cone. Denote r is the radius of its base, H- height. The lateral surface area is calculated by the formula: . It is equal to the area of ​​a quarter of a circle: . We get an equation with two unknowns r and l(generator of a cone). In this case, the generatrix is ​​equal to the radius of a quarter of a circle R, so we get the following equation: , whence Knowing the radius of the base and generatrix, we find the height of the cone:

Answer: 2 cm, .

Example 3 A rectangular trapezoid with an acute angle of 45 O, a smaller base of 3 cm and an inclined side equal to , rotates around the side perpendicular to the bases. Find the volume of the obtained body of revolution.

Decision. Let's make a drawing (Fig. 37).

As a result of rotation, we get a truncated cone; in order to find its volume, we calculate the radius of the larger base and the height. in a trapeze O 1 O 2 AB we will spend AC^O 1 B. In we have: so this triangle is isosceles AC=BC\u003d 3 cm.

Answer:

Example 4 A triangle with sides 13 cm, 37 cm and 40 cm rotates around an external axis that is parallel to the larger side and is 3 cm away from it (the axis is located in the plane of the triangle). Find the surface area of ​​the resulting body of revolution.

Decision . Let's make a drawing (Fig. 38).

The surface of the resulting body of revolution consists of the side surfaces of two truncated cones and the side surface of the cylinder. In order to calculate these areas, it is necessary to know the radii of the bases of the cones and the cylinder ( BE and OC) forming cones ( BC and AC) and the height of the cylinder ( AB). The unknown is only CO. is the distance from the side of the triangle to the axis of rotation. Let's find DC. The area of ​​triangle ABC on one side is equal to the product of half of side AB and the height drawn to it DC, on the other hand, knowing all the sides of the triangle, we calculate its area using Heron's formula.

When studying the material of the topic, you need to learn:

types of bodies of revolution;

definitions of bodies of revolution;

definitions of elements of bodies of revolution;

concepts of development of a cylinder and a cone;

definition and calculation of the lateral and full surface of the cylinder and cone;

definition of the tangent plane to the sphere and its properties;

the concept of the surface area of ​​a sphere;

definition of a polyhedron inscribed in a sphere and described around it.

In the process of solving problems, the following skills are tested:

depict bodies of revolution;

Calculate elements of bodies of revolution;

depict sections of bodies;

Calculate the area of ​​the lateral and full surface of the cylinder and cone;

Write an equation for a sphere.

Questions of the theoretical test

Option 1

1. The concept of a cylindrical surface and its elements. Formulate the definition of a cylinder and its elements.

2. Derive a formula for calculating the surface area of ​​a sphere.

3. Find the ratio of the lateral surface area and the axial section of the cone.

Option 2

1. The concept of a conical surface. Formulate the definition of a cone and its elements.

2. Determine the position of the center of the sphere circumscribed about a regular quadrangular pyramid. Prove your claim.

3. Find the ratio of the area of ​​the lateral surface and the axial section of the cylinder.

Option 3

1. Formulate the definition of a truncated cone and its elements.

2. Determine the position of the center of the sphere inscribed in a regular triangular pyramid. Prove your claim.

3. Prove that the total surface of an equilateral cone is equal to the surface of a ball with a diameter of the height of the cone.

Option 4

1. Formulate the definitions of a sphere and a ball. Write down the equations of a sphere of radius R centered at the point O(0; 0; 0) and at the point A(x0; y0; z0).

2. Derive a formula for calculating the lateral surface of a cone.

3. Prove that the area of ​​the full surface of a cylinder is equal to the area of ​​the lateral surface of another cylinder of the same radius, whose height is equal to the sum of the radius and the height of this cylinder.

Independent work 17

Option 1

1. The area of ​​the axial section of the cylinder is 16. Find the area of ​​the section of this cylinder, which is parallel to the axis and located at a distance from it equal to half the radius of the base of the cylinder.

2. The semicircle is folded into a conical surface. Find the angle between the generatrix and the height of the cone.

3. The radii of two balls are 16 and 20 dm, the distance between their centers is 25 dm. Find the circumference of the circle where their surfaces intersect.

Option 2

1. The radius of the base of the cylinder is 26 cm, forming 4.8 dm. At what distance from the axis of the cylinder should a section be drawn that is parallel to the axis and has the shape of a square?

2. The radius of the sector is 3 m, its angle is 120°. The sector is folded into a conical surface. Find the radius of the base of the cone.

3. The diagonals of the rhombus are 30 and 40 cm. The spherical surface touches all sides of the rhombus. Find the distance from the center of the sphere to the plane of the rhombus if the radius of the sphere is 13 cm.

Option 3

1. The radius of the base of the cylinder is 12 cm. Find the distance between the axial section and the section with half the area.

2. The angle of development of the side surface of the cone is 120°. The generatrix of the cone is 15 cm. Calculate the diameter of the base of the cone.

3. A rhombus is superimposed on a ball with a radius of 10 cm so that each side of it, equal to 12.5 cm, touches the ball. The plane of the rhombus is 8 cm away from the center of the ball. Find the area of ​​the rhombus.

Option 4

1. Two mutually perpendicular sections are drawn through the generatrix of the cylinder, the areas of which are equal to 60 and 80 dm. Find the area of ​​the axial section.

2. The radius of the base of the cone is 12 cm, forming 40 cm. Calculate the sweep angle of this cone.

3. The sides of the triangle are 10 dm, 10 dm and 12 dm. Find the distance from the plane of the triangle to the center of the ball tangent to the sides of the triangle. The radius of the ball is 5 dm.

Independent work 18

Option 1

1. The diagonal of the axial section of the cylinder is 25% greater than the diameter of its base. Find the total area of ​​the cylinder if the distance between its centers is 15 cm.

2. Development of the side surface of the cylinder - a square with a side of 4 dm. Find the volume of the cylinder.

3. The diagonals of the axial section of the truncated cone are mutually perpendicular, the height of the cone is H, forming l. Find the lateral surface of the cone.

4. The radius of the base of the cone is 12 cm, forming 40 cm. Find the angle of development of the side surface of the cone.

5. Generator of a truncated cone 10 cm, base difference 6 cm, axial section area 112 cm2. Find the lateral surface of the cone.

6. A parallelogram whose sides are 21 cm and 89 cm and whose diagonal is 100 cm revolves around the smaller side. Find the volume of the body of revolution.

7. A right triangle with legs 16 and 12 cm revolves around the hypotenuse. Find the volume and area of ​​rotation.

Option 2

1. The side surface of the cylinder is half of its total surface. Find the total surface of the cylinder if the diagonal of the axial section is 10 in.

2. The total surface of the cylinder is 500 p cm2, the diameter of its base is 20 cm. Find the volume of the cylinder.

3. The generatrix of a truncated cone refers to its height as 41:40. The base radii are 24 and 6 cm. Find the lateral surface of the cone.

4. The angle of development of the side surface of the cone is 120°. The generatrix of the cone is 15 cm. Find the total surface of the cone.

5. Find the height of a truncated cone if its lateral surface is equal to the sum of the areas of the bases, and the radii of the bases are R and r.

6. An isosceles trapezoid with bases of 12 and 18 cm and an acute angle of 60 ° rotates around a smaller base. Find the surface and volume of the body of revolution.

7. A triangle with two sides equal to 5 cm and 8 cm, make an angle of 60 °, rotates around the largest side. Find the surface and volume of the body of revolution.

Independent work 19

Option 1

1. A right triangle with legs 16 and 12 cm revolves around the hypotenuse. Find the surface of the body of revolution.

2. The radii of the bases of the spherical belt are 63 and 39 cm, its height is 36 cm. Find the surface of the spherical belt.

3. The height of a regular triangular pyramid h. Lateral ribs are mutually perpendicular. Find the radius of the circumscribed sphere.

4. In a regular triangular truncated pyramid, the height is 17 cm, the radii of the circles described around the bases are 5 and 12 cm. Find the radius of the circumscribed ball.

5. A square with a side equal to a rotates around a perpendicular to the diagonal, drawn through its end. Find the surface of the resulting body.

Option 2

1. A triangle whose two sides are 5 and 8 cm, make an angle of 60 °, rotates around the largest side. Find the surface of the body of revolution.

2. The total surface of the spherical segment is equal to S. Determine the height of the segment if the radius of the ball is R.

3. The base of the pyramid is a regular triangle, the side of which is 3 dm. One of the side edges is 2 dm and perpendicular to the base. Find the radius of the circumscribed sphere.

4. The sides of the bases of a regular quadrangular truncated pyramid are 7 and 1 dm. The side edge is inclined to the base at an angle of 45°. Find the radius of the circumscribed sphere.

5. A regular hexagon with side a rotates around the external axis, which is parallel to the side and spaced from it by the length of the apothem. Find the surface of the resulting body.

Independent work 20

Option 1

1. The lateral edge of a regular triangular pyramid is equal to b and forms an angle a with the base plane. An equilateral cylinder is inscribed in a pyramid so that the plane of the base lies in the plane of the base of the pyramid. Find the volume of the cylinder.

2. The base of the pyramid is a regular triangle. One side edge is perpendicular to the base plane and is equal to l, and the other two form an angle a with the base plane. A straight prism is inscribed in the pyramid, three vertices of which lie on the lateral edges of the pyramid, and the other three lie on the base of the pyramid, the diagonal of the lateral face of the prism is with the plane of the base Ð b. Find the height of the prism.

3. In a regular quadrangular prism, the area of ​​the side face is equal to q. Find the area of ​​the diagonal section.

4. A plane perpendicular to the diameter of the ball divides it into parts of 3 and 9 cm. What parts is the volume of the ball divided into?

Option 2

1. The angle at the top of the axial section of the cone is 2b. The circumference of the base is c. Determine the area of ​​the lateral surface of the cone.

2. The diagonals of the axial section of a truncated cone are divided by the intersection point in a ratio of 2: 1, counting from the large base. The angle between the diagonals facing the base is a. The diagonal is l. Find the volume of the cone.

3. The lateral edge of a right parallelepiped is 5 cm, the sides of the base are 6 and 8 cm, one of the diagonals of the base is 12 cm. Find the diagonals of the parallelepiped.

4. What part of the volume of the ball is the volume of a spherical segment with a height of 0.1 of the ball diameter?

Option 3

1. The generatrix of the cone is equal to l and is inclined to the plane of the base at an angle a. Determine the total surface area of ​​the inscribed cube.

2. A square is inscribed in the base of the cone, the side of which is a. The plane passing through one of the sides of this square and the vertex of the cone, when intersecting with the surface of the cone, forms an isosceles triangle with an angle at the vertex equal to a. Find the volume of the cone.

3. The side of the base of a regular quadrangular prism is 15 cm, and the height is 20 cm. Find the shortest distance from the side of the base to the diagonal of the prism that does not intersect it.

4. Two equal balls are arranged so that the center of one lies on the surface of the other. How is the volume of the total part of the balls related to the volume of the whole ball?

Option 4

1. A right triangular prism with equal ribs is inscribed in a cone, the generatrix of which is inclined to the plane of the base at an angle a. Find the volume of the prism if the radius of the base of the cone is R.

2. The volume of the cone is V. A pyramid is inscribed in the cone, at the base of which lies an isosceles triangle with an angle a between the sides. Find the volume of the pyramid.

3. In a right parallelepiped, the lateral edge is 1 m, the sides of the base are 23 dm and 11 dm, the diagonals of the base are 2: 3. Find the areas of the diagonal sections.

4. On the side of the base a and the side edge b, find the full surface of a regular hexagonal prism.

. Cone. Basic concepts.

Definition. cone called a geometric figure obtained by rotating a right triangle around one of its legs. The leg, relative to which the rotation occurs - axis cone, numerically equal to its height; second leg - radius grounds; hypotenuse - generatrix (forms the lateral surface of the cone during rotation).

M- top of the cone O- base center MO- the axis of the cone, MO = H is the height of the cone, OA = OV =…= R is the radius of the base, AM= BM =…= l is the generatrix of a cone. Axial section of the cone isosceles triangle (e.g. triangle AMB). The section of a cone by a plane parallel to the base is a circle similar to the base.

The development of the surface of the cone consists of a circle and a sector of the circle.

. Frustum.

Definition. truncated cone called a geometric figure obtained by rotating a rectangular trapezoid around its smaller side. In other words: a truncated cone is the part of the cone enclosed between the base and the section of the cone parallel to the base. Axial section isosceles trapezoid (for example, ABB 1 BUT 1 ) .

B 1 A 1

. Volume and surface area of ​​a cone.

	truncated

Here R is the radius of the bottom base, r is the radius of the top base, H- height, l- generating.

Questions and tasks

A bag is folded from paper, having the shape of a cone with a base radius of 5 cm and a height of 10 cm. Determine the surface area of ​​the bag.

The generatrix of the cone is 2 cm, and the radius of the base is 1 cm. Explain whether the area of ​​\u200b\u200bits total surface is more or less than 6 cm 2.

Find the total surface area of ​​the cone if:

a) the radius of its base is 2, and the generatrix is ​​4;

b) the radius of the base is 3, and the height is 4;

c) the radius of the base is 4, and the angle of inclination of the generatrix to the base is 30 0 .

Find the volume of the cone if:

a) its base radius is 2 and its height is 3;

b) the radius of its base is 3, and the generatrix is ​​5;

c) the radius of the base is equal to 2, and the generatrix is ​​inclined to the plane of the base at an angle of 30°;

d) the radius of the base is 3, and the area of ​​the axial section is 12.

a and b (a < b) rotates first around one of them, and then around the other. Compare:

a) the area of ​​the side surfaces of the resulting cones;

b) the areas of the total surfaces of the resulting cones.

An isosceles right triangle with legs of length 2 is rotated around the hypotenuse. Find the area of ​​the resulting surface.

A right triangle with legs 3 and 4 is rotated around the hypotenuse. Find the area of ​​the resulting surface.

A right triangle with legs 6 cm and 8 cm revolves around the smaller leg. Calculate the areas of the lateral and full surfaces of the cone formed during this rotation.

Right triangle with legs a and b revolve around the hypotenuse. Find the volume of the resulting body of revolution.

A parallelogram with sides of 6 cm and 8 cm and an angle of 60 0 is rotated around a straight line containing the larger side of the parallelogram. Find the area of ​​the resulting surface.

The angle between the generatrix and the axis of the cone is 45°, the generatrix is ​​6.5 cm. Find the area of ​​the lateral surface of the cone.

The area of ​​the axial section of the cone is 0.6 cm². The height of the cone is 1.2 cm. Calculate the total surface area of ​​the cone.

Find the volume of a cone if its base area is Q and its lateral surface area is P.

The height of a cone is equal to the diameter of its base. Find the volume of a cone if its height is H.

Find the volume of a cone if its generatrix is ​​13 cm and the area of ​​the axial section is 60 cm².

The radii of the bases of the truncated cone are 3 m and 6 m, and the generatrix is ​​5 m. Find the volume of the truncated cone.

A cone with a base radius of 5 cm and a generatrix of 3 cm is considered. A section parallel to the base of the cone is drawn through a point of the generatrix located at a distance of 1 cm from the top. Complete the following tasks in sequence:

a) find the area of ​​this section;

b) find the area of ​​the lateral surface of this cone;

c) find the area of ​​the lateral surface of the cone cut off by the drawn plane;

d) find the area of ​​the lateral surface of the truncated cone cut off by the drawn plane;

e) find the total surface area of ​​this truncated cone.

Find the generatrix of a truncated cone if the radii of the bases are 3 cm and 6 cm, and the height is 4 cm.

The area of ​​the base of the cone is 12 cm², its height is 6 cm. Find the area of ​​its section, parallel to the base and drawn:

a) through the middle height;

b) at a distance of 2 cm from the top of the cone;

c) at a distance of 4 cm from the top of the cone.

Find the volumes of cones whose bases are the considered sections and whose vertex is the vertex of the given cone.

The area of ​​the base of the cone is 25 cm², and the height is 5 cm. A section parallel to the base is drawn at a distance of 1 cm from the top. Find the volume of the truncated cone cut off by the drawn section.

The height of the cone is 5 cm. At a distance of 2 cm from the top, it is crossed by a plane parallel to the base. Find the volume of the original cone if the volume of the smaller cone cut off from the original one is 24 cm³.

In a truncated cone, the height is known h, forming l and area S side surface. Find the area of ​​the axial section and the volume of the truncated cone.

As is known; when a point rotates around an axis, it moves in a plane perpendicular to the axis of rotation and describes a circle. To apply the rotation method in order to transform the drawing, we note the following four elements (Fig. 5.8):

axis of rotation (MN);

point rotation plane(pl. S is perpendicular (MN));

center of rotation;

radius of rotation (R; R= |OA|).

As an axis of rotation, straight lines, perpendicular or parallel to the projection planes are usually used. Consider rotation about axes perpendicular to the projection planes.

Point A rotation on the drawing about the axis MN, perpendicular to the plane H, shown in figure 5.9. Plane of rotation S is parallel to the H plane and is depicted on the frontal projection as follows S v . Horizontal projection about the center of rotation about coincides with the projection tp axes, and the horizontal projection oa rotation radius OA is its natural value. point rotation BUT in figure 5.9 is made by an angle φ counterclockwise so that in the new position of the point with projections a1", a1 the radius of rotation was parallel to the planeV When a point rotates around the vertical axis, its horizontal projection moves along the circle, and the frontal projection moves parallel to the x-axis and perpendicular to the axis of rotation.

If the point is rotated around an axis perpendicular to the V plane, then its frontal projection will move along the circle, and the horizontal projection will move parallel to the x-axis.

The rotation of a point around a projecting line is used in solving some problems, for example, in determining the natural size of a line segment. For this (Fig. 5.10), an axis of rotation with projections is sufficient t "p", tp choose so that it passes through one of the extreme points of the segment, for example, a point with projections b", b. Then when turning the point BUT angle φ into position A1 (OA1 || square V, oa, || x-axis) segment AB moves to position A1B, parallel to the plane V and, therefore, is projected onto it in full size. At the same time, the angle a of the slope of the segment will be projected in full size AB to plane H.

Rotation (rotation) of a point with projections b", b relative to the axis with projections t"p", tp, perpendicular to the plane V, shown in Figure 5.11. When rotating the point AT moved in the plane of rotation T (Th) to position with projections b1", b1 so that the radius of rotation OV become parallel to the plane H (o "b" || x-axis).

Application of the rotation method without indicating on the drawing the axes of rotation perpendicular to the projection planes.If you rotate a geometric figure around an axis perpendicular to the projection plane, then the projection on this plane does not change either in appearance or in size (only the position of the projection relative to the projection axis changes). Projections of points of a geometric figure on a plane parallel to the axis of rotation move along straight lines parallel to the axis of projection (with the exception of projections of points located on the axis of rotation), and the projection as a whole changes in shape and size. Therefore, it is possible to apply the rotation method without specifying the representation of the rotation axis. In that

case, without changing the size and shape of one of the projections of the geometric image, move this projection to the required position, and then build another projection as indicated above.

Figure 5.12 shows the use of the rotation method without specifying the axes to determine the actual size of the triangle abc, given by projections a"b"c", abc. To do this, two rotations of the plane in general position, in which the triangle is located, are performed so that after the first rotation this plane becomes perpendicular to the plane V, and after the second - parallel to the plane H. The first rotation around the axis perpendicular to the plane H, without specifying its position, was carried out using a horizontal with projections s"1", s-1 in the plane of the triangle. In this case, the horizontal projection aс rotated to match the projection direction. The horizontal projection of the triangle retains its shape and size, only its position changes. points A, B and C with such a rotation, they move in planes parallel to the plane H. Projections a1", c1, b1" a"a1", b"b1" and c"c1". The frontal projection of the triangle in the new position is the segment a1"b1"c1".

The second rotation, bringing the triangle to a position parallel to the plane H, is made around the axis of rotation perpendicular to the plane H (the position of the axis is also not indicated). The frontal projection at the second rotation retains the appearance and size obtained after the first rotation. points A1, D1 and C1 move in planes parallel to the plane V Projections a 2 , b 2 , c 2 are on horizontal lines of communication a, a 2, blb2, c1c2. Projection a2b2c 2 is the actual size of the given triangle.

When performing the considered rotations around axes perpendicular to the projection planes, these axes are not indicated, but they can be easily found. For example, if you draw segments aa1, b1b2 and draw perpendiculars through their midpoints, then the resulting point of intersection of these perpendiculars will be the horizontal projection of the axis of rotation perpendicular to the plane H.

The use of the rotation method without specifying the axes somewhat simplifies the construction, there is no overlap of one

section on another, but the drawing occupies a large area. (The considered case of rotation without depicting the axes of rotation is a special case of the method of plane-parallel movement.)

A method of rotation around straight lines parallel to projection planes.The natural size of a flat figure can be determined by rotating around an axis parallel to the projection plane, bringing the figure to a position parallel to the projection plane with one turn.

Figure 5.13 shows the definition of the size of a triangle with projections a"b"c", abc rotation around the horizontal.In this case, all points of the triangle(with the exception of those lying on the axis of rotation)rotate around an axis in circles in planes perpendicular to the axis.If the triangle takes a position parallel to the plane of projections, the radii of rotation of its points will be parallel to this plane, that is, they will be projected onto the plane H real size.

The horizontal with projections was taken as the axis of rotation s"1", s-1.

Point C on the axis of rotation remains fixed. To image the horizontal projection of the triangle after the rotation, it is necessary to find the position of the projections of its other two vertices. Vertices with projections a", a and b", b displacement triangle-

are in planes P and Q movement of these points. Horizontal projection about vertex rotation center BUT is the point of intersection of the horizontal projection s-1 axes of rotation with horizontal projection Ph.h. Its frontal projection is marked on it. o. Segments oa - horizontal, o "a" - frontal projection of the radius of rotation of the point BUT. life size oA point rotation radius BUT defined in the manner discussed in 2.3 (see Fig. 2.9), i.e., by constructing a right triangle. On the legs oa and aA \u003d o "2" a triangle is built oaa, its hypotenuse is equal to the radius of rotation of the point BUT.

From a projection about pivot point BUT in the direction of the trace Ph of the plane of its motion, we set aside the natural value of the radius of rotation. Marking the horizontal projection a, points A, rotated to the position of a triangle parallel to the plane N. Horizontal projection bt point AT in the rotated position we find as the point of intersection of the horizontal projection 1-аt with trace Q h . Horizontal projection a1cb1 expresses the natural value of A ABC, since after the rotation the plane of the triangle is parallel to the plane N. The frontal projection of the rotated triangle coincides with the frontal projection of the horizontal 1"s", i.e., it is a straight line segment.

If you want to rotate a flat geometric image to a position parallel to the plane V, then the frontal is chosen for the axis of rotation.

Rotate the plane around its trace until it coincides with the corresponding projection plane(this case is also called the combination method). If the plane is rotated around its trace until it coincides with the projection plane in which this trace is located, then the geometric images located in the plane will be displayed without distortion. This method is a special case of rotation around a horizontal or frontal, since the horizontal trace of the plane can be considered as the "zero" horizontal of the horizontal plane, and the frontal trace as the "zero" frontal.

Figure 5.14 shows a visual representation of the rotation of the plane of general position R around the horizontal track P h in the direction from the plane V to the viewer until aligned with the plane N. In plane alignment position R with plane

H straight line P Uq is a trace R and, aligned with plane N. Trace Ph how the axis of rotation does not change its position. Dot Rx intersection of traces also does not change its position. To build a combined position P L , a trace P v it is enough to find one more point, for example the point N, this trace (except for the point R x) in a position aligned with the plane N.

Point N describes an arc in a plane Q, perpendicular to the axis of rotation. Centre O this arc is the point of intersection of the plane Q with trace P h . Point N 0 on the plane H is the intersection point of the arc of radius ON in the plane Q with trace Q h . Drawing a straight line through P x and N 0, we get P U0 . Segment P X N does not change its length when the plane rotates; so point N0 can be obtained by crossing Q h with an arc described in a plane H, from point Р x with radius P X N.

To perform the considered constructions on the drawing (Fig. 5.15) on the trace R and arbitrary point selected N (it coincides with its projection P"). Through its horizontal projection P direct on, perpendicular to the axis of rotation - trace Ph.h. A point is found on this line N 0 , i.e. point N after alignment with the plane N. She was found in the distance P X N 0 \u003d P x n "from the point P x or at a distance oN 0 from the point o, equal to the radius of rotation of the point N. Radius length oN 0 = oN defined, for example, as the hypotenuse of a right triangle with legs on and nN (nN=nn"). Straight line P U0 , passing through points P x and N 0, - combined track position R i.

The combined position of the C0 point is similarly constructed C. Radius of rotation oC found as the hypotenuse of a rectangular

triangle with one leg oc, the other leg cc = s "1. The second version of the construction is made using the horizontal plane P with projections c"2", c -2. Using arc radius R x 2" matched position found 2o points 2 on the line Pv0, and in the combined position 20C0 a horizontal line through a point 2 0 parallel to the trace of Ph.

If it is required to combine the plane with the frontal plane of projections, then the plane should be rotated around its frontal trace.

We draw

6.1. Let be a proper prism. The transfer is given by the vector: a) 0.5AB; b) AO, where O is the center of the lower base. Draw the image of the prism during this translation. Draw the union and intersection of the original and resulting prisms.

6.2. Given a regular tetrahedron. Draw a tetrahedron, which is obtained from the given one as a result of: a) central symmetry about the middle of the height; b) mirror symmetry with respect to the plane passing through the middle of the height perpendicular to it; c) rotation by 60° around its height; d) a 90" rotation around the line joining the midpoints of its opposite edges. Draw the union and intersection of the original and resulting tetrahedra.

6.3. Given cube. Draw a cube, which is obtained from the given one as a result of: a) transfer to a vector directed along its diagonal, with a length of half this diagonal; b) central symmetry about a point located on its diagonal and dividing it in a ratio of 2: 1; c) mirror symmetry with respect to a plane that intersects it along a regular hexagon; d) rotate 90" around a straight line passing through the midpoints of two parallel edges that do not lie on the same face. Draw the union and intersection of the original and resulting cubes.

6.4. Draw the bodies that can be obtained by rotating the circle

6.5. Draw the bodies that are obtained by rotating: a) a cube around an edge; b) a cube around the diagonal; c) a regular tetrahedron around an edge; d) a cone around a straight line parallel to the axis and passing outside it.

Are planning

6.6. How to find the volume and surface area of ​​figures - unions and intersections - from tasks 6.1, 6.2?

6.7. How to find the volume and surface area of ​​the figures from Problem 6.5?

Introducing

6.8. Can the center of symmetry of a body not belong to it?

6.9. Two equal segments: a) parallel; b) have exactly one common point; c) interbreed. What movement can one of them display on the other?

6.10. Two segments are symmetrical to each other with respect to two planes. What will be the figure if their ends are connected in series by segments?

6.11. All possible planes are drawn through a straight line. This point is reflected from all these planes. What shape do all the obtained points form?

6.12. Is it true that: a) an inclined parallelepiped, two faces of which are perpendicular to the base, has a plane of symmetry; b) among the faces of a parallelepiped having a plane of symmetry, there are rectangles; c) is a parallelepiped having two planes of symmetry rectangular?

6.13. How to cut a cube into three equal pyramids?

Evaluate

6.14. A right triangle with hypotenuse d rotates around one of the legs. Under what condition will the volume of the body of revolution be the largest?

6.15. The perimeter of an isosceles triangle is P. This triangle rotates around the base. Which of these triangles gives the largest volume of the body of revolution?

We think

6.16. The center of a cube is reflected in the plane of each of its faces. Prove that the obtained points are the vertices of the octahedron. Is it possible to obtain other regular polyhedra in this way?

6.17. This ball contains:

a) a regular tetrahedron;

b) cube. The faces of this polyhedron were extended to the intersection with the sphere. What shapes is the sphere divided into? What shape is the ball divided into? How many of them are equal to each other?

Exploring

6.18. Is the movement of space such a transformation that puts a point with coordinates in correspondence with a point with coordinates:

6.19. A polyhedron has a center of symmetry, a center of an inscribed ball, a center of an inscribed ball, and a center of mass. How many of these points can coincide?

We enter the university

6.20. A chord is drawn from the end of the diameter of the ball so that the surface formed by rotating it around this diameter divides the volume of the ball into two equal parts. Determine the angle between the chord and the diameter.

6.21. An equilateral triangle with side a rotates around an external axis parallel to the side of the triangle and spaced from it at a distance equal to half the height of the triangle. Find the volume of the body of revolution.

6.22. The triangle rotates around the bisector AD. Prove that the areas of the surfaces described in this case by the sides AB and AC are related as the volumes obtained by rotating the parts ABD and

6.23. An isosceles triangle, the base of which is a, and the angle at the base a, rotates around a straight line passing through one of the ends of the base perpendicular to it. Find the surface area of ​​the resulting body of revolution.

6.24. The part of the square ABCD, which remains after a quarter of a circle with a centum at the vertex D and radii equal to the side of the square, is cut out of it, rotates around an axis passing through D parallel to the diagonal AC. Find the volume of the resulting body of revolution if the side of the square is a.

6.25. The area of ​​a rectangular trapezoid ABCD is equal to , the length of the height AB is equal to h, the value of the acute angle ADC of the trapezoid

equal to a. Point E is taken on the side of CD so that . Find the volume of the body obtained by rotating the quadrilateral ABED around the line AB.

6.26. Find the volume of the body obtained by rotating a regular hexagon around its side equal to a

6.27. Points A and B are given on the circle of a semicircle of radius R. If N is one of the ends of the diameter, and O is the center of the circle, then Determine the total surface area of ​​the body formed by the rotation of the circular sector AOB around the diameter.

6.28. Given a regular tetrahedron ABCD. Each of its vertices is symmetrically reflected relative to the plane of the face opposite to it, as a result of which the KLMN points are obtained, respectively. Find the ratio of the volumes of the original and resulting tetrahedra.

6.29. Segments are drawn in the tetrahedron connecting its vertices with the intersection points of the medians of opposite faces. They all intersect at the point O. The second tetrahedron is symmetrical to the first one with respect to the point O. The volume of the original tetrahedron is V. Find the volume of the common part of the two tetrahedra.

Answer: 0.5V.

6.30. The side of the base of a regular prism has a length a, and the side edge has a length of 1.125a. Point E is the middle of the edge AB, and point M lies on the segment EC and EM EC. The second prism is symmetrical to the prism with respect to a straight line. Find the volume of the common part of these prisms.

6.31. A regular tetrahedron of volume V is given. The second tetrahedron is obtained from the first one by rotating it through the angle

and around the straight line connecting the midpoints of the crossing edges of the tetrahedron. Find the volume of the common part of these two tetrahedra.

6.32. A cube with an edge a is rotated 90" around a straight line connecting the midpoints of two parallel edges that do not lie on the same face. Find the volume of the common part of the original cube and the rotated one.

6.33. A regular triangular pyramid with base side a is rotated around the axis of symmetry by an angle of 60. Determine the volume of the common part of the original and rotated pyramids if the side faces are right triangles.

6.34. A regular tetrahedron is inscribed in a ball of radius R. By turning it at an angle - around the height, a new tetrahedron is obtained, inscribed in a ball. Find the volume of the part of the sphere external to both tetrahedra.

6.35. A cone of revolution about an axis - a straight line perpendicular to its height and passing through the vertex. Find the cross-sectional area of ​​the resulting body of revolution by a plane passing through the axis of revolution if the generatrix of the cone is 5 and the height is 4.

TASKS To § 26

Complementing the theory

6.36. Prove that a plane passes into a plane parallel to it (if not into itself) as a result of:

a) transfer; b) central symmetry.

Are planning

6.37. In a cube, point O is the center of face ABCD. How to calculate the angle between line B, O and:

a) straight straight plane

d) plane

6.38. Let PABCD be a pyramid whose base is rhombus ABCD. RVCAVS). The area of ​​the RVS face is equal to S. Through the point K - the middle of the edge AD - a section is drawn parallel to the plane PAB. How to find its area?

6.39. Each side face of a regular tetrahedron has rotated around the edges of the base by the same angle to the outside. This resulted in a polyhedron with six vertices and equal edges. At what angle did the edges turn?

Introducing

6.40. Can two unequal cones have two equal circular sections with the same plane if they stand on the same plane on one side of it?

6.41. The two circles are centrally symmetrical and do not lie in the same plane. Is it true that they lie on the surface of: a) one sphere; b) one cylinder? What if these circles are mirror-symmetrical?

6.42. In which case are two equal:

a) a ball b) a cylinder; c) are the cones centrally symmetrical? Mirror symmetrical?

6.43. By what rotations can the ball be mapped onto itself?

6.44. By what turns is one of these figures mapped to the other if these figures are: a) two straight lines; b) two planes; c) two equal balls? Is there a rotation that maps the second figure onto the first one?

6.45. Do we always get a convex body by rotating a convex figure?

We think

6.46. Using translation properties, prove that: a) two perpendiculars to one plane are parallel; b) two planes perpendicular to one straight line are parallel; c) if a line is parallel to a line perpendicular to a plane, then it is perpendicular to the plane; d) the linear angles of a dihedral angle are equal to each other.

6.47. Prove that the union of two planes is a figure: a) centrally symmetrical; b) mirror-symmetrical.

6.48. The line b is obtained from the line a by reflection in the plane a. These lines have a common point. Prove that this point lies in the plane a.

6.49. In a ball of radius R, two planes are drawn through the center, forming an angle between them. How to find out in what ratio they broke the volume of the ball?

6.50. A plane is drawn through the bisector of an angle. Prove that the sides of an angle form equal angles with it.

Exploring

6.51. Is it possible to fill the entire space with equal parallelepipeds? Can this be done by other equal polyhedra?

6.52. Will the section of a centrally symmetrical body passing through the center of symmetry be centrally symmetrical?

6.53. The body is centrally symmetrical. Will its orthogonal projection be centrally symmetric? Would the opposite be true?

6.54. Each of the two bodies is centrally symmetrical. Will they be centrally symmetrical: a) unification; b) intersection?

6.55. A centrally symmetrical body is divided by a plane. One part of it turned out to be centrally symmetrical. Will there be another part of it?

6.56. Is there a polyhedron that has any preassigned number of symmetry planes?

TASKS To § 27

Complementing the theory

6.57. Prove that the composition of two reflections in intersecting planes is a rotation, and in two parallel planes is a translation.

6.58. Draw a figure that passes into itself as a result of: a) a screw; b) mirror turn; c) gliding reflection.

6.59. Let the cube As a result of some movement, it passes into another cube. Draw this other cube if the movement is: a) a screw with the axis of rotation passing through the centers of the faces

vector a, the rotation angle is equal to the mirror rotation on with the rotation axis , and reflection in a plane perpendicular to the straight line and passing through the center of the cube; c) gliding reflection, where the reflection occurs in a plane perpendicular to the diagonal of the cube and passing through the center of the cube, and the vector is equal to AC.

6.60. Let RABC be a regular tetrahedron. As a result of movement, it passes into another tetrahedron. Draw this other tetrahedron if the motion is like this:

a) a screw with an axis of rotation center of the base), a rotation angle of 60 "and a vector

b) mirror rotation with axis of rotation PQ, angle of rotation 60° and plane of reflection perpendicular to PQ and passing through the mid-height

c) grazing reflection with a reflection plane passing through PB and K - the middle of the AC, and a vector of 0.5 KV.

Introducing

6.61. Does the orientation of the basis preserve: a) translation; b) central symmetry; c) mirror symmetry; d) turn; e) screw; e) mirror rotation; g) gliding reflection?

6.62. Does the movement have fixed points, if this movement: a) transfer; b) central symmetry; c) mirror symmetry; d) turn; e) screw; e) mirror rotation; g) gliding reflection?

6.63. Given two equal isosceles triangles. What movements can they be combined if they have in common: a) the top of equal sides; b) the side of the base; c) lateral side; d) median to base; e) the middle line of the sides?

c) one of its heights to another;

d) a segment connecting the midpoints of opposite edges to another such segment;

e) the section by one plane of symmetry to another is the same;

f) a section that is a square to another one that is the same? Will the second figure be mapped onto the first in such a movement?

6.66. As a result of what movements is displayed on itself:

a) cut b) straight line; c) a circle; d) square; e) a regular polygon; e) rhombus; g) plane; h) dihedral angle?

6.67. As a result of what movements, the tetrahedron RABC is displayed on itself, in which: a) ; b)

6.68. The body is the union of two balls, but not a ball. What movements it is displayed on itself?

6.69. A quadrangular pyramid has: a) all side edges are equal and opposite flat angles at the top are equal;

b) all flat angles at the vertex are equal and opposite side edges are equal. With what movements can it be self-combined?

6.70. What movements reflect the antiprism on itself?

6.71. How to divide a cube into: a) 8 equal cubes; b) 6 equal pyramids; c) 3 equal pyramids; d) 4 equal triangular prisms?

6.72. How to divide a right triangular prism into 3 equal tetrahedra? Are any of them equal?

6.73. How to divide a parallelepiped into: a) 6 equal-sized pyramids; b) three equal pyramids? Are any of them equal?

6.74. In a ball with a radius of 1, three radii OA, OB, OS were drawn, of which every two are perpendicular. What part of the volume of the ball is limited by quarters of the great circles of the ball OAB, OAC, OBC and the surface? What part of the surface?

We think

6.75. Two regular quadrangular pyramids and have a common base ABCD. Point K is the middle of the edge, point L is the middle of the edge, point M is the point of intersection of the medians in the face, point N is the point of intersection of the medians in the face. Prove that:

e) the distance from point K to the plane is equal to the distance from point L to the plane of the RHVS.

Exploring

6.76. Take the composition of any two movements you know and find out: a) does it change the orientation of the plane; b) does it have fixed points?

6.77. How many fixed points can each movement you know have? How are they located? And how many fixed lines can it have? Planes?

6.78. The line b is obtained from the line a by some movement. Establish the location of these lines between themselves, if this movement is: a) a screw; b) mirror turn; c) mirror reflection.

Switching

6.79. A wire is wound on a cylinder with radius R and height H. How do you know its length?

6.80. You need to design a spiral staircase. How will you act?

6.81. Can you explain how a corner reflector works? It is composed of three pairwise perpendicular mirrors.