It is necessary to build a development of the faceted bodies and draw on the development the line of intersection of the prism and the pyramid.

To solve this problem in descriptive geometry, you need to know:

- information about the development of surfaces, methods of their construction and, in particular, the construction of developments of faceted bodies;

- one-to-one properties between a surface and its unfolding and methods for transferring points belonging to the surface to the unfolding;

- methods for determining the natural values ​​of geometric images (lines, planes, etc.).

Procedure for solving the Problem

The scan is called a flat figure, which is obtained by cutting and unbending the surface until it is completely aligned with the plane. All surface unfolds ( blanks, patterns) are built only from natural values.

1. Since the scans are built from natural values, we proceed to determine them, for which a tracing paper (graph paper or other paper) of A3 format is transferred task No. z with all points and lines of intersection of polyhedra.

2. To determine the natural values ​​of the edges and the base of the pyramid, we use right triangle method. Of course, others are possible, but in my opinion, this method is more intelligible for students. Its essence lies in the fact that “on the constructed right angle, the projection value of the straight line segment is plotted on one leg, and on the other, the difference in the coordinates of the ends of this segment, taken from the conjugate projection plane. Then the hypotenuse of the resulting right angle gives the natural value of this line segment..

Fig.4.1

Fig.4.2

Fig.4.3

3. So, in the free space of the drawing (Fig.4.1.a) making a right angle.

On the horizontal line of this angle, we set aside the projection value of the edge of the pyramid DA taken from the horizontal projection plane - lDA. On the vertical line of the right angle, we plot the difference in the coordinates of the points D’ andA’ taken from the frontal projection plane (along the axis z way down) - . Connecting the obtained points with a hypotenuse, we obtain the natural size of the edge of the pyramid | DA| .

Thus, we determine the natural values ​​\u200b\u200bof other edges of the pyramid D.B. and DC, as well as the base of the pyramid AB, BC, AC (fig.4.2), for which we construct the second right angle. Note that the definition of the natural size of the edge DC is made in those cases when it is given in projection on the original drawing. This is easily determined if we remember the rule: if a straight line on any projection plane is parallel to the coordinate axis, then on the conjugate plane it is projected in full size.

In particular, in the example of our problem, the frontal projection of the edge D’ C’ parallel to axis X, therefore, in the horizontal plane DC immediately expressed in natural size | DC| (fig.4.1).

Fig.4.4

4. Having determined the natural values ​​of the edges and the base of the pyramid, we proceed to the construction of a sweep ( fig.4.4). To do this, on a sheet of paper closer to the left side of the frame, we take an arbitrary point D considering that this is the top of the pyramid. Draw from a point D arbitrary straight line and set aside on it the natural size of the edge | DA| , getting a point BUT. Then from the point BUT, taking on the solution of the compass the full size of the base of the pyramid R=|AB| and placing the leg of the compass at the point BUT we make an arc. Next, we take on the solution of the compass the full size of the edge of the pyramid R=| D.B.| and placing the leg of the compass at the point D we make a second arc notch. At the intersection of arcs we get a point AT, connecting it with points A and D get the edge of the pyramid DAB. Similarly, we attach to the edge D.B. facet DBC, and to the edge DC- edge DCBUT.

To one side of the base, for example ATC, we attach the base of the pyramid also by the method of geometric serifs, taking the size of the sides on the compass solution BUTBandAFROM and making arc serifs from points BandC getting a point A(fig.4.4).

5. Building a sweep prism is simplified by the fact that in the original drawing in the horizontal plane of projections the base, and in the frontal plane - 85 mm high, it set at full size

To build a sweep, we mentally cut the prism along some edge, for example, along E, having fixed it on the plane, we will expand the other faces of the prism until it is completely aligned with the plane. It is quite obvious that we will get a rectangle whose length is the sum of the lengths of the sides of the base, and the height is the height of the prism - 85mm.

So, to build a sweep of the prism, we proceed:

- on the same format where the pyramid sweep is built, on the right side we draw a horizontal straight line and from an arbitrary point on it, for example E, successively lay off segments of the base of the prism EK, KG, GU, UE, taken from the horizontal projection plane;

- from points E, K, G, U, E we restore the perpendiculars, on which we set aside the height of the prism, taken from the frontal projection plane (85mm);

- connecting the obtained points with a straight line, we obtain a development of the side surface of the prism and to one of the sides of the base, for example, GU we attach the upper and lower bases using the method of geometric serifs, as was done when building the base of the pyramid.

Fig.4.5

6. To build a line of intersection on the development, we use the rule that "any point on the surface corresponds to a point on the development." Take, for example, the edge of a prism GU where the line of intersection with the points 1-2-3 ; . Set aside on the development of the base GU points 1,2,3 by distances taken from the horizontal projection plane. Restore the perpendiculars from these points and plot the heights of the points on them 1’ , 2’, 3’ , taken from the frontal projection plane - z 1 , z 2 andz 3 . Thus, we got points on the sweep 1, 2, 3, connecting which we get the first branch of the line of intersection.

All other points are transferred similarly. The constructed points are connected, getting the second branch of the line of intersection. Highlight in red - the desired line. Let us add that in case of incomplete intersection of faceted bodies, there will be one closed branch of the intersection line on the development of the prism.

7. The construction (transfer) of the intersection line on the development of the pyramid is carried out in the same way, but taking into account the following:

- since the sweeps are built from natural values, it is necessary to transfer the position of the points 1-8 lines of intersection of projections on the lines of edges of natural sizes of the pyramid. To do this, take, for example, the points 2 and 5 in the frontal projection of the rib DA we transfer them to the projection value of this right angle edge (fig.4.1) along communication lines parallel to the axis X, we get the required segments | D2| and |D5| ribs DA in natural values, which we set aside (transfer) to the development of the pyramid;

- all other points of the intersection line are transferred in the same way, including points 6 and 8 lying on the generators Dm and Dn why right angle (fig.4.3) the natural values ​​of these generators are determined, and then points are transferred to them 6 and 8;

- on the second right angle, where the natural values ​​\u200b\u200bof the base of the pyramid are determined, points are transferred mandn intersections of generators with the base, which are subsequently transferred to the development.

Thus, the points obtained on natural values 1-8 and transferred to the development, we connect in series with straight lines and finally we get the line of intersection of the pyramid on its development.

Section: Descriptive geometry /

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Definition.

This is a hexagon, the bases of which are two equal squares, and the side faces are equal rectangles.

Side rib is the common side of two adjacent side faces

Prism Height is a line segment perpendicular to the bases of the prism

Prism Diagonal- a segment connecting two vertices of the bases that do not belong to the same face

Diagonal plane- a plane that passes through the diagonal of the prism and its side edges

Diagonal section- the boundaries of the intersection of the prism and the diagonal plane. The diagonal section of a regular quadrangular prism is a rectangle

Perpendicular section (orthogonal section)- this is the intersection of a prism and a plane drawn perpendicular to its side edges

Elements of a regular quadrangular prism

The figure shows two regular quadrangular prisms, which are marked with the corresponding letters:

• Bases ABCD and A 1 B 1 C 1 D 1 are equal and parallel to each other
• Side faces AA 1 D 1 D, AA 1 B 1 B, BB 1 C 1 C and CC 1 D 1 D, each of which is a rectangle
• Lateral surface - the sum of the areas of all the side faces of the prism
• Total surface - the sum of the areas of all bases and side faces (the sum of the area of ​​the side surface and bases)
• Side ribs AA 1 , BB 1 , CC 1 and DD 1 .
• Diagonal B 1 D
• Base diagonal BD
• Diagonal section BB 1 D 1 D
• Perpendicular section A 2 B 2 C 2 D 2 .

Properties of a regular quadrangular prism

• The bases are two equal squares
• The bases are parallel to each other
• The sides are rectangles.
• Side faces are equal to each other
• Side faces are perpendicular to the bases
• Lateral ribs are parallel to each other and equal
• Perpendicular section perpendicular to all side ribs and parallel to the bases
• Perpendicular Section Angles - Right
• The diagonal section of a regular quadrangular prism is a rectangle
• Perpendicular (orthogonal section) parallel to the bases

Formulas for a regular quadrangular prism

Instructions for solving problems

When solving problems on the topic " regular quadrangular prism" implies that:

Correct prism- a prism at the base of which lies a regular polygon, and the side edges are perpendicular to the planes of the base. That is, a regular quadrangular prism contains at its base square. (see above the properties of a regular quadrangular prism) Note. This is part of the lesson with tasks in geometry (section solid geometry - prism). Here are the tasks that cause difficulties in solving. If you need to solve a problem in geometry, which is not here - write about it in the forum. To denote the action of extracting a square root in solving problems, the symbol is used√ .

A task.

In a regular quadrangular prism, the base area is 144 cm 2 and the height is 14 cm. Find the diagonal of the prism and the total surface area.

Solution.
A regular quadrilateral is a square.
Accordingly, the side of the base will be equal to

√144 = 12 cm.
Whence the diagonal of the base of a regular rectangular prism will be equal to
√(12 2 + 12 2 ) = √288 = 12√2

The diagonal of a regular prism forms a right triangle with the diagonal of the base and the height of the prism. Accordingly, according to the Pythagorean theorem, the diagonal of a given regular quadrangular prism will be equal to:
√((12√2) 2 + 14 2 ) = 22 cm

Answer: 22 cm

A task

Find the total surface area of ​​a regular quadrangular prism if its diagonal is 5 cm and the diagonal of the side face is 4 cm.

Solution.
Since the base of a regular quadrangular prism is a square, then the side of the base (denoted as a) is found by the Pythagorean theorem:

A 2 + a 2 = 5 2
2a 2 = 25
a = √12.5

The height of the side face (denoted as h) will then be equal to:

H 2 + 12.5 \u003d 4 2
h 2 + 12.5 = 16
h 2 \u003d 3.5
h = √3.5

The total surface area will be equal to the sum of the lateral surface area and twice the base area

S = 2a 2 + 4ah
S = 25 + 4√12.5 * √3.5
S = 25 + 4√43.75
S = 25 + 4√(175/4)
S = 25 + 4√(7*25/4)
S \u003d 25 + 10√7 ≈ 51.46 cm 2.

Answer: 25 + 10√7 ≈ 51.46 cm 2.

At the heart of a geometric body - a prism - are polygons, and each side face is a parallelogram. The uninitiated may have been a little scared. But if your child is asked to come to a lesson with a prism, you will naturally want to help him and explain how to make a paper prism.

Let's start by making a straight prism. In this prism, the side edges are perpendicular to the bases. The easiest to make with your own hands is a paper prism with three faces, since its bases are the simplest of polygons - triangles. Let's make a "correct" prism. Its bases are represented by equilateral triangles.

triangular prism

Let's think about the height of our triangular paper prism. Let's draw a rectangle with one side equal to the height, and the other equal to the length of the perimeter of the triangle at the base. The resulting rectangle is divided by parallel lines into three equal parts. From the corners of the rectangle located in the middle, we draw circles with a compass with a radius equal to the side of our triangle at the base. Where the circles intersect outside the original rectangle, put points and connect them to the centers of the circles. We should get the figure shown in the middle of the picture. Next, we cut out the figure with small allowances for gluing, bend along the existing straight lines and get the finished prism.

According to what template a paper prism with four faces is made, the diagram in the figure clearly demonstrates.

Hexagonal prism

An example of a blank for a five-sided prism is shown in the figure. Here the height of the pyramid is 10 cm, the length of the sides of the pentahedron at the base is 3 cm. hexagonal prism made of paper, but at its base lies a hexagon.

tilted prism

An inclined paper prism is shown in this figure. Its side faces are at an angle to the base. Such a prism can be made according to a scanning template.