Prism is geometric body, a polyhedron whose bases are equal polygons and whose lateral faces are parallelograms. For the uninitiated, this may sound somewhat intimidating. And when your child needs to bring a prism that you made at home to your geometry lesson, you are at a loss, not knowing how to help your beloved child. In fact, everything is not so difficult and, using our tips on how to make a prism, you will adequately cope with this problem.

How to make a prism out of paper

Let’s immediately agree that we will make a straight prism, that is, a prism whose side ribs will be perpendicular to the bases. Making an inclined prism from paper is very problematic (such models are usually made from wire).

We already know that at the bases of the prism there are two identical polygons. Therefore, we will begin our work with them. The simplest of polygons is a triangle. This means that we will first make the prism triangular.

How to make a triangular prism

We will need thick white paper for drawing, a pencil, a protractor, a compass, a ruler, scissors and glue.

We draw a triangle, we can use any triangle, but to make our prism especially beautiful, we will make an equilateral triangle. In geometry, such a prism is called “regular”. We choose at our discretion the size of the side of the triangle, say 10 cm. We put this segment on paper with a ruler and use a protractor to measure an angle of 60 ∗ from one end of our segment.

We draw an inclined line. Using a ruler, mark it 10 cm from the end of the segment. Thus, we have found the third vertex of the triangle. We connect this point with the ends of the initial segment and the equilateral triangle is ready. It can be cut out. We make the second triangle in the same way, or carefully trace the contours of the first one on paper. Well, we already have two reasons.

Making the side edges. We decide what height the prism will have. Let's say 20 cm. We draw a rectangle, the size of one side of which is the height of the prism (in our case - 20 cm), and the second side is equal to the size of the base side multiplied by the number of these sides (for us: 10 cm x 3 = 30 cm) .

On the long sides we make marks every 10 cm. We connect the opposite marks with straight lines. Then you will need to carefully bend the paper along them. These are the side edges of our prism. We mark narrow allowances for gluing along two long and one short sides of the rectangle (strips 1 cm wide are enough). We cut out the rectangle along with the allowances, carefully bend them along the markings. Bend the ribs.

Let's start assembly. Glue the rectangle along the side edge into a triangular pipe. Glue base triangles onto the folded allowances at the top and bottom. The prism is ready.

It’s probably not worth going into the details of the question of how to make a prism from cardboard. The entire assembly algorithm remains the same, just replace the paper with thin cardboard. By changing the number of sides of the base polygons, you can now make a pentagonal and hexagonal prism yourself.

Given:
The intersection of a pyramid and a prism
Necessary:
Construct a development of a straight prism and show on it the line of intersection of the prism with the pyramid.

Constructing a development of a straight prism is much easier than constructing a development of a pyramid.

Constructing a prism scan

The construction of a development of a straight prism is facilitated by the fact that all dimensions for development are taken from diagrams and we do not need to find the natural values ​​of the edges of the prism. Since a straight prism is given, the lateral edges of the prism are projected onto the frontal plane of projections in full size. The edges of the bases of a straight prism are parallel to the horizontal projection plane and are also projected onto it in full size.

Algorithm for constructing a prism scan

• We draw a horizontal line.
• From an arbitrary point G of this line we lay off segments GU, UE, EK, KG equal to the lengths of the sides of the base of the prism.
• From points G, U, ..., perpendiculars are restored and values ​​equal to the height of the prism are laid on them. The resulting points are connected by a straight line. Rectangle GG1G1G is a development of the lateral surface of the prism. To indicate the faces of the prism on the development, perpendiculars are drawn from points U, E, K.
• To obtain a complete development of the surface of the prism, polygons of its bases are attached to the development of the surface.

To construct on the development the line of intersection of the prism with the pyramid of closed broken lines 1, 2, 3 and 4, 5, 6, 7, 8, we use vertical straight lines.

More details in the video tutorial on descriptive geometry in AutoCAD

A prism is a three-dimensional figure, a polyhedron, of which there are many types: positive and irregular, straight and oblique. According to the figure lying at the base, the prism can be from triangular to polygonal. It’s easier for everyone to make a straight prism, but an inclined one requires a little more work.

You will need

• – compass;
• - ruler;
• - pencil;
• - scissors;
• - glue;
• – paper or cardboard.

Instructions

1. Draw the bases of the prism, in this case they will be 2 hexagons. To draw a regular hexagon, use a compass. Draw a circle with it, and using the same radius, divide the circle into six parts (for a regular hexagon, the sides are equal to the radius of the circumscribed circle). The resulting figure resembles a honeycomb cell. Draw an irregular hexagon at random, but with the help of a ruler.

2. Now start designing the “pattern”. The walls of the prism are parallelograms, and you need to draw them. In the straight model, the parallelogram would be a light rectangle. And its width will invariably be equal to the side of the hexagon lying at the base of the prism. With the correct figure at the base, all faces of the prism will be equal to each other. If it is incorrect, the entire side of the hexagon will correspond to only one parallelogram (one side face), suitable in size. At the same time, keep an eye on the sequence of edge sizes.

3. On a horizontal straight line, stepwise lay out 6 segments equal to the side of the base of the hexagon. From the obtained points, draw perpendicular lines of the required height. Connect the ends of the perpendiculars with a 2nd horizontal line. You now have 6 rectangles joined together.

4. Attach 2 previously constructed hexagons to the bottom and top sides of one of the rectangles. To any base if it is positive, and to the corresponding length if the hexagon is irregular. Outline the silhouette with a solid line and the fold lines inside the figure with a dotted line. You have obtained a development of the surface of a straight prism.

5. To create an inclined prism, leave the base the same. Draw a parallelogram side, which will be one of the faces. There should be six such faces, as you remember. In order to now draw the development of an inclined prism, it is necessary to arrange six parallelograms in the following order: three in ascending order, so that their oblique sides form one line, then three in descending order with the same condition. The steepness of the resulting line is directly proportional to the degree of inclination of the prism.

6. To the five rectangles in the development, draw small trapezoidal overlaps on the short sides to glue the figure, as well as on one free long side. Cut out the blank for the prism along with the overlaps and glue the model together.

A prism is a device that separates typical light into individual colors: scarlet, orange, yellow, green, blue, indigo, violet. It is a translucent object, with a flat surface that refracts light waves depending on their wavelengths and, as a result, allows light to be seen in different colors. Do prism easy enough on your own.

You will need

• Two sheets of paper
• Foil
• Cup
• CD
• Coffee table
• Flashlight
• Pin

Instructions

1. A prism can be made from a simple glass. Fill the glass a little more than half full with water. Place the glass on the edge of the coffee table so that about half of the bottom of the glass hangs in the air. At the same time, make sure that the glass is stable on the table.

2. Place two sheets of paper, one after another, next to the coffee table. Turn on the flashlight and shine rays of light through the glass so that it falls on the paper.

3. Adjust the position of the flashlight and paper until you see a rainbow on the sheets - this is how your beam of light is decomposed into spectra.

Video on the topic

The basic skill of an artist in academic drawing is the knowledge to depict on a plane the simplest volumetric geometric shapes - a cube, prism, cylinder, cone, pyramid and sphere. Having mastered this skill, you can build more complex, combined volumetric forms of architectural and other objects. A prism is a polyhedron whose two faces (bases) have identical shapes and are parallel to each other. The lateral faces of the prism are parallelograms. Depending on the number of side faces, prisms can be 3-, tetrahedral, etc.

You will need

• – drawing paper;
• – primitive pencils;
• – easel;
• – a prism or an object shaped like a prism (wooden block, box, casket, part children's construction set etc.), preferably white.

Instructions

1. Erect prism allowed by inscribing it either in a parallelepiped or in a cylinder. The main difficulty when drawing a prism is the positive construction of the shape of the 2 faces of its base. When drawing a prism lying on one of the side faces, an additional difficulty appears in observing the laws of perspective, because in such an arrangement the perspective reduction of the side faces becomes noticeable.

2. When drawing a vertical prism, start by marking its central axis - a vertical line drawn in the middle of the sheet. On the axis line, mark the center of the top (visible) edge of the base and draw a horizontal line through this point. Determine the ratio of the height and width of the prism using the sighting method: look at nature, covering one eye, and, holding a pencil in an outstretched hand on the tier of the eyes, mark with your finger on the pencil the width of the prism visible from your point of view and mentally lay this distance along the line of the height of the prism a certain number times (as many times as possible).

3. Measuring the segments with a pencil closer to the drawing, mark the width and height of the prism with dots on the 2 previously drawn lines, observing the resulting ratio. Draw an ellipse around the center of the top edge. Try to accurately convey its imaginary form, looking at nature. Draw approximately the same ellipse (but less flattened) in the plane of the lower edge of the base of the prism. Combine the resulting ellipses with two vertical lines.

4. Now on the upper ellipse it is necessary to mark the segments of intersection of the side faces and its bases. Looking at nature, mark the points - the vertices of the polygon - lying at the base of the prism, as you see them, and unite them step by step. From these points, draw lines down to the intersection with the lower ellipse. Combine the resulting intersection points as well. During subsequent drawing, the edges visible from the selected point of view are erased or shaded, therefore, draw all auxiliary construction lines without pressure.

5. Lying on her side prism draw with the help of an auxiliary parallelepiped. Focusing on nature, draw a parallelepiped, observing the principles of perspective - the lines of the lateral edges, when mentally extended to the horizon line, which is invariably located on the tier of the viewer’s eyes, converge at one point. Consequently, the (noticeable) edge far from us will be slightly smaller than the front one. When determining the aspect ratio of a parallelepiped, use the “arm's length” (or sight) method.

6. On the front and back square faces, mark the vertices of the polygons lying at the base of the prism and construct them. Combine these points in pairs on 2 faces - draw the side edges of the prism. Remove unnecessary lines. Highlight the lines of edges and corners of the prism that are closest to you in bolder colors, and indicate the distant ones with light lines.

7. Looking at nature, determine the angle of incidence of light, the clearest, most shaded edges and, with the help of shading of varying intensity, convey these light relationships in the drawing. Draw a shadow falling from the object. Emphasize the contact line between the prism and the table with the darkest line. Please note that the light reflected from the table surface (reflex) falls on the most shaded face of the prism from below and slightly illuminates it. When applying shading to this face, take this result into account and apply a less saturated tone in the place of the reflex.

Video on the topic

A prism is a polyhedron formed by any final number of faces, two of which - the bases - must certainly be parallel. Any straight line drawn perpendicular to the bases contains a segment connecting them, called the height of the prism. If all the side faces are adjacent to both bases at an angle of 90°, the prism is called straight .

You will need

• Prism drawing, pencil, ruler.

Instructions

1. IN straight prism Every lateral edge is, by definition, perpendicular to the base. And the distance between the parallel planes of the side faces is identical at every point, including at those points where the side edge adjoins them. From these two circumstances it follows that the length of the edge of each lateral face straight the prism is equal to the height of this volumetric figure. This means that if you have a drawing that depicts such a polyhedron, it already contains segments (edges of the lateral faces), all of which can be designated as the height of the prism. If this is not prohibited by the conditions of the task, simply designate every lateral edge as a height, and the problem will be solved.

2. If you want to draw a height in the drawing that does not coincide with the side edges, draw a segment parallel to any of these edges connecting the bases. It is not always possible to do this “by eye”; therefore, build two auxiliary diagonals on the side faces - combine a pair of any angles on the top and a corresponding pair on the bottom base. After this, measure any comfortable distance on the upper diagonal and put a point - this will be the intersection of the height with the upper base. On the lower diagonal, measure exactly the same distance and place a second point - the intersection of the height with the lower base. Connect these points with a segment, and construct the height straight the prism will be completed.

3. A prism can be depicted taking into account perspective, that is, the lengths of identical edges of the figure can have different lengths in the drawing, the side faces can adjoin the bases at different and not strictly right angles, etc. In this case, in order to correctly maintain the proportions, proceed in the same way as described in the previous step, but place the points on the upper and lower diagonals correctly in their middles.

In detail - how to fold a sheet of paper and cut out a beautiful snowflake.

You will need

• A sheet of paper, I have an ordinary A4 sheet, it’s better to take huge napkins
• Scissors

Instructions

1. Fold the sheet in half crosswise

2. Now double it, just to find the middle

3. We wrap the edges of the paper folded in half, one by one - you can see it as in the photo

4. We make sure that the leaf bends evenly and the ends reach the folds.

5. Now we fold the resulting envelope in half. It is necessary to practice in order to ensure that the outer edge of the sheet reaches exactly to the fold.

6. While you don’t have the skill, it’s better to draw the approximate silhouette of a snowflake in advance.

7. Carefully cut out according to the silhouette.

8. We carefully unwrap it.

Note!
Remember that it is impossible to make a through cut; the snowflake will fall apart.

Helpful advice
The thinner the paper, the easier it is to cut out the snowflake. You can also make snowflakes from foil.

Note!
When developing an inclined prism, do not draw its edges at too great an angle; on the contrary, the model will be unstable.

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- this is a 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 lateral edges

Diagonal section- the boundaries of the intersection of the prism and the diagonal plane. The diagonal cross 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 lateral edges

Elements of a regular quadrangular prism

The figure shows two regular quadrangular prisms, which are indicated by the corresponding letters:

• The 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 lateral faces of the prism
• Total surface - the sum of the areas of all bases and side faces (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 side faces are rectangles
• The side edges are equal to each other
• The side faces are perpendicular to the bases
• The lateral ribs are parallel to each other and equal
• Perpendicular section perpendicular to all side ribs and parallel to the bases
• Angles of perpendicular section - straight
• The diagonal cross 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" means 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 properties of a regular quadrangular prism above) Note. This is part of a lesson with geometry problems (section stereometry - prism). Here are problems that are difficult to solve. If you need to solve a geometry problem that is not here, write about it in the forum. To denote the action of extracting the square root in solving problems, the symbol is used√ .

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

√144 = 12 cm.
From where 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

Task

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

Solution.
Since the base of a regular quadrangular prism is a square, we find the side of the base (denoted as a) using 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 = 4 2
h 2 + 12.5 = 16
h 2 = 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 = 25 + 10√7 ≈ 51.46 cm 2.

Answer: 25 + 10√7 ≈ 51.46 cm 2.

It is necessary to construct developments of faceted bodies and mark the intersection of the prism and the pyramid on the development.

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

— information about developments of surfaces, methods of their construction and, in particular, construction of developments of faceted bodies;

— one-to-one properties between a surface and its development and methods of transferring points belonging to the surface to the development;

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

Procedure for solving the Problem

It's called a sweep a flat figure that is obtained by cutting and bending the surface until it is completely aligned with the plane. All surface developments ( blanks, patterns) are constructed only from natural quantities.

1. Since the developments are constructed from natural quantities, we proceed to their determination, for which we use tracing paper (graph paper or other paper) of A3 format to transfer task No. 3 with all the points and lines of intersection of the polyhedra.

2. To determine the natural values ​​of the edges and base of the pyramid, we use right triangle method. Of course, others are possible, but in my opinion, this method is more understandable for students. Its essence is that “on the constructed right angle, the projection value of the straight line segment is plotted on one side, and on the other, the difference in 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 the given line segment.”.

Fig.4.1

Fig.4.2

Fig.4.3

3. So on free space drawing (Fig.4.1.a) we build a right angle.

Along the horizontal line of this angle we plot the projection value of the edge of the pyramid D.A. taken from the horizontal projection plane - lDA. Along the vertical line of the right angle we plot the difference in the coordinates of the points D’ AndA’ , taken from the frontal plane of projections (along the axis z down) - . By connecting the resulting points with the hypotenuse, we obtain the actual size of the edge of the pyramid | D.A.| .

In this way we determine the natural values ​​of the other edges of the pyramid D.B. And DC, as well as the base of the pyramid AB, BC, AS (Fig.4.2), for which we construct a second right angle. Note that determining the natural size of an edge DC is carried out in cases where it is given projectionally in 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 natural size.”

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

Fig.4.4

4. Having determined the natural values ​​of the edges and base of the pyramid, we proceed to construct the development ( Fig.4.4). To do this, take an arbitrary point on a sheet of paper closer to the left side of the frame D believing that this is the top of the pyramid. We carry out from the point D an arbitrary straight line and plot the natural size of the edge on it | D.A.| , getting a point A. Then from the point A, using a compass solution to measure the actual size of the base of the pyramid R=|AB| and placing the leg of the compass at the point A we make an arc notch. Next, take the actual size of the edge of the pyramid using a compass solution 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 IN, connecting it with points A and D we get the edge of the pyramid DAB. Similarly, we attach to the edge D.B. edge DBC, and to the edge DC- edge DCA.

To one side of the base, for example INC, we attach the base of the pyramid also using the method of geometric serifs, taking the dimensions of the sides on the compass solution ABAndAWITH and making arc serifs from points BAndC getting the point A(Fig.4.4).

5. Constructing a sweep the prism is simplified by the fact that in the original drawing in the horizontal plane of projections the base, and in the frontal plane - with a height of 85 mm, it set immediately in natural size

To construct a scan, we mentally cut the prism along some edge, for example along E Having fixed it on the plane, we will unfold the other faces of the prism until they are 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 construct a prism scan we do:

- on the same format where the pyramid is constructed, draw a horizontal straight line on the right side and from an arbitrary point on it, for example E, sequentially lay out the segments of the base of the prism E.K., KG, G.U., UE, taken from the horizontal plane of projections;

- from dots E, K, G, U, E we restore the perpendiculars on which we plot the height of the prism taken from the frontal plane of the projections (85mm);

— connecting the obtained points with a straight line, we obtain a development of the lateral surface of the prism and to one of the sides of the base, for example, G.U. we attach the upper and lower bases using the geometric serif method, as we did when building the base of the pyramid.

Fig.4.5

6. To construct an intersection line on a development, we use the rule that “any point on the surface corresponds to a point on the development.” Take, for example, the face of a prism G.U., where the line of intersection with the points lies 1-2-3 ; . Let's put the bases on the development G.U. points 1,2,3 by distances taken from the horizontal projection plane. Let us restore perpendiculars from these points and plot the heights of the points on them 1’ , 2’, 3’ , taken from the frontal plane of projection – z 1 , z 2 Andz 3 . Thus, we got points on the scan 1, 2, 3, connecting which we get the first branch of the intersection line.

All other points are transferred similarly. The constructed points are connected, obtaining the second branch of the intersection line. Highlight the desired line in red. 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 pyramid development is carried out in the same way, but taking into account the following:

— since scans are built from natural values, it is necessary to transfer the position of the points 1-8 the lines of intersection of projections on the lines of edges of natural dimensions of the pyramid. To do this, take, for example, the points 2 and 5 in the frontal projection of the rib D.A. let us transfer them to the projection value of this edge of the right angle (Fig.4.1) along communication lines parallel to the axis X, we obtain the required segments | D2| and |D5| ribs D.A. in natural quantities, which we put 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 at a right angle (Fig.4.3) the natural values ​​of these generators are determined, and then the points are transferred to them 6 and 8;

- on the second right angle, where the natural values ​​of the base of the pyramid are determined, the points are transferred mAndn intersections of generatrices 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 sequentially with straight lines and finally obtain the line of intersection of the pyramid on its development.

Section: Descriptive Geometry /