Triangular hexagonal prism from paper scheme. Volume and surface area of ​​a regular quadrangular prism. Algorithm for constructing a prism scan

Given:
Intersection of pyramid and prism
Necessary:
Build a sweep of a straight prism and show on it the line of intersection of the prism with the pyramid.

Building a straight prism sweep is much easier than a pyramid sweep.

Construction of a prism sweep

The construction of a sweep of a straight prism is facilitated by the fact that all dimensions for the sweep are taken from diagrams and we do not need to find the natural dimensions of the edges of the prism. Since a straight prism is given, the lateral edges of the prism are projected onto the frontal projection plane in full size. The edges of the bases of a straight prism are parallel to the horizontal plane of projections 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 set aside the 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 quantities 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 side surface of the prism. To indicate on the development of the faces of the prism from the points U, E, K, perpendiculars are restored.
• To obtain a complete development of the surface of the prism, the polygons of its bases are attached to the development of the surface.

To build on the scan 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

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 how high ours will be triangular prism from paper. 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. Similarly, a hexagonal prism of paper can be made, 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.

AT school curriculum in the course of solid geometry, the study of three-dimensional figures usually begins with a simple geometric body - a prism polyhedron. The role of its bases is performed by 2 equal polygons lying in parallel planes. A special case is a regular quadrangular prism. Its bases are 2 identical regular quadrilaterals, to which the sides are perpendicular, having the shape of parallelograms (or rectangles if the prism is not inclined).

What does a prism look like

A regular quadrangular prism is a hexagon, at the bases of which there are 2 squares, and the side faces are represented by rectangles. Another name for this geometric figure- a straight parallelepiped.

The figure, which depicts a quadrangular prism, is shown below.

You can also see in the picture the most important elements that make up a geometric body. They are commonly referred to as:

Sometimes in problems in geometry you can find the concept of a section. The definition will sound like this: a section is all points of a volumetric body that belong to the cutting plane. The section is perpendicular (crosses the edges of the figure at an angle of 90 degrees). For a rectangular prism, a diagonal section is also considered ( maximum amount sections that can be built - 2) passing through 2 edges and diagonals of the base.

If the section is drawn in such a way that the cutting plane is not parallel to either the bases or the side faces, the result is a truncated prism.

Various ratios and formulas are used to find the reduced prismatic elements. Some of them are known from the course of planimetry (for example, to find the area of ​​the base of a prism, it is enough to recall the formula for the area of ​​a square).

Surface area and volume

To determine the volume of a prism using the formula, you need to know the area of ​​\u200b\u200bits base and height:

V = Sprim h

Since the base of a regular tetrahedral prism is a square with side a, You can write the formula in a more detailed form:

V = a² h

If we are talking about a cube - a regular prism with equal length, width and height, the volume is calculated as follows:

To understand how to find the lateral surface area of ​​a prism, you need to imagine its sweep.

It can be seen from the drawing that the side surface is made up of 4 equal rectangles. Its area is calculated as the product of the perimeter of the base and the height of the figure:

Sside = Pos h

Since the perimeter of a square is P = 4a, the formula takes the form:

Sside = 4a h

For cube:

Sside = 4a²

To calculate the total surface area of ​​a prism, add 2 base areas to the side area:

Sfull = Sside + 2Sbase

As applied to a quadrangular regular prism, the formula has the form:

Sfull = 4a h + 2a²

For the surface area of ​​a cube:

Sfull = 6a²

Knowing the volume or surface area, you can calculate the individual elements of a geometric body.

Finding prism elements

Often there are problems in which the volume is given or the value of the lateral surface area is known, where it is necessary to determine the length of the side of the base or the height. In such cases, formulas can be derived:

• base side length: a = Sside / 4h = √(V / h);
• height or side rib length: h = Sside / 4a = V / a²;
• base area: Sprim = V / h;
• side face area: Side gr = Sside / 4.

To determine how much area a diagonal section has, you need to know the length of the diagonal and the height of the figure. For a square d = a√2. Therefore:

Sdiag = ah√2

To calculate the diagonal of the prism, the formula is used:

dprize = √(2a² + h²)

To understand how to apply the above ratios, you can practice and solve a few simple tasks.

Examples of problems with solutions

Here are some of the tasks that appear in the state final exams in mathematics.

Exercise 1.

Sand is poured into a box shaped like a regular quadrangular prism. The height of its level is 10 cm. What will the level of sand be if you move it into a container of the same shape, but with a base length 2 times longer?

It should be argued as follows. The amount of sand in the first and second containers did not change, i.e., its volume in them is the same. You can define the length of the base as a. In this case, for the first box, the volume of the substance will be:

V₁ = ha² = 10a²

For the second box, the length of the base is 2a, but the height of the sand level is unknown:

V₂ = h(2a)² = 4ha²

Because the V₁ = V₂, the expressions can be equated:

10a² = 4ha²

After reducing both sides of the equation by a², we get:

As a result new level sand will be h = 10 / 4 = 2.5 cm.

Task 2.

ABCDA₁B₁C₁D₁ is a regular prism. It is known that BD = AB₁ = 6√2. Find the total surface area of ​​the body.

To make it easier to understand which elements are known, you can draw a figure.

Since we are talking about a regular prism, we can conclude that the base is a square with a diagonal of 6√2. The diagonal of the side face has the same value, therefore, the side face also has the shape of a square equal to the base. It turns out that all three dimensions - length, width and height - are equal. We can conclude that ABCDA₁B₁C₁D₁ is a cube.

The length of any edge is determined through the known diagonal:

a = d / √2 = 6√2 / √2 = 6

The total surface area is found by the formula for the cube:

Sfull = 6a² = 6 6² = 216

Task 3.

The room is being renovated. It is known that its floor has the shape of a square with an area of ​​9 m². The height of the room is 2.5 m. What is the lowest cost of wallpapering a room if 1 m² costs 50 rubles?

Since the floor and ceiling are squares, that is, regular quadrilaterals, and its walls are perpendicular to horizontal surfaces, we can conclude that it is correct prism. It is necessary to determine the area of ​​its lateral surface.

The length of the room is a = √9 = 3 m.

The square will be covered with wallpaper Sside = 4 3 2.5 = 30 m².

The lowest cost of wallpaper for this room will be 50 30 = 1500 rubles.

Thus, to solve problems for a rectangular prism, it is enough to be able to calculate the area and perimeter of a square and a rectangle, as well as to know the formulas for finding the volume and surface area.

How to find the area of ​​a cube

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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 /