Let us separately draw the base of the prism, i.e. triangle ABC (Fig. 307, a), and build it up to a rectangle, for which we draw a straight line KM through vertex B || AC and from points A and C we lower perpendiculars AF and CE onto this line. We get rectangle ACEF. Drawing the height ВD of triangle ABC, we see that rectangle ACEF is divided into 4 right triangles. Moreover, \(\Delta\)ALL = \(\Delta\)BCD and \(\Delta\)BAF = \(\Delta\)BAD. This means that the area of rectangle ACEF is twice the area of triangle ABC, i.e. equal to 2S.
To this prism with base ABC we will attach prisms with bases ALL and BAF and height h(Fig. 307, b). We obtain a rectangular parallelepiped with an ACEF base.
If we dissect this parallelepiped with a plane passing through straight lines BD and BB’, we will see that the rectangular parallelepiped consists of 4 prisms with bases BCD, ALL, BAD and BAF.
Prisms with bases BCD and BC can be combined, since their bases are equal (\(\Delta\)BCD = \(\Delta\)BCE) and their side edges, which are perpendicular to the same plane, are also equal. This means that the volumes of these prisms are equal. The volumes of prisms with bases BAD and BAF are also equal.
Thus, it turns out that the volume of a given triangular prism with base ABC is half the volume of a rectangular parallelepiped with base ACEF.
We know that the volume of a rectangular parallelepiped is equal to the product of the area of its base and its height, i.e. in this case it is equal to 2S h. Hence the volume of this right triangular prism is equal to S h.
The volume of a right triangular prism is equal to the product of the area of its base and its height.
2. Volume of a right polygonal prism.
To find the volume of a right polygonal prism, for example a pentagonal one, with base area S and height h, let's divide it into triangular prisms (Fig. 308).
Denoting the base areas of triangular prisms by S 1, S 2 and S 3, and the volume of a given polygonal prism by V, we obtain:
V = S 1 h+ S 2 h+ S 3 h, or
V = (S 1 + S 2 + S 3) h.
And finally: V = S h.
In the same way, the formula for the volume of a right prism with any polygon at its base is derived.
Means, The volume of any right prism is equal to the product of the area of its base and its height.
Prism volume
Theorem. The volume of a prism is equal to the product of the area of the base and the height.
First we prove this theorem for a triangular prism, and then for a polygonal one.
1) Let us draw (Fig. 95) through edge AA 1 of the triangular prism ABCA 1 B 1 C 1 a plane parallel to face BB 1 C 1 C, and through edge CC 1 a plane parallel to face AA 1 B 1 B; then we will continue the planes of both bases of the prism until they intersect with the drawn planes.
Then we get a parallelepiped BD 1, which is divided by the diagonal plane AA 1 C 1 C into two triangular prisms (one of which is this one). Let us prove that these prisms are equal in size. To do this, we draw a perpendicular section abcd. The cross-section will produce a parallelogram whose diagonal ac is divided into two equal triangles. This prism is equal in size to a straight prism whose base is \(\Delta\) abc, and the height is edge AA 1. Another triangular prism is equal in area to a straight line whose base is \(\Delta\) adc, and the height is edge AA 1. But two straight prisms with equal bases and equal heights are equal (because when inserted they are combined), which means that the prisms ABCA 1 B 1 C 1 and ADCA 1 D 1 C 1 are equal in size. It follows from this that the volume of this prism is half the volume of the parallelepiped BD 1; therefore, denoting the height of the prism by H, we get:
$$ V_(\Delta ex.) = \frac(S_(ABCD)\cdot H)(2) = \frac(S_(ABCD))(2)\cdot H = S_(ABC)\cdot H $$
2) Let us draw diagonal planes AA 1 C 1 C and AA 1 D 1 D through the edge AA 1 of the polygonal prism (Fig. 96).
Then this prism will be cut into several triangular prisms. The sum of the volumes of these prisms constitutes the required volume. If we denote the areas of their bases by b 1 , b 2 , b 3, and the total height through H, we get:
volume of polygonal prism = b 1H+ b 2H+ b 3 H =( b 1 + b 2 + b 3) H =
= (area ABCDE) H.
Consequence. If V, B and H are numbers expressing in the corresponding units the volume, base area and height of the prism, then, according to what has been proven, we can write:
Other materialsIn the school curriculum for a stereometry course, the study of three-dimensional figures usually begins with a simple geometric body - the polyhedron of a prism. 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 quadrangles, 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, the bases of which are 2 squares, and the side faces are represented by rectangles. Another name for this geometric figure is a straight parallelepiped.
A drawing showing a quadrangular prism is shown below.
You can also see in the picture the most important elements that make up a geometric body. These include:
Sometimes in geometry problems you can come across the concept of a section. The definition will sound like this: a section is all the points of a volumetric body belonging to a cutting plane. The section can be perpendicular (intersects the edges of the figure at an angle of 90 degrees). For a rectangular prism, a diagonal section is also considered (the maximum number of sections that can be constructed is 2), passing through 2 edges and the 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.
To find the reduced prismatic elements, various relations and formulas are used. Some of them are known from the planimetry course (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 its base and height:
V = Sbas h
Since the base of a regular tetrahedral prism is a square with side a, You can write the formula in 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 development.
From the drawing it can be seen 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 = Posn h
Taking into account that the perimeter of the square is equal to P = 4a, the formula takes the form:
Sside = 4a h
For cube:
Sside = 4a²
To calculate the total surface area of the prism, you need to add 2 base areas to the lateral area:
Sfull = Sside + 2Smain
In relation to a quadrangular regular prism, the formula looks like:
Stotal = 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, the formulas can be derived:
- base side length: a = Sside / 4h = √(V / h);
- height or side rib length: h = Sside / 4a = V / a²;
- base area: Sbas = V / h;
- side face area: Side gr = Sside / 4.
To determine how much area the 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 a prism, use the formula:
dprize = √(2a² + h²)
To understand how to apply the given relationships, you can practice and solve several simple tasks.
Examples of problems with solutions
Here are some tasks found on 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 sand level be if you move it into a container of the same shape, but with a base twice as long?
It should be reasoned 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 denote the length of the base by 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, we can equate the expressions:
10a² = 4ha²
After reducing both sides of the equation by a², we get:
As a result, the new sand level will be h = 10 / 4 = 2.5 cm.
Task 2.
ABCDA₁B₁C₁D₁ is a correct 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 at the base there is a square with a diagonal of 6√2. The diagonal of the side face has the same size, 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 a known diagonal:
a = d / √2 = 6√2 / √2 = 6
The total surface area is found using the formula for a 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, i.e. regular quadrangles, and its walls are perpendicular to horizontal surfaces, we can conclude that it is a regular prism. It is necessary to determine the area of its lateral surface.
The length of the room is a = √9 = 3 m.
The area 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 involving a rectangular prism, it is enough to be able to calculate the area and perimeter of a square and rectangle, as well as to know the formulas for finding the volume and surface area.
How to find the area of a cube
Prism volume. Problem solving
Geometry is the most powerful means for sharpening our mental faculties and enabling us to think and reason correctly.
G. Galileo
The purpose of the lesson:
- teach solving problems on calculating the volume of prisms, summarize and systematize the information students have about a prism and its elements, develop the ability to solve problems of increased complexity;
- develop logical thinking, the ability to work independently, skills of mutual control and self-control, the ability to speak and listen;
- develop the habit of constant employment in some useful activity, fostering responsiveness, hard work, and accuracy.
Lesson type: lesson on applying knowledge, skills and abilities.
Equipment: control cards, media projector, presentation “Lesson. Prism Volume”, computers.
During the classes
- Lateral ribs of the prism (Fig. 2).
- The lateral surface of the prism (Figure 2, Figure 5).
- The height of the prism (Fig. 3, Fig. 4).
- Straight prism (Figure 2,3,4).
- An inclined prism (Figure 5).
- The correct prism (Fig. 2, Fig. 3).
- Diagonal section of the prism (Figure 2).
- Diagonal of the prism (Figure 2).
- Perpendicular section of the prism (Fig. 3, Fig. 4).
- The lateral surface area of the prism.
- The total surface area of the prism.
- Prism volume.
- HOMEWORK CHECK (8 min)
- Collaboration between teacher and class (2-3 min.).
- PHYSICAL MINUTE (3 min)
- PROBLEM SOLVING (10 min)
- Students working independently on the test at the computer
Exchange notebooks, check the solution on the slides and mark it (mark 10 if the problem has been compiled)
Make up a problem based on the picture and solve it. The student defends the problem he has compiled at the board. Figure 6 and Figure 7.
Chapter 2,§3
Problem.2. The lengths of all edges of a regular triangular prism are equal to each other. Calculate the volume of the prism if its surface area is cm 2 (Fig. 8)
Chapter 2,§3
Problem 5. The base of the right prism ABCA 1B 1C1 is a right triangle ABC (angle ABC=90°), AB=4cm. Calculate the volume of the prism if the radius of the circle circumscribed about triangle ABC is 2.5 cm and the height of the prism is 10 cm. (Figure 9).
Chapter2,§3
Problem 29. The length of the side of the base of a regular quadrangular prism is 3 cm. The diagonal of the prism forms an angle of 30° with the plane of the side face. Calculate the volume of the prism (Figure 10).
Purpose: summing up the results of the theoretical warm-up (students grade each other), learning how to solve problems on the topic.
At this stage, the teacher organizes frontal work on repeating methods for solving planimetric problems and planimetric formulas. The class is divided into two groups, some solve problems, others work at the computer. Then they change. Students are asked to solve all No. 8 (orally), No. 9 (orally). Then they divide into groups and proceed to solve problems No. 14, No. 30, No. 32.
Chapter 2, §3, pages 66-67
Problem 8. All edges of a regular triangular prism are equal to each other. Find the volume of the prism if the cross-sectional area of the plane passing through the edge of the lower base and the middle of the side of the upper base is equal to cm (Fig. 11).
Chapter 2,§3, page 66-67
Problem 9. The base of a straight prism is a square, and its side edges are twice the size of the side of the base. Calculate the volume of the prism if the radius of the circle described near the cross section of the prism by a plane passing through the side of the base and the middle of the opposite side edge is equal to cm (Fig. 12)
Chapter 2,§3, page 66-67
Problem 14 The base of a straight prism is a rhombus, one of the diagonals of which is equal to its side. Calculate the perimeter of the section with a plane passing through the major diagonal of the lower base, if the volume of the prism is equal and all side faces are squares (Fig. 13).
Chapter 2,§3, page 66-67
Problem 30 ABCA 1 B 1 C 1 is a regular triangular prism, all edges of which are equal to each other, the point is the middle of edge BB 1. Calculate the radius of the circle inscribed in the section of the prism by the AOS plane, if the volume of the prism is equal to (Fig. 14).
Chapter 2,§3, page 66-67
Problem 32.In a regular quadrangular prism, the sum of the areas of the bases is equal to the area of the lateral surface. Calculate the volume of the prism if the diameter of the circle described near the cross section of the prism by a plane passing through the two vertices of the lower base and the opposite vertex of the upper base is 6 cm (Fig. 15).
While solving problems, students compare their answers with those shown by the teacher. This is a sample solution to a problem with detailed comments... Individual work of a teacher with “strong” students (10 min.).
1. The side of the base of a regular triangular prism is equal to , and the height is 5. Find the volume of the prism.
1) 152) 45 3) 104) 125) 18
2. Choose the correct statement.
1) The volume of a right prism whose base is a right triangle is equal to the product of the area of the base and the height.
2) The volume of a regular triangular prism is calculated by the formula V = 0.25a 2 h - where a is the side of the base, h is the height of the prism.
3) The volume of a straight prism is equal to half the product of the area of the base and the height.
4) The volume of a regular quadrangular prism is calculated by the formula V = a 2 h-where a is the side of the base, h is the height of the prism.
5) The volume of a regular hexagonal prism is calculated by the formula V = 1.5a 2 h, where a is the side of the base, h is the height of the prism.
3. The side of the base of a regular triangular prism is equal to . A plane is drawn through the side of the lower base and the opposite vertex of the upper base, which passes at an angle of 45° to the base. Find the volume of the prism.
1) 92) 9 3) 4,54) 2,255) 1,125
4. The base of a right prism is a rhombus, the side of which is 13, and one of the diagonals is 24. Find the volume of the prism if the diagonal of the side face is 14.
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DIRECT PRISM. SURFACE AND VOLUME OF A DIRECT PRISM.
§ 68. VOLUME OF A DIRECT PRISM.
1. Volume of a right triangular prism.
Suppose we need to find the volume of a right triangular prism, the base area of which is equal to S, and the height is equal to h= AA" = = BB" = SS" (drawing 306).
Let us separately draw the base of the prism, i.e. triangle ABC (Fig. 307, a), and build it up to a rectangle, for which we draw a straight line KM through vertex B || AC and from points A and C we lower perpendiculars AF and CE onto this line. We get rectangle ACEF. Drawing the height ВD of triangle ABC, we see that rectangle ACEF is divided into 4 right triangles. Moreover /\ ALL = /\ BCD and /\ VAF = /\ VAD. This means that the area of rectangle ACEF is twice the area of triangle ABC, i.e. equal to 2S.
To this prism with base ABC we will attach prisms with bases ALL and BAF and height h(Figure 307, b). We obtain a rectangular parallelepiped with a base
ACEF.
If we dissect this parallelepiped with a plane passing through straight lines BD and BB", we will see that the rectangular parallelepiped consists of 4 prisms with bases
BCD, ALL, BAD and BAF.
Prisms with bases BCD and VSE can be combined, since their bases are equal ( /\ ВСD = /\ BSE) and their side edges are also equal, which are perpendicular to the same plane. This means that the volumes of these prisms are equal. The volumes of prisms with bases BAD and BAF are also equal.
Thus, it turns out that the volume of a given triangular prism with a base
ABC is half the volume of a rectangular parallelepiped with base ACEF.
We know that the volume of a rectangular parallelepiped is equal to the product of the area of its base and its height, i.e. in this case it is equal to 2S h. Hence the volume of this right triangular prism is equal to S h.
The volume of a right triangular prism is equal to the product of the area of its base and its height.
2. Volume of a right polygonal prism.
To find the volume of a right polygonal prism, for example a pentagonal one, with base area S and height h, let's divide it into triangular prisms (Fig. 308).
Denoting the base areas of triangular prisms by S 1, S 2 and S 3, and the volume of a given polygonal prism by V, we obtain:
V = S 1 h+ S 2 h+ S 3 h, or
V = (S 1 + S 2 + S 3) h.
And finally: V = S h.
In the same way, the formula for the volume of a right prism with any polygon at its base is derived.
Means, The volume of any right prism is equal to the product of the area of its base and its height.
Exercises.
1. Calculate the volume of a straight prism with a parallelogram at its base using the following data:
2. Calculate the volume of a straight prism with a triangle at its base using the following data:
3. Calculate the volume of a straight prism having at its base an equilateral triangle with a side of 12 cm (32 cm, 40 cm). Prism height 60 cm.
4. Calculate the volume of a straight prism that has a right triangle at its base with legs of 12 cm and 8 cm (16 cm and 7 cm; 9 m and 6 m). The height of the prism is 0.3 m.
5. Calculate the volume of a straight prism that has a trapezoid at its base with parallel sides of 18 cm and 14 cm and a height of 7.5 cm. The height of the prism is 40 cm.
6. Calculate the volume of your classroom (physical education hall, your room).
7. The total surface of the cube is 150 cm 2 (294 cm 2, 864 cm 2). Calculate the volume of this cube.
8. The length of a building brick is 25.0 cm, its width is 12.0 cm, its thickness is 6.5 cm. a) Calculate its volume, b) Determine its weight if 1 cubic centimeter of brick weighs 1.6 g.
9. How many pieces of building bricks will be needed to build a solid brick wall in the shape of a rectangular parallelepiped 12 m long, 0.6 m wide and 10 m high? (Brick dimensions from exercise 8.)
10. The length of a cleanly cut board is 4.5 m, width - 35 cm, thickness - 6 cm. a) Calculate the volume b) Determine its weight if a cubic decimeter of the board weighs 0.6 kg.
11. How many tons of hay can be stacked in a hayloft covered with a gable roof (Fig. 309), if the length of the hayloft is 12 m, the width is 8 m, the height is 3.5 m and the height of the roof ridge is 1.5 m? (Take the specific gravity of hay as 0.2.)
12. It is required to dig a ditch 0.8 km long; in section, the ditch should have the shape of a trapezoid with bases of 0.9 m and 0.4 m, and the depth of the ditch should be 0.5 m (drawing 310). How many cubic meters of earth will have to be removed?