The height of the base of a regular triangular prism. Regular quadrangular prism

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, ability to work independently, skills of mutual control and self-control, ability to speak and listen;
• develop the habit of constant employment, in some way useful thing, education of responsiveness, hard work, 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.

1. HOMEWORK CHECK (8 min)

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 a straight 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).



2. Collaboration teachers with the class (2-3 min.).

Purpose: summing up the theoretical warm-up (students give marks each other), studying ways to solve problems on a topic.

3. PHYSICAL MINUTE (3 min)
4. PROBLEM SOLVING (10 min)

On 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 are correct 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 the problem with detailed comments... Individual work teachers with “strong” students (10 min.).

5. Independent work students working on a test at the computer

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.

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 triangle. Moreover /\ ALL = /\ BCD and /\ VAF = /\ VAD. This means that the area of ​​the rectangle ACEF is doubled more area 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 rectangular parallelepiped with ACEF base.

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 with a base 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. Full surface cube is equal to 150 cm 2 (294 cm 2, 864 cm 2). Calculate the volume of this cube.

8. Length building bricks- 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 brick weighs 1.6 g.

9. How many pieces of building bricks will be needed to build a solid brick wall, having 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? ( Specific gravity take 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?

Schoolchildren who are preparing for passing the Unified State Exam in mathematics, you should definitely learn how to solve problems to find the area of ​​a straight line and correct prism. Many years of practice confirm the fact that many students consider such geometry tasks to be quite difficult.

At the same time, high school students with any level of training should be able to find the area and volume of a regular and straight prism. Only in this case will they be able to count on receiving competitive scores based on the results of passing the Unified State Exam.

Key Points to Remember

• If the lateral edges of a prism are perpendicular to the base, it is called a straight line. All side faces of this figure are rectangles. The height of a straight prism coincides with its edge.
• A regular prism is one whose side edges are perpendicular to the base in which the regular polygon is located. Side faces of this figure are equal rectangles. A correct prism is always straight.

Preparing for the unified state exam together with Shkolkovo is the key to your success!

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Specialists educational project“Shkolkovo” proposes to go from simple to complex: first we give theory, basic formulas, theorems and elementary problems with solutions, and then gradually move on to expert-level tasks.

Basic information is systematized and clearly presented in the “Theoretical Information” section. If you have already managed to repeat the necessary material, we recommend that you practice solving problems on finding the area and volume of a right prism. The "Catalog" section presents large selection exercises varying degrees difficulties.

Try to calculate the area of ​​a straight and regular prism or right now. Analyze any task. If it does not cause any difficulties, you can safely move on to expert-level exercises. And if certain difficulties do arise, we recommend that you regularly prepare for the Unified State Exam online together with the Shkolkovo mathematical portal, and tasks on the topic “Straight and Regular Prism” will be easy for you.

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’ = CC’ (Fig. 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, \(\Delta\)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 materials

Job type: 8
Theme: Prism

Condition

In a regular triangular prism ABCA_1B_1C_1, the sides of the base are 4 and the side edges are 10. Find the cross-sectional area of ​​the prism by the plane passing through the midpoints of the edges AB, AC, A_1B_1 and A_1C_1.

Show solution

Solution

Consider the following figure.

The segment MN is midline triangle A_1B_1C_1, therefore MN = \frac12 B_1C_1=2. Likewise, KL=\frac12BC=2. In addition, MK = NL = 10. It follows that the quadrilateral MNLK is a parallelogram. Since MK\parallel AA_1, then MK\perp ABC and MK\perp KL. Therefore, the quadrilateral MNLK is a rectangle. S_(MNLK) = MK\cdot KL = 10\cdot 2 = 20.

Answer

Job type: 8
Theme: Prism

Condition

The volume of a regular quadrangular prism ABCDA_1B_1C_1D_1 is 24 . Point K is the middle of edge CC_1. Find the volume of the pyramid KBCD.

Show solution

Solution

According to the condition, KC is the height of the pyramid KBCD. CC_1 is the height of the prism ABCDA_1B_1C_1D_1 .

Since K is the midpoint of CC_1, then KC=\frac12CC_1. Let CC_1=H , then KC=\frac12H. Note also that S_(BCD)=\frac12S_(ABCD). Then, V_(KBCD)= \frac13S_(BCD)\cdot\frac(H)(2)= \frac13\cdot\frac12S_(ABCD)\cdot\frac(H)(2)= \frac(1)(12)\cdot S_(ABCD)\cdot H= \frac(1)(12)V_(ABCDA_1B_1C_1D_1). Hence, V_(KBCD)=\frac(1)(12)\cdot24=2.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 8
Theme: Prism

Condition

Find the lateral surface area of ​​a regular hexagonal prism whose base side is 6 and height is 8.

Show solution

Solution

The area of ​​the lateral surface of the prism is found by the formula S side. = P basic · h = 6a\cdot h, where P basic. and h are, respectively, the perimeter of the base and the height of the prism, equal to 8, and a is the side of a regular hexagon, equal to 6. Therefore, S side. = 6\cdot 6\cdot 8 = 288.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 8
Theme: Prism

Condition

Water was poured into a vessel shaped like a regular triangular prism. The water level reaches 40 cm. At what height will the water level be if it is poured into another vessel of the same shape, whose side of the base is twice as large as the first? Express your answer in centimeters.

Show solution

Solution

Let a be the side of the base of the first vessel, then 2 a is the side of the base of the second vessel. By condition, the volume of liquid V in the first and second vessels is the same. Let us denote by H the level to which the liquid has risen in the second vessel. Then V= \frac12\cdot a^2\cdot\sin60^(\circ)\cdot40= \frac(a^2\sqrt3)(4)\cdot40, And, V=\frac((2a)^2\sqrt3)(4)\cdot H. From here \frac(a^2\sqrt3)(4)\cdot40=\frac((2a)^2\sqrt3)(4)\cdot H, 40=4H, H=10.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 8
Theme: Prism

Condition

In a regular hexagonal prism ABCDEFA_1B_1C_1D_1E_1F_1 all edges are equal to 2. Find the distance between points A and E_1.

Show solution

Solution

Triangle AEE_1 is rectangular, since edge EE_1 is perpendicular to the plane of the base of the prism, angle AEE_1 will be a right angle.

Then, by the Pythagorean theorem, AE_1^2 = AE^2 + EE_1^2. Let's find AE from triangle AFE using the cosine theorem. Each interior angle of a regular hexagon is 120^(\circ). Then AE^2= AF^2+FE^2-2\cdot AF\cdot FE\cdot\cos120^(\circ)= 2^2+2^2-2\cdot2\cdot2\cdot\left (-\frac12 \right).

Hence, AE^2=4+4+4=12,

AE_1^2=12+4=16,

AE_1=4.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 8
Theme: Prism

Condition

Find the lateral surface area of ​​a straight prism, at the base of which lies a rhombus with diagonals equal to 4\sqrt5 and 8, and a side edge equal to 5.

Show solution

Solution

The area of ​​the lateral surface of a straight prism is found by the formula S side. = P basic · h = 4a\cdot h, where P basic. and h, respectively, the perimeter of the base and the height of the prism, equal to 5, and a is the side of the rhombus. Let's find the side of the rhombus using the fact that the diagonals of the rhombus ABCD are mutually perpendicular and bisected by the point of intersection.