In physics, a triangular prism made of glass is often used to study the spectrum of white light because it can resolve it into its individual components. In this article we will consider the volume formula
What is a triangular prism?
Before giving the volume formula, let's consider the properties of this figure.
To get this, you need to take a triangle of any shape and move it parallel to itself to some distance. The vertices of the triangle in the initial and final positions should be connected by straight segments. Received volumetric figure called a triangular prism. It consists of five sides. Two of them are called bases: they are parallel and equal to each other. The bases of the prism in question are triangles. The three remaining sides are parallelograms.
In addition to the sides, the prism in question is characterized by six vertices (three for each base) and nine edges (6 edges lie in the planes of the bases and 3 edges are formed by the intersection of the sides). If the side edges are perpendicular to the bases, then such a prism is called rectangular.
Difference triangular prism from all other figures of this class is that it is always convex (four-, five-, ..., n-gonal prisms may also be concave).
This rectangular figure, which is based on equilateral triangle.
Volume of a general triangular prism
How to find the volume of a triangular prism? Formula in general view similar to that for any type of prism. It has the following mathematical notation:
Here h is the height of the figure, that is, the distance between its bases, S o is the area of the triangle.
The value of S o can be found if some parameters for the triangle are known, for example, one side and two angles or two sides and one angle. The area of a triangle is equal to half the product of its height and the length of the side by which this height is lowered.
As for the height h of the figure, it is easiest to find for rectangular prism. IN the latter case h coincides with the length of the side edge.
Volume of a regular triangular prism
General formula volume of a triangular prism, which is given in the previous section of the article, can be used to calculate the corresponding value for a regular triangular prism. Since its base is an equilateral triangle, its area is equal to:
Anyone can get this formula if they remember that in an equilateral triangle all angles are equal to each other and amount to 60 o. Here the symbol a is the length of the side of the triangle.
The height h is the length of the edge. It is in no way connected with the base of a regular prism and can take arbitrary values. As a result, the formula for the volume of a triangular prism is the right kind looks like that:
Having calculated the root, you can rewrite this formula as follows:
Thus, to find the volume of a regular prism with triangular base, it is necessary to square the side of the base, multiply this value by the height and multiply the resulting value by 0.433.
Job type: 8
Theme: Prism
Condition
In a regular triangular prism ABCA_1B_1C_1 the sides of the base are 4, and side ribs are equal to 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 solutionSolution
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.
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.
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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 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.
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 the right 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 solutionSolution
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.
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Then, by the Pythagorean theorem, AE_1^2 = AE^2 + EE_1^2. Let's find AE from triangle AFE using the cosine theorem. Every internal corner 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.
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.
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 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(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 rectangular parallelepiped with ACEF base.
We know that the volume of a rectangular parallelepiped equal to the product area of its base by height, i.e. in in this case 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 line polygonal prism, for example pentagonal, 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 divisible by two equal triangle. 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 equally And equal heights are equal (because when nested 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 materialsDefinition.
This is a hexagon whose bases are two equal square, and the side faces are equal rectangles
Side rib- This common side 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. Diagonal section of the correct 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
- Side surface- the sum of the areas of all lateral faces of the prism
- Full 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
- 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 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 regular polygon, and the side ribs 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 indicate the action of retrieving square root the symbol is used in solving problems√ .
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
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 with the diagonal of the base and the height of the prism right triangle. 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.
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 with a trapezoid at its base parallel sides 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(gym, 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 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? ( 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?