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 materialsSchoolchildren who are preparing to take the Unified State Exam in mathematics should definitely learn how to solve problems on finding the area of a straight and regular 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. The side faces of this figure are equal rectangles. A correct prism is always straight.
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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.
What is the volume of a prism and how to find it
The volume of a prism is the product of the area of its base and its height.
However, we know that at the base of the prism there can be a triangle, a square or some other polyhedron.
Therefore, to find the volume of a prism, you simply need to calculate the area of the base of the prism, and then multiply this area by its height.
That is, if there is a triangle at the base of the prism, then first you need to find the area of the triangle. If the base of the prism is a square or other polygon, then first you need to look for the area of the square or other polygon.
It should be remembered that the height of the prism is the perpendicular drawn to the bases of the prism.
What is a prism
Now let's remember the definition of a prism.
A prism is a polygon, two faces (bases) of which are in parallel planes, and all edges located outside these faces are parallel.
To put it simply:
A prism is any geometric figure that has two equal bases and flat faces.
The name of a prism depends on the shape of its base. When the base of a prism is a triangle, then such a prism is called triangular. A polyhedral prism is a geometric figure whose base is a polyhedron. Also, a prism is a type of cylinder.
What types of prisms are there?
If we look at the picture above, we will see that prisms are straight, regular and oblique.
Exercise
1. Which prism is called correct?
2. Why is it called that?
3. What is the name of a prism whose bases are regular polygons?
4. What is the height of this figure?
5. What is a prism whose edges are not perpendicular called?
6. Define a triangular prism.
7. Can a prism be a parallelepiped?
8. What geometric figure is called a semiregular polygon?
What elements does a prism consist of?
A prism consists of elements such as a lower and upper base, side faces, edges and vertices.
Both bases of the prism lie in planes and are parallel to each other.
The side faces of the pyramid are parallelograms.
The lateral surface of a pyramid is the sum of its lateral faces.
The common sides of the side faces are nothing more than the side edges of a given figure.
The height of the pyramid is the segment connecting the planes of the bases and perpendicular to them.
Prism properties
A geometric figure, like a prism, has a number of properties. Let's take a closer look at these properties:
Firstly, the bases of a prism are equal polygons;
Secondly, the side faces of a prism are presented in the form of a parallelogram;
Thirdly, this geometric figure has parallel and equal edges;
Fourthly, the total surface area of the prism is:
Now let's look at the theorem, which provides the formula used to calculate the lateral surface area and proof.
Have you ever thought about such an interesting fact that not only a geometric body, but also other objects around us can be a prism? Even an ordinary snowflake, depending on the temperature, can turn into an ice prism, taking the shape of a hexagonal figure.
But calcite crystals have such a unique phenomenon as breaking up into fragments and taking on the shape of a parallelepiped. And what’s most amazing is that no matter how small the calcite crystals are crushed into, the result is always the same: they turn into tiny parallelepipeds.
It turns out that the prism has gained popularity not only in mathematics, demonstrating its geometric body, but also in the field of art, since it is the basis of paintings created by such great artists as P. Picasso, Braque, Griss and others.
In 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
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. The resulting volumetric figure is 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.
The difference between a triangular prism and all other figures of this class is that it is always convex (four-, five-, ..., n-gonal prisms can also be concave).
This is a rectangular figure with an equilateral triangle at its base.
Volume of a general triangular prism
How to find the volume of a triangular prism? The formula in general is similar to that for a prism of any type. 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 it for a rectangular prism. In the latter case, h coincides with the length of the side edge.
Volume of a regular triangular prism
The general formula for the 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 of the correct type looks like this:
Having calculated the root, you can rewrite this formula as follows:
Thus, to find the volume of a regular prism with a 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.
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