The base area of ​​a regular hexagonal prism is 3. The largest diagonal of a regular hexagonal prism, having length d, makes an angle α with the lateral edge of the prism


Different prisms are different from each other. At the same time, they have a lot in common. To find the area of ​​the base of the prism, you will need to understand what type it has.

General theory

A prism is any polyhedron whose sides have the shape of a parallelogram. Moreover, its base can be any polyhedron - from a triangle to an n-gon. Moreover, the bases of the prism are always equal to each other. What does not apply to the side faces is that they can vary significantly in size.

When solving problems, not only the area of ​​the base of the prism is encountered. It may require knowledge of the lateral surface, that is, all the faces that are not bases. The complete surface will be the union of all the faces that make up the prism.

Sometimes problems involve height. It is perpendicular to the bases. The diagonal of a polyhedron is a segment that connects in pairs any two vertices that do not belong to the same face.

It should be noted that the base area of ​​a straight or inclined prism does not depend on the angle between them and the side faces. If they have the same figures on the top and bottom faces, then their areas will be equal.

Triangular prism

It has at its base a figure with three vertices, that is, a triangle. As you know, it can be different. If so, it is enough to remember that its area is determined by half the product of the legs.

The mathematical notation looks like this: S = ½ av.

To find out the area of ​​the base in general, the formulas are useful: Heron and the one in which half of the side is taken by the height drawn to it.

The first formula should be written as follows: S = √(р (р-а) (р-в) (р-с)). This notation contains a semi-perimeter (p), that is, the sum of three sides divided by two.

Second: S = ½ n a * a.

If you want to find out the area of ​​the base of a triangular prism, which is regular, then the triangle turns out to be equilateral. There is a formula for it: S = ¼ a 2 * √3.

Quadrangular prism

Its base is any of the known quadrangles. It can be a rectangle or square, parallelepiped or rhombus. In each case, in order to calculate the area of ​​the base of the prism, you will need your own formula.

If the base is a rectangle, then its area is determined as follows: S = ab, where a, b are the sides of the rectangle.

When it comes to a quadrangular prism, the area of ​​the base of a regular prism is calculated using the formula for a square. Because it is he who lies at the foundation. S = a 2.

In the case when the base is a parallelepiped, the following equality will be needed: S = a * n a. It happens that the side of a parallelepiped and one of the angles are given. Then, to calculate the height, you will need to use an additional formula: n a = b * sin A. Moreover, angle A is adjacent to side “b”, and height n is opposite to this angle.

If there is a rhombus at the base of the prism, then to determine its area you will need the same formula as for a parallelogram (since it is a special case of it). But you can also use this: S = ½ d 1 d 2. Here d 1 and d 2 are two diagonals of the rhombus.

Regular pentagonal prism

This case involves dividing the polygon into triangles, the areas of which are easier to find out. Although it happens that figures can have a different number of vertices.

Since the base of the prism is a regular pentagon, it can be divided into five equilateral triangles. Then the area of ​​the base of the prism is equal to the area of ​​one such triangle (the formula can be seen above), multiplied by five.

Regular hexagonal prism

Using the principle described for a pentagonal prism, it is possible to divide the hexagon of the base into 6 equilateral triangles. The formula for the base area of ​​such a prism is similar to the previous one. Only it should be multiplied by six.

The formula will look like this: S = 3/2 a 2 * √3.

Tasks

No. 1. Given a regular straight line, its diagonal is 22 cm, the height of the polyhedron is 14 cm. Calculate the area of ​​the base of the prism and the entire surface.

Solution. The base of the prism is a square, but its side is unknown. You can find its value from the diagonal of the square (x), which is related to the diagonal of the prism (d) and its height (h). x 2 = d 2 - n 2. On the other hand, this segment “x” is the hypotenuse in a triangle whose legs are equal to the side of the square. That is, x 2 = a 2 + a 2. Thus it turns out that a 2 = (d 2 - n 2)/2.

Substitute the number 22 instead of d, and replace “n” with its value - 14, it turns out that the side of the square is 12 cm. Now just find out the area of ​​the base: 12 * 12 = 144 cm 2.

To find out the area of ​​the entire surface, you need to add twice the base area and quadruple the side area. The latter can be easily found using the formula for a rectangle: multiply the height of the polyhedron and the side of the base. That is, 14 and 12, this number will be equal to 168 cm 2. The total surface area of ​​the prism turns out to be 960 cm 2.

Answer. The area of ​​the base of the prism is 144 cm 2. The entire surface is 960 cm 2.

No. 2. Given At the base there is a triangle with a side of 6 cm. In this case, the diagonal of the side face is 10 cm. Calculate the areas: the base and the side surface.

Solution. Since the prism is regular, its base is an equilateral triangle. Therefore, its area turns out to be 6 squared, multiplied by ¼ and the square root of 3. A simple calculation leads to the result: 9√3 cm 2. This is the area of ​​one base of the prism.

All side faces are the same and are rectangles with sides of 6 and 10 cm. To calculate their areas, just multiply these numbers. Then multiply them by three, because the prism has exactly that many side faces. Then the area of ​​the lateral surface of the wound turns out to be 180 cm 2.

Answer. Areas: base - 9√3 cm 2, lateral surface of the prism - 180 cm 2.

Regular hexagonal prism- a prism, at the bases of which there are two regular hexagons, and all the side faces are strictly perpendicular to these bases.

  • A B C D E F A1 B1 C1 D1 E1 F1 - regular hexagonal prism
  • a- length of the side of the base of the prism
  • h- length of the side edge of the prism
  • Smain- area of ​​the prism base
  • Sside .- area of ​​the lateral face of the prism
  • Sfull- total surface area of ​​the prism
  • Vprisms- prism volume

Prism base area

At the bases of the prism there are regular hexagons with sides a. According to the properties of a regular hexagon, the area of ​​the bases of the prism is equal to

This way

Smain= 3 3 2 a2


Thus it turns out that SA B C D E F= SA1 B1 C1 D1 E1 F1 = 3 3 2 a2

Total surface area of ​​the prism

The total surface area of ​​a prism is the sum of the areas of the lateral faces of the prism and the areas of its bases. Each of the lateral faces of the prism is a rectangle with sides a And h. Therefore, according to the properties of the rectangle

Sside .= a ⋅ h

A prism has six side faces and two bases, therefore, its total surface area is equal to

Sfull= 6 ⋅ Sside .+ 2 ⋅ Smain= 6 ⋅ a ⋅ h + 2 ⋅ 3 3 2 a2

Prism volume

The volume of a prism is calculated as the product of the area of ​​its base and its height. The height of a regular prism is any of its lateral edges, for example, the edge A A1 . At the base of a regular hexagonal prism there is a regular hexagon, the area of ​​which is known to us. We get

Vprisms= Smain⋅A A1 = 3 3 2 a2 ⋅ h

Regular hexagon at prism bases

We consider the regular hexagon ABCDEF lying at the base of the prism.

We draw segments AD, BE and CF. Let the intersection of these segments be point O.

According to the properties of a regular hexagon, triangles AOB, BOC, COD, DOE, EOF, FOA are regular triangles. It follows that

A O = O D = E O = O B = C O = O F = a

We draw a segment AE intersecting with a segment CF at point M. The triangle AEO is isosceles, in it A O = O E = a , ∠ E O A = 120 . According to the properties of an isosceles triangle.

A E = a ⋅ 2 (1 − cos E O A )− − − − − − − − − − − − = 3 ⋅ a

Similarly, we come to the conclusion that A C = C E = 3 ⋅ a, F M = M O = 1 2 ⋅ a.

We find E A1

In a triangleA E A1 :

  • A A1 = h
  • A E = 3 ⋅ a- as we just found out
  • ∠ E A A1 = 90

A E A1

E A1 = A A2 1 +A E2 − − − − − − − − − − = h2 + 3 ⋅ a2 − − − − − − − −

If h = a, so then E A1 = 2 ⋅ a

F B1 = A C1 = B D1 = C E1 = D F1 = h2 + 3 ⋅ a2 − − − − − − − − .

We findEB 1

In a triangle B E B1 :

  • B B1 = h
  • B E = 2 ⋅ a- because E O = O B = a
  • ∠ E B B1 = 90 - according to the properties of the correct straightness

Thus, it turns out that the triangle B E B1 rectangular. According to the properties of a right triangle

E B1 = B B2 1 +B E2 − − − − − − − − − − = h2 + 4 ⋅ a2 − − − − − − − −

If h = a, so then

E B1 = 5 ⋅ a

After similar reasoning we obtain that F C1 = A D1 = B E1 = C F1 = D A1 = h2 + 4 ⋅ a2 − − − − − − − − .

We find O F1

In a triangle F O F1 :

  • F F1 = h
  • F O = a
  • ∠ O F F1 = 90 - according to the properties of a regular prism

Thus, it turns out that the triangle F O F1 rectangular. According to the properties of a right triangle

O F1 = F F2 1 + O F2 − − − − − − − − − − = h2 + a2 − − − − − −

If h = a, so then

The site has already discussed some types of problems in stereometry, which are included in a single bank of tasks for the mathematics exam.For example, tasks about .

A prism is called regular if its sides are perpendicular to the bases and a regular polygon lies at the bases. That is, a regular prism is a straight prism with a regular polygon at its base.

A regular hexagonal prism has a regular hexagon at the base, the side faces are rectangles.

In this article you will find problems to solve a prism, the base of which is a regular hexagon. There are no special features or difficulties in the solution. What's the point? Given a regular hexagonal prism, you need to calculate the distance between two vertices or find a given angle. The problems are actually simple; in the end, the solution comes down to finding an element in a right triangle.

The Pythagorean theorem is used and. Knowledge of the definitions of trigonometric functions in a right triangle is required.

Be sure to look at the information about the regular hexagon in.You will also need the skill of extracting a large number of them. You can solve polyhedra, they also calculated the distance between vertices and angles.

Briefly: what is a regular hexagon?

It is known that in a regular hexagon the sides are equal. In addition, the angles between the sides are also equal.

*Opposite sides are parallel.

Additional Information

The radius of a circle circumscribed about a regular hexagon is equal to its side. *This is confirmed very simply: if we connect the opposite vertices of a hexagon, we get six equal equilateral triangles. Why equilateral?

Each triangle has an angle with its vertex lying in the center equal to 60 0 (360:6=60). Since the two sides of a triangle having a common vertex in the center are equal (these are the radii of the circumscribed circle), then each angle at the base of such an isosceles triangle is also equal to 60 degrees.

That is, a regular hexagon, figuratively speaking, consists of six equal equilateral triangles.

What other fact should be noted that is useful for solving problems? The vertex angle of a hexagon (the angle between its adjacent sides) is 120 degrees.

*We deliberately did not touch upon the formulas for a regular N-gon. We will consider these formulas in detail in the future; they are simply not needed here.

Let's consider the tasks:

272533. In a regular hexagonal prism ABCDEFA 1 B 1 C 1 D 1 E 1 F 1 all edges are equal 48. Find the distance between points A and E 1 .

Consider right triangle AA 1 E 1 . According to the Pythagorean theorem:

*The angle between the sides of a regular hexagon is 120 degrees.

Section AE 1 is the hypotenuse, AA 1 and A 1 E 1 legs. Rib AA 1 we know. Catet A 1 E 1 we can find using using .

Theorem: The square of any side of a triangle is equal to the sum of the squares of its two other sides without twice the product of these sides by the cosine of the angle between them.

Hence

According to the Pythagorean theorem:

Answer: 96

*Please note that squaring 48 is not necessary.

In a regular hexagonal prism ABCDEFA 1 B 1 C 1 D 1 E 1 F 1 all edges are 35. Find the distance between points B and E.

It is said that all edges are equal to 35, that is, the side of the hexagon lying at the base is equal to 35. And also, as already said, the radius of the circle described around it is equal to the same number.

Thus,

Answer: 70

273353. In a regular hexagonal prism ABCDEFA 1 B 1 C 1 D 1 E 1 F 1 all edges are equal to forty roots of five. Find the distance between points B and E 1.

Consider right triangle BB 1 E 1 . According to the Pythagorean theorem:

Segment B 1 E 1 is equal to two radii of the circle circumscribed about a regular hexagon, and its radius is equal to the side of the hexagon, that is

Thus,


Answer: 200

273683. In a regular hexagonal prism ABCDEFA 1 B 1 C 1 D 1 E 1 F 1 all edges are equal to 45. Find the tangent of the angle AD 1 D.

Consider a right triangle ADD 1 in which AD equal to the diameter of a circle circumscribed around the base. It is known that the radius of a circle circumscribed around a regular hexagon is equal to its side.

Thus,

Answer: 2

In a regular hexagonal prism ABCDEFA 1 B 1 C 1 D 1 E 1 F 1 all edges are equal 23. Find the angle DAB. Give your answer in degrees.

Consider a regular hexagon:

In it, the angles between the sides are 120°. Means,

The length of the edge itself does not matter; it does not affect the angle.

Answer: 60

In a regular hexagonal prism ABCDEFA 1 B 1 C 1 D 1 E 1 F 1 all edges are equal to 10. Find the angle AC 1 C. Give the answer in degrees.

Consider the right triangle AC 1 C:

Let's find A.C.. In a regular hexagon, the angles between its sides are equal to 120 degrees, then according to the cosine theorem for a triangleABC:


Thus,

So angle AC 1 C is equal to 60 degrees.

Answer: 60

274453. In a regular hexagonal prism ABCDEFA 1 B 1 C 1 D 1 E 1 F 1 all edges are equal to 10. Find the angle AC 1 C. Give the answer in degrees.

From each vertex of a prism, for example from vertex A 1 (Fig.), three diagonals can be drawn (A 1 E, A 1 D, A 1 C).

They are projected onto the plane ABCDEF by the diagonals of the base (AE, AD, AC). Of the inclined ones A 1 E, A 1 D, A 1 C, the largest is the one with the largest projection. Consequently, the largest of the three diagonals taken is A 1 D (in the prism there are also diagonals equal to A 1 D, but there are no larger ones).

From triangle A 1 AD, where ∠DA 1 A = α and A 1 D = d , we find H=AA 1 = d cos α ,
AD= d sin α .

The area of ​​an equilateral triangle AOB is equal to 1/4 AO 2 √3. Hence,

S ocn. = 6 1/4 AO 2 √3 = 6 1/4 (AD/2) 2 √3.

Volume V = S H = 3√ 3 / 8 AD 2 AA 1

Answer: 3√ 3 / 8 d 3 sin 2 α cos α .

Comment . To depict a regular hexagon (the base of a prism), you can construct an arbitrary parallelogram BCDO. Laying out the segments OA = OD, OF = OC and OE = OB on the continuations of lines DO, CO, BO, we obtain the hexagon ABCDEF. Point O represents the center.




In the fifth century BC, the ancient Greek philosopher Zeno of Elea formulated his famous aporias, the most famous of which is the “Achilles and the Tortoise” aporia. Here's what it sounds like:

Let's say Achilles runs ten times faster than the tortoise and is a thousand steps behind it. During the time it takes Achilles to run this distance, the tortoise will crawl a hundred steps in the same direction. When Achilles runs a hundred steps, the tortoise crawls another ten steps, and so on. The process will continue ad infinitum, Achilles will never catch up with the tortoise.

This reasoning became a logical shock for all subsequent generations. Aristotle, Diogenes, Kant, Hegel, Hilbert... They all considered Zeno's aporia in one way or another. The shock was so strong that " ... discussions continue to this day; the scientific community has not yet been able to come to a common opinion on the essence of paradoxes ... mathematical analysis, set theory, new physical and philosophical approaches were involved in the study of the issue; none of them became a generally accepted solution to the problem..."[Wikipedia, "Zeno's Aporia". Everyone understands that they are being fooled, but no one understands what the deception consists of.

From a mathematical point of view, Zeno in his aporia clearly demonstrated the transition from quantity to . This transition implies application instead of permanent ones. As far as I understand, the mathematical apparatus for using variable units of measurement has either not yet been developed, or it has not been applied to Zeno’s aporia. Applying our usual logic leads us into a trap. We, due to the inertia of thinking, apply constant units of time to the reciprocal value. From a physical point of view, this looks like time slowing down until it stops completely at the moment when Achilles catches up with the turtle. If time stops, Achilles can no longer outrun the tortoise.

If we turn our usual logic around, everything falls into place. Achilles runs at a constant speed. Each subsequent segment of his path is ten times shorter than the previous one. Accordingly, the time spent on overcoming it is ten times less than the previous one. If we apply the concept of “infinity” in this situation, then it would be correct to say “Achilles will catch up with the turtle infinitely quickly.”

How to avoid this logical trap? Remain in constant units of time and do not switch to reciprocal units. In Zeno's language it looks like this:

In the time it takes Achilles to run a thousand steps, the tortoise will crawl a hundred steps in the same direction. During the next time interval equal to the first, Achilles will run another thousand steps, and the tortoise will crawl a hundred steps. Now Achilles is eight hundred steps ahead of the tortoise.

This approach adequately describes reality without any logical paradoxes. But this is not a complete solution to the problem. Einstein’s statement about the irresistibility of the speed of light is very similar to Zeno’s aporia “Achilles and the Tortoise”. We still have to study, rethink and solve this problem. And the solution must be sought not in infinitely large numbers, but in units of measurement.

Another interesting aporia of Zeno tells about a flying arrow:

A flying arrow is motionless, since at every moment of time it is at rest, and since it is at rest at every moment of time, it is always at rest.

In this aporia, the logical paradox is overcome very simply - it is enough to clarify that at each moment of time a flying arrow is at rest at different points in space, which, in fact, is motion. Another point needs to be noted here. From one photograph of a car on the road it is impossible to determine either the fact of its movement or the distance to it. To determine whether a car is moving, you need two photographs taken from the same point at different points in time, but you cannot determine the distance from them. To determine the distance to a car, you need two photographs taken from different points in space at one point in time, but from them you cannot determine the fact of movement (of course, you still need additional data for calculations, trigonometry will help you). What I want to draw special attention to is that two points in time and two points in space are different things that should not be confused, because they provide different opportunities for research.

Wednesday, July 4, 2018

The differences between set and multiset are described very well on Wikipedia. Let's see.

As you can see, “there cannot be two identical elements in a set,” but if there are identical elements in a set, such a set is called a “multiset.” Reasonable beings will never understand such absurd logic. This is the level of talking parrots and trained monkeys, who have no intelligence from the word “completely”. Mathematicians act as ordinary trainers, preaching to us their absurd ideas.

Once upon a time, the engineers who built the bridge were in a boat under the bridge while testing the bridge. If the bridge collapsed, the mediocre engineer died under the rubble of his creation. If the bridge could withstand the load, the talented engineer built other bridges.

No matter how mathematicians hide behind the phrase “mind me, I’m in the house,” or rather, “mathematics studies abstract concepts,” there is one umbilical cord that inextricably connects them with reality. This umbilical cord is money. Let us apply mathematical set theory to mathematicians themselves.

We studied mathematics very well and now we are sitting at the cash register, giving out salaries. So a mathematician comes to us for his money. We count out the entire amount to him and lay it out on our table in different piles, into which we put bills of the same denomination. Then we take one bill from each pile and give the mathematician his “mathematical set of salary.” Let us explain to the mathematician that he will receive the remaining bills only when he proves that a set without identical elements is not equal to a set with identical elements. This is where the fun begins.

First of all, the logic of the deputies will work: “This can be applied to others, but not to me!” Then they will begin to reassure us that bills of the same denomination have different bill numbers, which means they cannot be considered the same elements. Okay, let's count salaries in coins - there are no numbers on the coins. Here the mathematician will begin to frantically remember physics: different coins have different amounts of dirt, the crystal structure and arrangement of atoms is unique for each coin...

And now I have the most interesting question: where is the line beyond which the elements of a multiset turn into elements of a set and vice versa? Such a line does not exist - everything is decided by shamans, science is not even close to lying here.

Look here. We select football stadiums with the same field area. The areas of the fields are the same - which means we have a multiset. But if we look at the names of these same stadiums, we get many, because the names are different. As you can see, the same set of elements is both a set and a multiset. Which is correct? And here the mathematician-shaman-sharpist pulls out an ace of trumps from his sleeve and begins to tell us either about a set or a multiset. In any case, he will convince us that he is right.

To understand how modern shamans operate with set theory, tying it to reality, it is enough to answer one question: how do the elements of one set differ from the elements of another set? I'll show you, without any "conceivable as not a single whole" or "not conceivable as a single whole."

Sunday, March 18, 2018

The sum of the digits of a number is a dance of shamans with a tambourine, which has nothing to do with mathematics. Yes, in mathematics lessons we are taught to find the sum of the digits of a number and use it, but that’s why they are shamans, to teach their descendants their skills and wisdom, otherwise shamans will simply die out.

Do you need proof? Open Wikipedia and try to find the page "Sum of digits of a number." She doesn't exist. There is no formula in mathematics that can be used to find the sum of the digits of any number. After all, numbers are graphic symbols with which we write numbers, and in the language of mathematics the task sounds like this: “Find the sum of graphic symbols representing any number.” Mathematicians cannot solve this problem, but shamans can do it easily.

Let's figure out what and how we do in order to find the sum of the digits of a given number. And so, let us have the number 12345. What needs to be done in order to find the sum of the digits of this number? Let's consider all the steps in order.

1. Write down the number on a piece of paper. What have we done? We have converted the number into a graphical number symbol. This is not a mathematical operation.

2. We cut one resulting picture into several pictures containing individual numbers. Cutting a picture is not a mathematical operation.

3. Convert individual graphic symbols into numbers. This is not a mathematical operation.

4. Add the resulting numbers. Now this is mathematics.

The sum of the digits of the number 12345 is 15. These are the “cutting and sewing courses” taught by shamans that mathematicians use. But that is not all.

From a mathematical point of view, it does not matter in which number system we write a number. So, in different number systems the sum of the digits of the same number will be different. In mathematics, the number system is indicated as a subscript to the right of the number. With the large number 12345, I don’t want to fool my head, let’s consider the number 26 from the article about. Let's write this number in binary, octal, decimal and hexadecimal number systems. We won't look at every step under a microscope; we've already done that. Let's look at the result.

As you can see, in different number systems the sum of the digits of the same number is different. This result has nothing to do with mathematics. It’s the same as if you determined the area of ​​a rectangle in meters and centimeters, you would get completely different results.

Zero looks the same in all number systems and has no sum of digits. This is another argument in favor of the fact that. Question for mathematicians: how is something that is not a number designated in mathematics? What, for mathematicians nothing exists except numbers? I can allow this for shamans, but not for scientists. Reality is not just about numbers.

The result obtained should be considered as proof that number systems are units of measurement for numbers. After all, we cannot compare numbers with different units of measurement. If the same actions with different units of measurement of the same quantity lead to different results after comparing them, then this has nothing to do with mathematics.

What is real mathematics? This is when the result of a mathematical operation does not depend on the size of the number, the unit of measurement used and on who performs this action.

Sign on the door He opens the door and says:

Oh! Isn't this the women's restroom?
- Young woman! This is a laboratory for the study of the indephilic holiness of souls during their ascension to heaven! Halo on top and arrow up. What other toilet?

Female... The halo on top and the arrow down are male.

If such a work of design art flashes before your eyes several times a day,

Then it’s not surprising that you suddenly find a strange icon in your car:

Personally, I make an effort to see minus four degrees in a pooping person (one picture) (a composition of several pictures: a minus sign, the number four, a designation of degrees). And I don’t think this girl is a fool who doesn’t know physics. She just has a strong stereotype of perceiving graphic images. And mathematicians teach us this all the time. Here's an example.

1A is not “minus four degrees” or “one a”. This is "pooping man" or the number "twenty-six" in hexadecimal notation. Those people who constantly work in this number system automatically perceive a number and a letter as one graphic symbol.

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