Often pupils indignantly ask: "How can I use it in my life?". On any subject of each subject. The theme of the volume of the parallelepiped is not an exception. And here you can just say: "It will come in handy."

How, for example, can I find out if the package will fit in the mail box? Of course, you can choose the right method by trial and error. And if there is no such possibility? Then the calculation will come to the rescue. Knowing the capacity of the box, you can calculate the volume of the parcel (at least approximately) and answer the question posed.

## Parallelepiped and its types

If you literally translate its name from the ancient Greek, it turns out that this is a figure consisting of parallel planes. There are such equivalent definitions of a parallelepiped:

• a prism with a base in the form of a parallelogram;
• a polyhedron, each face of which is a parallelogram.

Its species are distinguished depending on which figure lies at its base and how the lateral ribs are directed. In the general case, we speak ofinclined parallelepiped. whose base and all faces are parallelograms. If the previous type of side faces become rectangles, then it will need to be called alreadydirect. Andrectangular  and the base also has angles of 90 °.

And the latter in geometry try to represent so that it is noticeable that all the edges are parallel. Here, by the way, the main difference between mathematicians and artists is observed. Last it is important to transfer the body in accordance with the law of perspective. And in this case the parallelism of the ribs is completely invisible.

## Formulas for an inclined parallelepiped

The first and second for the areas:

The third is to calculate the volume of a parallelepiped:

Since the base is a parallelogram, it will be necessary to use the corresponding expressions to calculate its area.

## Formulas for a rectangular parallelepiped

Similarly to the first point - two formulas for areas:

And one more for volume:

Condition. Given a cuboid whose volume you want to find. The well-known diagonal — 18 cm – and the fact that it forms the angles of 30 and 45 degrees with the plane of the side faces and side edge, respectively.

Decision.  To answer the question of the problem, you will need to know all sides in three right-angled triangles. They will give the necessary values ​​of the edges, for which you need to calculate the volume.

First you need to find out where the angle is at 30º. To do this, we must draw the diagonal of the lateral face from the same vertex from which the main diagonal of the parallelogram is drawn. The angle between them will be what you need.

The first triangle, which will give one of the values ​​of the sides of the base, will be as follows. It contains the desired side and the two diagonals drawn. It is rectangular. Now we need to use the ratio of the opposite leg (the side of the base) and the hypotenuse (diagonal). It is equal to the sine of 30º. That is, the unknown side of the base will be defined as a diagonal multiplied by a sine of 30º or ½. Let it be denoted by the letter "a".

It is easy to calculate: a = 18 * 1 = 9 (cm).

The second will be a triangle containing a known diagonal and an edge, with which it forms 45º. It is also rectangular, and you can again use the ratio of the leg to the hypotenuse. In other words, the lateral edges to the diagonal. It is equal to the cosine of 45 °. That is, "c" is calculated as the product of the diagonal by the cosine of 45 °.

In the same triangle you need to find another cathet. This is necessary in order to then calculate the third unknown - "in". Let it be denoted by the letter "x". It is easy to calculate by the Pythagorean theorem:

x = √ (18 2 - (9√2) 2) = 9√2 (cm).

Now we need to consider another rectangular triangle. It contains already known sides "c", "x" and the one that you need to count, "in":

in = √ ((9√2) 2 - 9 2 = 9 (cm).

All three values ​​are known. You can use the formula for the volume and count it:

V = 9 * 9 * 9√2 = 729√2 (cm 3).

Answer:  the volume of the parallelepiped is 729√2 cm 3.

Condition. You want to find the volume of a parallelepiped. It is known the sides of a parallelogram, which lies at the base, 3 and 6 cm, and its acute angle of 45°. The lateral edge has a slope to the base is 30 ° and equal to 4 cm.

Decision.  To answer the question of the problem, it is necessary to take the formula, which was written for the volume of the inclined parallelepiped. But both quantities are unknown in it.

The base area, that is, the parallelogram, will be determined by the formula in which it is necessary to multiply the known sides and the sine of the acute angle between them.

Sabout  = 3 * 6 sin 45º = 18 * (√2) / 2 = 9 √2 (cm 2).

The second unknown quantity is the height. It can be drawn from any of the four vertices above the base. It can be found from a rectangular triangle, in which the height is a leg, and the lateral edge - the hypotenuse. At the same angle of 30º lies opposite the unknown height. Hence, we can use the ratio of the leg to the hypotenuse.

Now all the values ​​are known and you can calculate the volume:

V = 9 √2 * 2 = 18 √2 (cm 3).

Answer:  the volume is 18 √2 cm 3.

Condition. Find the volume of the parallelepiped if you know that he is straight. Side of its base form a parallelogram and is equal to 2 and 3 refer to the Acute angle between them is 60º. The smaller the diagonal of the parallelepiped is equal to the major diagonal of the base.

Decision.  In order to know the volume of a parallelepiped, we use the formula with the base area and height. Both quantities are unknown, but they are easy to calculate. The first of these is the height.

Since the smaller diagonal of the parallelepiped coincides in size with a larger base, they can be denoted by a single letter d. The greater angle of the parallelogram is 120º, since it forms 180º with a sharp angle. Let the second diagonal of the base be denoted by the letter "x". Now for two diagonals of the base, cosine theorems can be written:

d 2 = a 2 + in 2 - 2 a cos 120 °,

x 2 = a 2 + in 2 - 2 a cos 60 °.

Finding values ​​without squares does not make sense, since then they will be raised again to the second power. After substituting the data, we get:

d 2 = 2 2 + 3 2 - 2 * 2 * 3 cos 120º = 4 + 9 + 12 * 1 = 19,

x 2 = a 2 + a 2 - 2 a cos 60º = 4 + 9 - 12 * 1 = 7.

Now the height, which is the lateral edge of the parallelepiped, will be a leg in the triangle. The hypotenuse is the known diagonal of the body, and the second leg is the "x". We can write down the Pythagorean Theorem:

n 2 = d 2 - x 2 = 19 - 7 = 12.

Hence: n = √12 = 2√3 (cm).

Now the second unknown quantity is the area of ​​the base. It can be calculated by the formula mentioned in the second problem.

Sabout  = 2 * 3 sin 60º = 6 * √3 / 2 = 3√3 (cm 2).

Combining all of the volume formula, we get:

V = 3√3 * 2√3 = 18 (cm 3).

Answer: V = 18 cm 3.

Condition. It is required to know the volume of a parallelepiped corresponding to such conditions: the base is a square with a side of 5 cm; the lateral faces are rhombuses; One of the vertices above the base is equidistant from all vertices lying in the base.

Decision.  First you need to understand the condition. There are no questions about the first item about the square. The second, about rhombuses, makes it clear that the parallelepiped is inclined. And all its edges are equal to 5 cm, because the sides of the diamond are the same. And from the third it becomes clear that the three diagonals drawn from it are equal. These are two that lie on the side faces, and the latter is inside the parallelepiped. And these diagonals are equal to an edge, that is, they also have a length of 5 cm.

To determine the volume, we need a formula written for an inclined box. There are no known quantities in it again. However, the area of ​​the base can be easily calculated, because this is a square.

The situation with height is a bit more complicated. It will be such in three figures: a parallelepiped, a quadrangular pyramid and an isosceles triangle. The last circumstance also needs to be taken advantage of.

Because it height, is the leg in a right triangle. The hypotenuse in this will known edge, and a second leg equal to half the diagonal of the square (height – it is also the median). And the diagonal of the base to find just:

d = √ (2 * 5 2) = 5√2 (cm).

The height will need to be counted as the difference between the second degree of the edge and the square of the half of the diagonal and do not forget then to extract the square root:

n = √ (5 2 - (5/2 * √2) 2) = √ (25 - 25/2) = √ (25/2) = 2.5 √2 (cm).

It remains to calculate the volume:

V = 25 * 2.5 √2 = 62.5 √2 (cm 3).