Often students indignantly ask: "How can I use this in life?" On any subject of every subject. The issue of parallelepiped volume is no exception. And here you can just say: "It is useful."
How, for example, to find out whether the package will fit in the mail box? Of course, it is possible by trial and error to choose the appropriate one. And if there is no such possibility? Then calculations will come to the rescue. Knowing the capacity of the box, you can calculate the volume of the package (at least approximately) and answer the question posed.
Parallelepiped and its types
If you literally translate its name from ancient Greek, it turns out that this is a figure consisting of parallel planes. There are such equivalent definitions of a parallelepiped:
- prism with a base in the form of a parallelogram;
- polyhedron, each face of which is a parallelogram.
Its types stand out depending on which figure lies at its base and how the side edges are directed. In general, they talk aboutinclined parallelepiped. whose base and all faces are parallelograms. If in the previous view the side faces become rectangles, then it will need to be calledstraight. And yrectangular and the base also has 90º angles.
And the latter in geometry are trying to portray 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 a body with observance of the law of prospect. And in this case, the parallelism of the edges is completely invisible.
Formulas for oblique parallelepiped
The first and second for space:
The third in order to calculate the volume of the parallelepiped:
Since the base is a parallelogram, then to calculate its area, you will need to use the appropriate expressions.
Formulas for rectangular parallelepiped
Similar to the first item - 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 need to know all sides in three right triangles. They will give the necessary values of the edges, which need to be counted volume.
First you need to find out where the angle is at 30º. To do this, you need to draw a diagonal of the side face from the same vertex where the main diagonal of the parallelogram was drawn from. The angle between them will be what is needed.
The first triangle, which will give one of the values of the sides of the base, will be next. It contains the desired side and two diagonally drawn. It is rectangular. Now you need to use the ratio of the opposite side (base side) and 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 the sine of 30º or ½. Let it be marked with the letter “a”.
It is easy to count: a = 18 * ½ = 9 (cm).
The second is a triangle containing a known diagonal and an edge with which it forms 45º. It is also rectangular, and you can again use the attitude of the leg to the hypotenuse. In other words, the side edge to the diagonal. It is equal to the cosine of 45º. That is, “c” is calculated as the product of the diagonal by the cosine of 45º.
In the same triangle is required to find another leg. This is necessary in order to then count the third unknown - "in". Let it be marked with 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 right triangle. It contains the already known sides "c", "x" and the one you need to count, "c":
c = √ ((9√2) 2 - 9 2 = 9 (cm).
All three quantities 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, you need to take the formula that was written for the volume of the inclined parallelepiped. But both quantities are unknown in it.
The area of the base, that is, the parallelogram, will be determined by the formula in which you need 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 is height. It can be drawn from any of the four vertices above the base. It can be found from a right-angled triangle, in which the height is the leg and the lateral edge is the hypotenuse. In this case, an angle of 30º lies opposite an unknown height. So, you can use the attitude of the leg to the hypotenuse.
Now all the values are known and the volume can be calculated:
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 find out the volume of the parallelepiped, we use the formula with the area of the base and height. Both quantities are unknown, but they are not difficult to calculate. The first one is height.
Since the smaller diagonal of the parallelepiped coincides in size with a larger base, they can be designated by one letter d. The larger angle of the parallelogram is 120º, since it forms 180º with a sharp one. Let the second diagonal of the base be marked with the letter “x”. Now for the two base diagonals, we can write cosine theorems:
d 2 = a 2 + in 2 - 2av cos 120º,
x 2 = a 2 + in 2 - 2av cos 60º.
Finding values without squares does not make sense, since then they will be raised to the second power again. After data substitution is obtained:
d 2 = 2 2 + 3 2 - 2 * 2 * 3 cos 120º = 4 + 9 + 12 * ½ = 19,
x 2 = a 2 + in 2 - 2av cos 60º = 4 + 9 - 12 * ½ = 7.
Now the height, which is also the side edge of the parallelepiped, will be the leg in the triangle. The hypotenuse will be the known diagonal of the body, and the second leg will be the “x”. You can write the Pythagorean Theorem:
n 2 = d 2 - x 2 = 19 - 7 = 12.
From here: n = √12 = 2√3 (cm).
Now the second unknown quantity is the area of the base. It can be counted by the formula mentioned in the second problem.
Sabout = 2 * 3 sin 60º = 6 * √3 / 2 = 3√3 (cm 2).
Combining all in the formula of volume, we get:
V = 3√3 * 2√3 = 18 (cm 3).
Answer: V = 18 cm 3.
Condition. It is required to know the volume of the parallelepiped that meets these conditions: base - a square with a side of 5 cm; the side faces are rhombus; one of the vertices above the base is equidistant from all the vertices at the base.
Decision. First you need to deal with the condition. With the first paragraph about the square there are no questions. The second, about rhombuses, makes it clear that the parallelepiped is oblique. Moreover, all its edges are 5 cm, since the sides of the rhombus 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 the edge, that is, they also have a length of 5 cm.
To determine the volume, you will need a formula written for an inclined parallelepiped. Again, there are no known values. However, the area of the base is easy to calculate because it is square.
A little more difficult is the case with height. It will be such in three figures: a parallelepiped, a quadrangular pyramid and an isosceles triangle. The last circumstance and need to take advantage.
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 of the second degree of the edge and the square of the half of the diagonal and then do not forget to extract the square root:
n = √ (5 2 - (5/2 * √2) 2) = √ (25 - 25/2) = √ (25/2) = 2.5 √2 (cm).
It remains to count the volume:
V = 25 * 2.5 √2 = 62.5 √2 (cm 3).
Answer: 62.5 √2 (cm 3).