Any geometric body can be characterized by surface area (S) and volume (V). Area and volume are not the same thing at all. An object can have a relatively small V and a large S, for example, this is how the human brain works. Calculate these indicators for simple geometric shapes much simpler.
Parallelepiped: definition, types and properties
A parallelepiped is a quadrangular prism with a parallelogram at its base. Why might you need a formula for finding the volume of a figure? Books, packaging boxes and many other things from Everyday life. Rooms in residential and office buildings are usually rectangular parallelepipeds. To install ventilation, air conditioning and determine the number of heating elements in a room, it is necessary to calculate the volume of the room.
The figure has 6 faces - parallelograms and 12 edges; two randomly selected faces are called bases. A parallelepiped can be of several types. The differences are due to the angles between adjacent edges. The formulas for finding the Vs of different polygons are slightly different.
If the 6 faces of a geometric figure are rectangles, then it is also called rectangular. Cube is special case a parallelepiped in which all 6 faces are equal squares. In this case, to find V, you need to find out the length of only one side and raise it to the third power.
To solve problems, you will need knowledge not only of ready-made formulas, but also of the properties of the figure. The list of basic properties of a rectangular prism is small and very easy to understand:
- The opposite sides of the figure are equal and parallel. This means that the ribs located opposite are the same in length and angle of inclination.
- All lateral faces of a right parallelepiped are rectangles.
- The four main diagonals of a geometric figure intersect at one point and are divided in half by it.
- The square of the diagonal of a parallelepiped is equal to the sum of the squares of the dimensions of the figure (follows from the Pythagorean theorem).
Pythagorean theorem states that the sum of the areas of squares built on the sides of a right triangle is equal to the area of a triangle built on the hypotenuse of the same triangle.
The proof of the last property can be seen in the image below. The process of solving the problem is simple and does not require detailed explanations.
Formula for the volume of a rectangular parallelepiped
The formula for finding for all types of geometric figures is the same: V=S*h, where V is the required volume, S is the area of the base of the parallelepiped, h is the height lowered from the opposite vertex and perpendicular to the base. In a rectangle, h coincides with one of the sides of the figure, so to find the volume of a rectangular prism, you need to multiply three dimensions.
Volume is usually expressed in cm3. Knowing all three values of a, b and c, finding the volume of a figure is not at all difficult. The most common type of problem in the Unified State Exam is finding the volume or diagonal of a parallelepiped. Solve many typical Unified State Exam assignments It’s impossible without the formula for the volume of a rectangle. An example of a task and the design of its solution is shown in the figure below.
Note 1. The surface area of a rectangular prism can be found by multiplying by 2 the sum of the areas of the three faces of the figure: the base (ab) and two adjacent side faces (bc + ac).
Note 2. The surface area of the side faces can be easily determined by multiplying the perimeter of the base by the height of the parallelepiped.
Based on the first property of parallelepipeds AB = A1B1, and face B1D1 = BD. According to corollaries of the Pythagorean theorem, the sum of all angles in right triangle is equal to 180°, and the leg lying opposite the angle of 30° is equal to the hypotenuse. Applying this knowledge to a triangle, we can easily find the length of sides AB and AD. Then we multiply the obtained values and calculate the volume of the parallelepiped.
Formula for finding the volume of an inclined parallelepiped
To find the volume of an inclined parallelepiped, it is necessary to multiply the area of the base of the figure by the height lowered to the given base from the opposite corner.
Thus, the required V can be represented in the form of h - the number of sheets with a base area S, so the volume of the deck consists of the Vs of all cards.
Examples of problem solving
The tasks of the single exam must be completed within a certain time. Typical tasks, as a rule, do not contain large quantity calculations and complex fractions. Often a student is asked how to find the volume of an irregular geometric figure. In such cases, you should remember the simple rule that the total volume is equal to the sum of the Vs of the component parts.
As you can see from the example in the image above, there is nothing difficult in solving such problems. Tasks from more complex sections require knowledge of the Pythagorean theorem and its consequences, as well as the formula for the length of the diagonal of a figure. To successfully solve test tasks, it is enough to familiarize yourself with samples of typical problems in advance.
Measure all required distances in meters. The volume of many three-dimensional figures can be easily calculated using the appropriate formulas. However, all values substituted into formulas must be measured in meters. Therefore, before plugging values into the formula, make sure that they are all measured in meters, or that you have converted other units of measurement to meters.
- 1 mm = 0.001 m
- 1 cm = 0.01 m
- 1 km = 1000 m
To calculate the volume of rectangular figures (cuboid, cube), use the formula: volume = L × W × H(length times width times height). This formula can be considered as the product of the surface area of one of the faces of the figure and the edge perpendicular to this face.
- For example, let’s calculate the volume of a room with a length of 4 m, a width of 3 m and a height of 2.5 m. To do this, simply multiply the length by the width and by the height:
- 4 × 3 × 2.5
- = 12 × 2.5
- = 30. The volume of this room is 30 m 3.
- A cube is a three-dimensional figure with all sides equal. Thus, the formula for calculating the volume of a cube can be written as: volume = L 3 (or W 3, or H 3).
To calculate the volume of figures in the form of a cylinder, use the formula: pi× R 2 × H. Calculating the volume of a cylinder comes down to multiplying the area of the circular base by the height (or length) of the cylinder. Find the area of the circular base by multiplying pi (3.14) by the square of the radius of the circle (R) (radius is the distance from the center of the circle to any point lying on this circle). Then multiply the result by the height of the cylinder (H) and you will find the volume of the cylinder. All values are measured in meters.
- For example, let's calculate the volume of a well with a diameter of 1.5 m and a depth of 10 m. Divide the diameter by 2 to get the radius: 1.5/2 = 0.75 m.
- (3.14) × 0.75 2 × 10
- = (3.14) × 0.5625 × 10
- = 17.66. The volume of the well is 17.66 m 3.
To calculate the volume of a ball, use the formula: 4/3 x pi× R 3 . That is, you only need to know the radius (R) of the ball.
- For example, let's calculate the volume hot air balloon with a diameter of 10 m. Divide the diameter by 2 to get the radius: 10/2=5 m.
- 4/3 x pi × (5) 3
- = 4/3 x (3.14) × 125
- = 4.189 × 125
- = 523.6. The volume of the balloon is 523.6 m 3.
To calculate the volume of cone-shaped figures, use the formula: 1/3 x pi× R 2 × H. The volume of a cone is equal to 1/3 of the volume of a cylinder, which has the same height and radius.
- For example, let's calculate the volume of an ice cream cone with a radius of 3 cm and a height of 15 cm. Converting to meters, we get: 0.03 m and 0.15 m, respectively.
- 1/3 x (3.14) × 0.03 2 × 0.15
- = 1/3 x (3.14) × 0.0009 × 0.15
- = 1/3 × 0.0004239
- = 0.000141. The volume of an ice cream cone is 0.000141 m 3.
To calculate the volume of irregular shapes, use several formulas. To do this, try to break the figure into several figures of the correct shape. Then find the volume of each such figure and add up the results.
- For example, let's calculate the volume of a small granary. The warehouse has a cylindrical body with a height of 12 m and a radius of 1.5 m. The warehouse also has a conical roof with a height of 1 m. By calculating the volume of the roof separately and the volume of the body separately, we can find the total volume of the granary:
- pi × R 2 × H + 1/3 x pi × R 2 × H
- (3.14) × 1.5 2 × 12 + 1/3 x (3.14) × 1.5 2 × 1
- = (3.14) × 2.25 × 12 + 1/3 x (3.14) × 2.25 × 1
- = (3.14) × 27 + 1/3 x (3.14) × 2.25
- = 84,822 + 2,356
- = 87.178. The volume of the granary is equal to 87.178 m 3.
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The video course “Get an A” includes all the topics necessary to successfully pass the Unified State Exam in mathematics with 60-65 points. Completely all tasks 1-13 of the Profile Unified State Exam in mathematics. Also suitable for passing the Basic Unified State Examination in mathematics. If you want to pass the Unified State Exam with 90-100 points, you need to solve part 1 in 30 minutes and without mistakes!
Preparation course for the Unified State Exam for grades 10-11, as well as for teachers. Everything you need to solve Part 1 of the Unified State Exam in mathematics (the first 12 problems) and Problem 13 (trigonometry). And this is more than 70 points on the Unified State Exam, and neither a 100-point student nor a humanities student can do without them.
All the necessary theory. Quick solutions, pitfalls and secrets of the Unified State Exam. All current tasks of part 1 from the FIPI Task Bank have been analyzed. The course fully complies with the requirements of the Unified State Exam 2018.
The course contains 5 large topics, 2.5 hours each. Each topic is given from scratch, simply and clearly.
Hundreds of Unified State Exam tasks. Word problems and probability theory. Simple and easy to remember algorithms for solving problems. Geometry. Theory, reference material, analysis of all types of Unified State Examination tasks. Stereometry. Tricky solutions, useful cheat sheets, development of spatial imagination. Trigonometry from scratch to problem 13. Understanding instead of cramming. Clear explanations of complex concepts. Algebra. Roots, powers and logarithms, function and derivative. A basis for solving complex problems of Part 2 of the Unified State Exam.
And the ancient Egyptians used methods for calculating the areas of various figures, similar to our methods.
In my books "Beginnings" the famous ancient Greek mathematician Euclid described quite big number methods for calculating the areas of many geometric figures. The first manuscripts in Rus' containing geometric information were written in the 16th century. They describe the rules for finding the areas of figures of various shapes.
Today with the help modern methods you can find the area of any figure with great accuracy.
Let's consider one of the simplest figures - a rectangle - and the formula for finding its area.
Rectangle area formula
Let's consider a figure (Fig. 1), which consists of $8$ squares with sides of $1$ cm. The area of one square with a side of $1$ cm is called a square centimeter and is written $1\ cm^2$.
The area of this figure (Fig. 1) will be equal to $8\cm^2$.
The area of a figure that can be divided into several squares with a side of $1\ cm$ (for example, $p$) will be equal to $p\ cm^2$.
In other words, the area of the figure will be equal to so many $cm^2$, into how many squares with side $1\ cm$ this figure can be divided.
Let's consider a rectangle (Fig. 2), which consists of $3$ stripes, each of which is divided into $5$ squares with a side of $1\ cm$. the entire rectangle consists of $5\cdot 3=15$ such squares, and its area is $15\cm^2$.
Picture 1.
Figure 2.
The area of figures is usually denoted by the letter $S$.
To find the area of a rectangle, you need to multiply its length by its width.
If we denote its length by the letter $a$, and its width by the letter $b$, then the formula for the area of a rectangle will look like:
Definition 1
The figures are called equal if, when superimposed on one another, the figures coincide. Equal figures have equal areas and equal perimeters.
The area of a figure can be found as the sum of the areas of its parts.
Example 1
For example, in Figure $3$, rectangle $ABCD$ is divided into two parts by line $KLMN$. The area of one part is $12\ cm^2$, and the other is $9\ cm^2$. Then the area of the rectangle $ABCD$ will be equal to $12\ cm^2+9\ cm^2=21\ cm^2$. Find the area of the rectangle using the formula:
As you can see, the areas found by both methods are equal.
Figure 3.
Figure 4.
The line segment $AC$ divides the rectangle into two equal triangles: $ABC$ and $ADC$. This means that the area of each triangle is equal to half the area of the entire rectangle.
Definition 2
Rectangle with equal sides called square.
If we denote the side of a square with the letter $a$, then the area of the square will be found by the formula:
Hence the name square of the number $a$.
Example 2
For example, if the side of a square is $5$ cm, then its area is:
Volumes
With the development of trade and construction back in the days of ancient civilizations, the need arose to find volumes. In mathematics, there is a branch of geometry that deals with the study of spatial figures, called stereometry. Mentions of this separate branch of mathematics were found already in the $IV$ century BC.
Ancient mathematicians developed a method for calculating the volume of simple figures - a cube and a parallelepiped. All buildings of those times were of this shape. But later methods were found to calculate the volume of figures of more complex shapes.
Volume of a rectangular parallelepiped
If you fill a mold with wet sand and then turn it over, you get three-dimensional figure, which is characterized by volume. If you make several such figures using the same mold, you will get figures that have the same volume. If you fill the mold with water, then the volume of water and the volume of the sand figure will also be equal.
Figure 5.
You can compare the volumes of two vessels by filling one with water and pouring it into the second vessel. If the second vessel is completely filled, then the vessels have equal volumes. If water remains in the first, then the volume of the first vessel is greater than the volume of the second. If, when pouring water from the first vessel, it is not possible to completely fill the second vessel, then the volume of the first vessel is less than the volume of the second.
Volume is measured using the following units:
$mm^3$ -- cubic millimeter,
$cm^3$ -- cubic centimeter,
$dm^3$ -- cubic decimeter,
$m^3$ -- cubic meter,
$km^3$ -- cubic kilometer.
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