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.

A parallelogram is a quadrangular figure whose opposite sides are parallel and equal in pairs. Its opposite angles are also equal, and the point of intersection of the diagonals of the parallelogram divides them in half, being the center of symmetry of the figure. Special cases of a parallelogram are such geometric shapes as square, rectangle and rhombus. The area of ​​a parallelogram can be found in various ways, depending on what initial data is used to formulate the problem.

The key characteristic of a parallelogram, very often used when finding its area, is its height. The height of a parallelogram is usually called a perpendicular drawn from an arbitrary point on the opposite side to a straight segment forming that side.

1. In the simplest case, the area of ​​a parallelogram is defined as the product of its base and its height.
   

   S = DC ∙ h

   
   

   where S is the area of ​​the parallelogram;
   a - base;
   h is the height drawn to the given base.

   This formula is very easy to understand and remember if you look at the following figure.

   

   As you can see from this image, if we cut off an imaginary triangle to the left of the parallelogram and attach it to the right, the result will be a rectangle. As you know, the area of ​​a rectangle is found by multiplying its length by its height. Only in the case of a parallelogram will the length be the base, and the height of the rectangle will be the height of the parallelogram lowered to a given side.

2. The area of ​​a parallelogram can also be found by multiplying the lengths of two adjacent bases and the sine of the angle between them:
   

   S = AD∙AB∙sinα

   
   

   where AD, AB are adjacent bases forming an intersection point and an angle a between themselves;
   α is the angle between the bases AD and AB.

   

3. You can also find the area of ​​a parallelogram by dividing in half the product of the lengths of the diagonals of the parallelogram by the sine of the angle between them.
   

   S = ½∙AC∙BD∙sinβ

   
   

   where AC, BD are the diagonals of the parallelogram;
   β is the angle between the diagonals.

   

4. There is also a formula for finding the area of ​​a parallelogram through the radius of the circle inscribed in it. It is written as follows:

Parallelogram is a quadrilateral whose sides are parallel in pairs.

In this figure, opposite sides and angles are equal to each other. The diagonals of a parallelogram intersect at one point and bisect it. Formulas for the area of ​​a parallelogram allow you to find the value using the sides, height and diagonals. A parallelogram can also be presented in special cases. They are considered a rectangle, square and rhombus.
First, let's look at an example of calculating the area of ​​a parallelogram by height and the side to which it is lowered.

This case is considered a classic and does not require additional investigation. It’s better to consider the formula for calculating the area through two sides and the angle between them. The same method is used in calculations. If the sides and the angle between them are given, then the area is calculated as follows:

Suppose we are given a parallelogram with sides a = 4 cm, b = 6 cm. The angle between them is α = 30°. Let's find the area:

Area of ​​a parallelogram through diagonals

The formula for the area of ​​a parallelogram using the diagonals allows you to quickly find the value.
For calculations, you will need the size of the angle located between the diagonals.

Let's consider an example of calculating the area of ​​a parallelogram using diagonals. Let a parallelogram be given with diagonals D = 7 cm, d = 5 cm. The angle between them is α = 30°. Let's substitute the data into the formula:

An example of calculating the area of ​​a parallelogram through the diagonal gave us an excellent result - 8.75.

Knowing the formula for the area of ​​a parallelogram through the diagonal, you can solve many interesting problems. Let's look at one of them.

Task: Given a parallelogram with an area of ​​92 square meters. see Point F is located in the middle of its side BC. Let's find the area of ​​the trapezoid ADFB, which will lie in our parallelogram. First, let's draw everything we received according to the conditions.
Let's get to the solution:

According to our conditions, ah =92, and accordingly, the area of ​​our trapezoid will be equal to

Instructions

Remove insulation from the cable cores. Using a caliper, or preferably a micrometer (this will allow for a more accurate measurement), find the diameter of the core. You will get the value in millimeters. Then calculate the cross-sectional area. To do this, multiply the coefficient 0.25 by the number π≈3.14 and the value of the diameter d squared S=0.25∙π∙d². Multiply this value by the number of cable cores. Knowing the length of the wire, its cross-section and the material from which it is made, calculate its resistance.

For example, if you need to find the cross-section of a copper cable with 4 cores, and the measurement of the core diameter gives a value of 2 mm, find its cross-sectional area. To do this, calculate the cross-sectional area of ​​one core. It will be equal to S=0.25∙3.14∙2²=3.14 mm². Then determine the cross-section of the entire cable for this cross-section of one core, multiply by their number in our example it is 3.14∙4=12.56 mm².

Now you can find out the maximum current that can flow through it, or its resistance if the length is known. Calculate the maximum current for a copper cable from the ratio of 8 A per 1 mm². Then the maximum value of the current that can pass through the cable taken in the example is 8∙12.56 = 100.5 A. Keep in mind that for this ratio it is 5 A per 1 mm².

For example, the cable length is 200 m. In order to find its resistance, multiply the copper resistivity ρ in Ohm∙mm²/m by the cable length l and divide by its cross-sectional area S (R=ρ∙l/S). Having made the substitution, you will get R=0.0175∙200/12.56≈0.279 Ohm, which will lead to very small losses of electricity when transmitting it through such a cable.

Sources:

• how to find out the cable cross-section

If a variable, sequence, or function has an infinite number of values ​​that vary according to some law, it may tend to o to the limit number, which is the limit sequences. Limits can be calculated in various ways.

You will need

• - the concept of numerical sequence and function;
• - ability to take derivatives;
• - ability to transform and shorten expressions;
• - calculator.

Instructions

To calculate the limit, substitute the limit value of the argument into its expression. Try the calculation. If this is possible, then the value with the substituted value is the desired one. Example: Find the values ​​of the limit with a common term (3 x?-2)/(2 x?+7) if x > 3. Substitute the limit into the expression sequences (3 3?-2)/(2 3?+7)=(27-2)/(18+7)=1.

If there is uncertainty when trying to substitute, choose a way to resolve it. This can be done by transforming the expressions in which . Having made the reductions, you will get the result. Example: Sequence (x+vx)/(x-vx), when x > 0. Direct substitution results in uncertainty 0/0. Get rid of it by removing the common factor from the numerator and denominator. In this case it will be vx. Get (vx (vx+1))/(vx (vx-1))= (vx+1)/(vx-1). Now the substitution field will get 1/(-1)=-1.

When it is impossible to reduce due to uncertainty (especially if the sequence contains irrational expressions), multiply its numerator and denominator by the conjugate expression in order to remove from the denominator. Example: Sequence x/(v(x+1)-1). The value of the variable x > 0. Multiply the numerator and denominator by the conjugate expression (v(x+1)+1). Get (x (v(x+1)+1))/((v(x+1)-1) (v(x+1)+1))=(x (v(x+1)+1) )/(x+1-1)= (x (v(x+1)+1))/x=v(x+1)+1. After substitution, you get =v(0+1)+1=1+1=2.

For uncertainties like 0/0 or?/? use L'Hopital's rule. For this, the numerator and denominator sequences imagine as functions, take from them . The limit of their relations will be equal to the limit of the relations of the functions themselves. Example: Find the limit sequences ln(x)/vx, for x > ?. Direct substitution gives uncertainty?/?. Take the derivatives of the numerator and denominator and get (1/x)/(1/2 vx)=2/vx=0.

To reveal uncertainties, use the first wonderful limit sin(x)/x=1 for x>0, or the second wonderful limit (1+1/x)^x=exp for x>?. Example: Find the limit sequences sin(5 x)/(3 x) for x>0. Transform the expression sin(5 x)/(3/5 5 x) multiply the denominator 5/3 (sin(5 x)/(5 x)) using the first limit you get 5/3 1=5/3.

Example: Find the limit (1+1/(5 x))^(6 x) for x>?. Multiply and divide powers by 5 x. Get the expression ((1+1/(5 x))^(5 x)) ^(6 x)/(5 x). Applying the rule of the second remarkable limit, you get exp^(6 x)/(5 x)=exp.

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Tip 9: How to find the axial cross-sectional area of ​​a truncated cone

To solve this problem, you need to remember what a truncated cone is and what properties it has. Be sure to make a drawing. This will allow you to determine what geometric figure the section represents. It is quite possible that after this, solving the problem will no longer be difficult for you.

Instructions

A round cone is a body obtained by rotating a triangle around one of its legs. Straight lines emanating from the apex cone and intersecting its base are called generators. If all generators are equal, then the cone is straight. At the base of the round cone lies a circle. The perpendicular dropped to the base from the vertex is the height cone. At the round straight cone the height coincides with its axis. The axis is a straight line connecting to the center of the base. If the horizontal cutting plane of a circular cone, then its upper base is a circle.

Since it is not specified in the problem statement that it is the cone that is given in this case, we can conclude that this is a straight truncated cone, the horizontal section of which is parallel to the base. Its axial section, i.e. vertical plane, which through the axis of the round cone, is an equilateral trapezoid. All axial sections round straight cone are equal to each other. Therefore, to find square axial sections, you need to find square trapezoid, the bases of which are the diameters of the bases of a truncated cone, and the lateral sides are its constituents. Frustum height cone is also the height of the trapezoid.

The area of ​​a trapezoid is determined by the formula: S = ½(a+b) h, where S – square trapezoid; a – the size of the lower base of the trapezoid; b – the size of its upper base; h – the height of the trapezoid.

Since the condition does not specify which ones are given, it is possible that the diameters of both bases of the truncated cone known: AD = d1 – diameter of the lower base of the truncated cone;BC = d2 – diameter of its upper base; EH = h1 – height cone.Thus, square axial sections truncated cone is defined: S1 = ½ (d1+d2) h1

Sources:

• area of ​​a truncated cone

The regulatory documents for the design of electrical networks indicate the cross-sections of the wires, but only the cores can be measured with a caliper. These quantities are interrelated and can be converted from one to another.

Instructions

To translate what is specified in a regulatory document section single-core wire in its diameter, use the following formula: D=2sqrt(S/π), where D is diameter, mm; S - conductor cross-section, mm2 (electricians call it “squares”).

A flexible stranded wire consists of many thin strands twisted together and placed in a common insulating sheath. This allows it not to break during frequent movements, which is connected to the source with its help. To find the diameter of one core of such a conductor (this is what can be measured with a caliper), first find the cross-section of this core: s=S/n, where s is the cross-section of one core, mm2; S - total wire cross-section (indicated in the regulations); n is the number of cores. Then convert the core cross-section to diameter as indicated above.

Printed circuit boards use flat conductors. Instead of diameter, they have thickness and width. The first value is taken from the technical data of the foil material in advance. Knowing it, you can find the width by . To do this, use the following formula: W=S/h, where W is the conductor, mm; S - conductor cross-section, mm2; h - conductor thickness, mm.

Square conductors are relatively rare. Its cross-section must be converted either to a side or to the diagonal of a square (both can be measured with a caliper). sides is calculated as follows: L=sqrt(S), where L is side length, mm; S is the cross-section of the conductor, mm2. To find out the diagonal from the side length, make the following calculations: d=sqrt(2(L^2)), where d is the diagonal of the square, mm; L - side length, mm.

If there is no conductor whose cross-section exactly matches the required one, use another one with a larger, but in no case smaller, cross-section. Select the type of conductor and the type of its insulation depending on the conditions of use.

note

Before measuring the conductor with a caliper, remove the supply voltage and make sure there is no voltage using a voltmeter.

Sources:

• diameter translation

For example, the diameter of the base of a straight cylinder is 8 cm, and its is 10 cm. Determine square its lateral surface. Calculate Radius cylinder. It is equal to R=8/2=4 cm. Generator of the straight line cylinder equal to its height, that is, L = 10 cm. For calculations, use a single formula, it is more convenient. Then S=2∙π∙R∙(R+L), substitute the corresponding numerical values ​​S=2∙3.14∙4∙(4+10)=351.68 cm².

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The cross section is formed at a right angle to the longitudinal axis. Moreover, the cross-section of various geometric shapes can be represented in different shapes. For example, a parallelogram’s cross-section in appearance resembles a rectangle or square, a cylinder’s cross-section resembles a rectangle or a circle, etc.

You will need

• - calculator;
• – initial data.

Instructions

1. In order to find the cross-sectional area of ​​a parallelogram, you need to know the value of its base and height. If, for example, only the length and width of the base are known, then find the diagonal by applying the Pythagorean theorem (the square of the length of the hypotenuse in a right triangle is equal to the sum of the squares of the legs: a2 + b2 = c2). In view of this, c = sqrt (a2 + b2).

2. Having found the value of the diagonal, substitute it into the formula S= c*h, where h is the height of the parallelogram. The resulting result will be the cross-sectional area of ​​the parallelogram.

3. If the section runs along 2 bases, then calculate its area using the formula: S=a*b.

4. To calculate the axial cross-sectional area of ​​a cylinder running perpendicular to the base (provided that one side of this rectangle is equal to the radius of the base, and the other to the height of the cylinder), use the formula S = 2R*h, in which R is the value of the radius of the circle (base), S is the cross-sectional area, and h is the height of the cylinder.

5. If, according to the conditions of the problem, the section does not pass through the axis of rotation of the cylinder, but is parallel to its bases, then the side of the rectangle will not be equal to the diameter of the base circle.

6. Independently calculate the unknown side by constructing the circle of the base of the cylinder, drawing perpendiculars from the side of the rectangle (sectional plane) to the circle, and calculating the size of the chord (using the Pythagorean theorem). Later, substitute the resulting value into S = 2a*h (2a is the value of the chord) and calculate the cross-sectional area.

7. The cross-sectional area of ​​the ball is determined by the formula S = ?R2. Please note that if the distance from the center of the geometric figure to the plane coincides with the plane, then the cross-sectional area will be zero, because the ball touches the plane at only one point.

If you suddenly begin to notice that the bones on your huge toes have enlarged, that it hurts to wear shoes (exclusively in the summer), this means that you have transverse flat feet. In this case, you should immediately consult an orthopedic doctor. Don’t hesitate, the sooner the treatment begins, the better the tea.

Instructions

1. The expert will examine you and recommend one of the main methods of treating transverse flatfoot. The first of them is conservative, it is suitable only for the treatment of the first degree of the disease. The method itself consists of reducing weight, reducing static load, giving up “heels” and uncomfortable shoes. In addition, with conservative treatment, the patient is prescribed physiotherapeutic procedures, physical therapy, and massage. The doctor may also recommend wearing insoles with special orthopedic cushions.

2. Another method (surgical) is used to treat transverse flatfoot of the 2nd and third degree. There are more than four hundred variations included in it, however, none of them eliminates the main cause of the disease - weakness of the muscle-ligamentous unit. In extreme cases, surgical ligation may be necessary, that is, a muscle tendon transplant or plastic surgery of the joint capsule. After such an operation, the patient must wear shoes only with individual insoles and insoles with a Seitz roller, as well as with arch supports.

3. You should not give up traditional medicine recipes. Here is one of them: take a 10% iodine solution and apply it to the thumb bone. This will help relieve inflammation and stop the growth of cartilage tissue. True, be careful with iodine, do not use a powerfully concentrated solution, on the contrary, you risk getting a skin burn. The same recommendation can be given for compresses with the addition of vinegar essence. By the way, modern medicine offers a huge selection of ointments and gels that can relieve joint inflammation and improve tissue nutrition. However, do not purchase similar products on your own; consult your doctor.

Helpful advice
Be careful not to stay in your shoes for too long, give your feet a break. It is worth noting that the shoes you purchase should be comfortable and breathable.

Tip 3: Section of a parallelepiped: how to calculate its area

A lot of problems are based on the properties of polyhedra. The edges of volumetric figures, as well as certain points on them, lie in different planes. If one of these planes is drawn through a parallelepiped at a certain angle, then the part of the plane lying within the polyhedron and dividing it into parts will be its cross section .

You will need

• - ruler
• - pencil

Instructions

1. Build a parallelepiped. Remember that its base and each of its faces must be a parallelogram. This means that you need to construct the polyhedron so that all opposite edges are parallel. If the condition says to construct a section of a rectangular parallelepiped, then make its edges rectangular. A straight parallelepiped has rectangular only 4 lateral faces. If the side faces parallelepiped are not perpendicular to the base, then such a polyhedron is called inclined. If you want to construct a section of a cube, first draw a rectangular parallelepiped with equal dimensions. Then all six of its faces will be squares. Name all vertices for ease of notation.

2. Mark two points that will belong to the section plane. Occasionally their location is indicated in the problem: the distance from the nearest vertex, the end of a segment drawn according to certain conditions. Now draw a straight line through points lying in the same plane.

3. Find lines at the intersection of the cutting plane with the faces parallelepiped. To perform this step, locate the points where the line lying in the cutting plane parallelepiped, intersects with a straight line belonging to the face parallelepiped. These lines must be in the same plane.

4. Complete the section parallelepiped. Remember that its plane must intersect parallel faces parallelepiped along parallel lines.

5. Construct a cutting plane in accordance with the initial data in the problem. There are several probabilities of constructing a section plane passing: - perpendicular to a given straight line through a given point; - perpendicular to a given plane through a given straight line; - parallel to two intersecting lines through a given point; - parallel to another given straight line through another given line; - parallel to a given plane through a given point. Using such initial data, construct a section according to the thesis described above.

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Note!
In order to construct a section of a parallelepiped, it is necessary to determine the points of intersection of the section plane with the edges of the parallelepiped, and then combine these points with segments. Please note that only connect points that lie in the plane of one face. Intersect the parallel faces of the parallelepiped with a cutting plane along parallel segments. If in the face plane only one point belongs to the cutting plane, construct an additional such point. To do this, find the intersection points of the constructed lines with those lines that lie in the required faces.

Helpful advice
The parallelepiped has 6 faces. Its sections can produce triangles, quadrangles, pentagons and figures with six angles. A plane, including a secant plane, is defined by: - ​​three points; - a straight line and one point; - two lines parallel to each other; - two straight lines intersecting each other.

Field orientation is the main component of many professions. Maps and compasses are used for this. To determine the direction on a map to a specific object, a directional angle and magnetic azimuths are used.

You will need

• Compass or compass, sharpened pencil, ruler, protractor.

Instructions

1. The directional angle in geodesy is the angle between the line passing through a given point in the direction of the target and a line parallel to the abscissa axis, reporting from the northern direction of the abscissa axis. It is counted from left to right (in the direction of the arrow) from 0° to 360°.

2. It is more comfortable for everyone to determine the directional angle on the map. Using a pencil and a ruler, draw a line through the centers of the symbols of the starting point and landmark. The length of the drawn line, for ease of measurement, must exceed the radius of the protractor. After this, align the center of the protractor with the point where the lines intersect and rotate it so that the zero on the protractor coincides with the vertical grid line on the map (or a line parallel to it). Count the angle values ​​in a clockwise direction. The average error in measuring the directional angle with a protractor is from 15/ to 1°.

3. Occasionally, magnetic azimuths are used to calculate directional angles. Magnetic azimuth is a flat horizontal angle formed by a line directed towards the target and the north direction of the magnetic meridian. It also counts from 0° to 360° clockwise. Magnetic azimuths are measured on the ground with the help of a compass or compass. The compass needle, or rather its magnetic field, interacts with the magnetic field of the area and shows the direction of the magnetic meridian.

4. Next, you need to determine the direction correction (the sum of the convergence of the meridians and the magnetic declination). Magnetic declination is the angle between the magnetic and geographic meridians at a given point. The convergence of meridians is the angle between the tangent drawn to the meridian of a given point and the tangent to the surface of the ellipsoid of revolution drawn at the same point, parallel to the original meridian. The direction correction is also counted from the north direction of the coordinate grid in a clockwise direction. The direction correction is considered positive if the arrow deviates to the right (east) and negative if it deviates to the left (west). The magnetic azimuth measured with the support of a compass on the ground can be converted into a directional angle by adding a direction correction to it, observing the sign of the correction observantly.

Note!
Many maps often indicate the values ​​of meridian convergence (also called Gaussian convergence) and direction corrections

Helpful advice
Pay special attention to the direction of reference and consider all the signs.

A parallelogram is a convex quadrangular geometric figure in which pairs of opposite sides have identical lengths. Also, pairs of angles at opposite vertices have identical values. The entire segment connecting two opposite sides and perpendicular to all of them can be called the height of this quadrilateral. Knowing the lengths of sides, angles and heights in various combinations of these parameters allows you to calculate the area of ​​a parallelogram.

Instructions

1. If the value of the angle at each vertex of the parallelogram (?) and the length of the adjacent sides (a and b) are known, then the area of ​​the figure (S) can be calculated using the trigonometric function - the sine. Multiply the known lengths of the sides by the sine of the given angle: S=a*b*sin(?). Say, if the angle is 30°, and the lengths of the sides are 15.5 and 8.25 centimeters, then the area of ​​the figure will be 63.9375 cm?, because 15.5*8.25*sin(30°)=127.875*0 .5=63.9375.

2. If we know the length (a) of 2 parallel sides (they are identical by definition) and the height (h) of each of these sides (they are also identical), then these data are sufficient to calculate the area (S) of such a quadrilateral. Multiply the famous side length by the height: S=a*h. Let's say, if the length of the opposite sides is 12.25 centimeters and the height is 5.75 centimeters, then the area of ​​the parallelogram will be equal to 70.07 cm?, because 12.25 * 5.75 = 70.07.

3. If the lengths of the sides are unknown, but there is data on the lengths of the diagonals of the parallelogram (e and f) and the size of the angle between them (?), then these parameters are sufficient to calculate the area (S) of the figure. Find half of the product of the known lengths of the diagonals and the sine of the angle between them: S=?*e*f*sin(?). Say, if the lengths of the diagonals are 20.25 and 15.75 centimeters, and the angle between them is 25°, then the area of ​​the polygon is approximately 134.7888 cm?, because 20.25*15.75*sin(25°)? 318.9375*0.42261?134.7888.

4. When making calculations, use, say, a calculator combined with the search function in the Nigma search engine. It is convenient because it allows you to calculate the area of ​​a parallelogram by entering the entire sequence of mathematical operations in one line. Let's say, to calculate the area with the data given in the last step, enter 20.25*15.75*sin(25) into the search query and click the button to send data to the server. The server will return the calculated area value accurate to 12 decimal places (134.788811853924).

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The line of intersection of a surface with a plane belongs simultaneously to the surface and the cutting plane. The line of intersection of a cylindrical surface with a cutting plane parallel to the straight generatrix is ​​a straight line. If the cutting plane is perpendicular to the axis of the surface of rotation, the section will be a circle. In the general case, the line of intersection of a cylindrical surface with a cutting plane is a curved line.

You will need

• Pencil, ruler, triangle, patterns, compass, meter.

Instructions

1. Example: construct a section of a cylinder using a frontal projecting plane?(?₂). In this example, the section line is constructed at the points of intersection of the generatrices of the cylinder with the cutting plane?.

2. On the general projection plane P₂, the section line coincides with the projection of the cutting plane?₂ in the form of a straight line. Designate the points of intersection of the generatrices of the cylinder with the projection?₂ 1₂, 2₂, etc. to points 10₂ and 11₂.

3. On the P₁ plane, the projection of a cylinder is a circle. Points 1₂, 2₂, etc. marked on the section plane?₂ with the help of the projection connection lines are designed on the sketch of this circle. Mark their horizontal projections symmetrically about the horizontal axis of the circle.

4. Thus, the projections of the desired section are determined: on the P₂ plane - a straight line (points 1₂, 2₂...10₂); on the P₁ plane – a circle (points 1₁, 2₁…10₁).

5. Using two projections, construct the natural size of the section of this cylinder by the frontal projecting plane?. To do this, use the method of replacing projection planes. Draw a new plane P₄ parallel to the projection plane?₂. On this new x₂₄ axis, mark point 1₀. Distances between points 1₂ – 2₂, 2₂ – 4₂, etc. from the general projection of the section, place it on the x₂₄ axis, draw thin lines of projection connection perpendicular to the x₂₄ axis. In this method, the plane P₄ is replaced by the plane P₁, therefore, from the horizontal projection, transfer the dimensions from the axis to the points to the axis of the plane P₄.

6. Let's say, on P₁ for points 2 and 3 this will be the distance from 2₁ and 3₁ to the axis (point A), etc.

7. When constructing a section, you need to especially note the location of the so-called reference points. These include points lying on the projection silhouette (points 1, 10, 11), on the projection of the outermost generatrices of the surface (points 6 and 7), visibility points, etc.

8. Laying aside the indicated distances from the horizontal projection, you get points 2₀, 3₀, 6₀, 7₀, 10₀, 11₀. After this, for greater accuracy of construction, the remaining intermediate points are determined.

9. By combining all the points with a curved scythe, you will obtain the desired natural size of the section of the cylinder by the frontally projecting plane.

As usual, every cable consists of several cores, which in cross-section represent a circle. The conductivity of the cable proportionally depends on the area of ​​this cross-section. If it is too small, the cable may burn out, which is one of the main causes of fires in the modern world.

You will need

• – cable with unknown cross-section;
• – caliper or micrometer;
• – table of resistivities of substances.

Instructions

1. Take the cable whose cross-section needs to be determined. Most often, it consists of 2-4 cores, which are insulated from each other with special materials. These cores have identical diameters. Occasionally you may come across a cable, one core of which is thinner than the rest - it is prepared for grounding.

2. Remove insulation from the cable cores. Using a caliper, or preferably a micrometer (this will allow you to make more accurate measurements), determine the diameter of the core. You will get the value in millimeters. After this, calculate the cross-sectional area. To do this, multiply the indicator 0.25 by the number??3.14 and the value of the diameter d squared S=0.25???d?. Multiply this value by the number of cable cores. Knowing the length of the wire, its cross-section and the material from which it is made, calculate its resistance.

3. Say, if you need to find the cross-section of a copper cable of 4 cores, and the measurement of the core diameter gives a value of 2 mm, find the cross-sectional area. To do this, calculate the cross-sectional area of ​​one core. It will be equal to S=0.25?3.14?2?=3.14 mm?. After this, determine the cross-section of each cable; for this, multiply the cross-section of one core by their number; in our example it is 3.14? 4 = 12.56 mm?.

4. Now you can find out the highest current, the one that can flow through it, or its resistance, if the length is known. Calculate the highest current for a copper cable from the ratio of 8 A per 1 mm?. Then the limiting value of the current that can pass through the cable taken in the example is 8? 12.56 = 100.5 A. Consider that for an aluminum cable this ratio is 5 A per 1 mm?.

5. Let's say the cable length is 200 m. In order to find its resistance, multiply the resistivity of copper? in Om? mm?/m, by the cable length l and divide by its cross-sectional area S (R=??l/S). Having made the substitution, you get R=0.0175?200/12.56?0.279 Ohm, which will lead to very small losses of electricity when transmitting it through such a cable.

If a variable, sequence or function has an unlimited number of values ​​that change according to some law, it may tend to to the limit number, which is the limit sequences. Limits can be calculated using different methods.

You will need

• – representation of a numerical sequence and function;
• – knowledge of taking derivatives;
• – knowledge to transform and reduce expressions;
• - calculator.

Instructions

1. To calculate the limit, substitute the limiting value of the argument into its expression. Try the calculation. If this is acceptable, then the value of the expression with the substituted value is the desired number. Example: Detect Limit Values sequences with the universal term (3 x?-2)/(2 x?+7), if x > 3. Substitute the limit into the expression sequences (3 3?-2)/(2 3?+7)=(27-2)/(18+7)=1.

2. If there is ambiguity when trying to substitute, choose a method that can be used to resolve it. This can be done by transforming the expressions in which the sequence is written. After making the reductions, you will get the result. Example: Sequence (x+vx)/(x-vx), when x > 0. Direct substitution results in ambiguity 0/0. Get rid of it by transferring the universal factor from the numerator and denominator. In this case it will be vx. Get (vx (vx+1))/(vx (vx-1))= (vx+1)/(vx-1). Now the substitution field will get 1/(-1)=-1.

3. When, due to uncertainty, it is impossible to reduce a fraction (exclusively if the sequence contains irrational expressions), multiply its numerator and denominator by the conjugated expression in order to remove the irrationality from the denominator. Example: Sequence x/(v(x+1)-1). The value of the variable x > 0. Multiply the numerator and denominator by the conjugate expression (v(x+1)+1). Get (x (v(x+1)+1))/((v(x+1)-1) (v(x+1)+1))=(x (v(x+1)+1) )/(x+1-1)= (x (v(x+1)+1))/x=v(x+1)+1. After substitution, you get =v(0+1)+1=1+1=2.

4. For uncertainties like 0/0 or?/? use L'Hopital's rule. For this, the numerator and denominator sequences imagine them as functions, take derivatives from them. The limit of their relations will be equal to the limit of the relations of the functions themselves. Example: Detect Limit sequences ln(x)/vx, for x > ?. Direct substitution gives ambiguity?/?. Take the derivatives of the numerator and denominator and get (1/x)/(1/2 vx)=2/vx=0.

5. To resolve uncertainties, use the first delightful limit sin(x)/x=1 for x>0, or the second delightful limit (1+1/x)^x=exp for x>?. Example: Detect Limit sequences sin(5 x)/(3 x) for x>0. Transform the expression sin(5 x)/(3/5 5 x) multiply the denominator 5/3 (sin(5 x)/(5 x)) applying the 1st amazing limit get 5/3 1=5/3.

6. Example: Find the limit (1+1/(5 x))^(6 x) for x>?. Multiply and divide the exponent by 5 x. Get the expression ((1+1/(5 x))^(5 x)) ^(6 x)/(5 x). Applying the second delightful limit rule gives exp^(6 x)/(5 x)=exp.

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Tip 9: How to Find the Axial Area of ​​a Frustum of a Cone

In order to solve this problem, you need to remember what a truncated cone is and what properties it has. Be sure to make a drawing. This will allow you to determine what geometric figure the section represents cone. It is absolutely possible that after this, solving the problem will no longer present any difficulties for you.

Instructions

1. A round cone is a body obtained by rotating a triangle around one of its legs. Straight lines emanating from the apex cone and intersecting its base are called generators. If all generators are equal, then the cone is straight. At the base of the round cone lies a circle. The perpendicular dropped to the base from the vertex is the height cone. At the round straight cone the height coincides with its axis. An axis is a straight line connecting the top to the center of the base. If the horizontal cutting plane of a circular cone parallel to the base, then its upper base is a circle.

2. Since the problem statement does not specify which cone is given in this case, we can conclude that it is a round straight truncated cone, the horizontal section of which is parallel to the base. Its axial section, i.e. vertical plane that passes through the axis of a circular truncated cone, is an equilateral trapezoid. All axial sections round straight cone are equal to each other. Consequently, in order to detect the area of ​​the axial sections, it is required to find the area of ​​a trapezoid whose bases are the diameters of the bases of the truncated cone, and the lateral sides are its constituents. Frustum height cone is simultaneously the height of the trapezoid.

3. The area of ​​a trapezoid is determined by the formula: S = ?(a+b) h, where S is the area of ​​the trapezoid; a is the value of the lower base of the trapezoid; b is the value of its upper base; h is the height of the trapezoid.

4. Since the condition does not specify what exact values ​​are given, it can be assumed that the diameters of both bases and the height of the truncated cone famous: AD = d1 – diameter of the lower base of the truncated cone;BC = d2 – diameter of its upper base; EH = h1 – height cone.Thus, the area of ​​the axial sections truncated cone is determined: S1 = ? (d1+d2) h1

The regulatory documents for the design of electrical networks indicate the cross-sections of the wires, and with a caliper you can only measure diameter veins. These quantities are interrelated and can be converted from one to another.

Instructions

1. In order to translate what is specified in the regulatory document section single-core wire in its diameter, use the following formula: D=2sqrt(S/?), where D – diameter, mm; S – conductor cross-section, mm2 (it is square millimeters that electricians abbreviate as “squares”).

2. An elastic stranded wire consists of many thin strands twisted together and placed in an overall insulating sheath. This allows it not to break down during frequent movements of the load, which is connected to the power source with its help. In order to determine the diameter of one core of such a conductor (this is what can be measured with a caliper), first find the cross-section of this core: s = S/n, where s is the cross-section of one core, mm2; S – total wire cross-section (indicated in regulatory documents); n is the number of cores. After this, convert the cross-section of the core to diameter, as indicated above.

3. Printed circuit boards use flat conductors. Instead of diameter, they have thickness and width. The first value is known in advance from the technical data of the foil material. Knowing it, you can determine the cross-sectional width. To do this, use the following formula: W=S/h, where W is the width of the conductor, mm; S – conductor cross-section, mm2; h – conductor thickness, mm.

4. Square conductors are relatively rare. Its cross-section must be converted either to the length of the side or to the diagonal of the square (you can measure both with a caliper). The side length is calculated as follows: L=sqrt(S), where L – side length, mm; S – conductor cross-section, mm2. To then find out the diagonal from the side length, make the following calculations: d=sqrt(2(L^2)), where d – diagonal of the square, mm; L – side length, mm.

5. If there is no conductor whose cross-section correctly matches the required one, use another one with a larger, but in no case smaller, cross-section. Select the type of conductor and the type of its insulation depending on the conditions of use.

Note!
Before measuring the conductor with a caliper, remove the supply voltage and make sure there is no voltage using a voltmeter.

Calculate square circle unthinkable, tea is a line, the representation of area for it is not defined. But it is possible to calculate square circle bounded by this circumference. To solve the problem you need to know the radius.

Instructions

1. A circle of radius R is a geometric locus of points on the plane such that the distance from the center of the circle to them does not exceed the radius. The boundary of a circle - a circle - is the geometric locus of points, the distance from which to the center is equal to the radius R.

2. Area is a collation of a flat figure. Conventionally, it can be said that it shows how much space a figure occupies on a plane. In general case, square is found by taking the definite integral of the function y(x).

3. If you know the radius of a circle, find it square according to the formula S=? R?, where S – square, ? – number “pi”, R – radius. The number "pi" is a transcendental irrational number, a constant equal to approximately 3.14. It expresses the ratio of length circle to diameter length: ?=L/D=L/2R.

4. Example. The circle has a radius of 2 cm. Calculate square circle bounded by this circle. Solution. If we apply the formula to find the area of ​​a circle through the radius, then S=? R?=? 2?=4??3.14 2??12.56 (cm?). Occasionally number? do not substitute, leaving the result in the form S=4?. This result is less visual (it is difficult to imagine the number “pi”), but mathematically more accurate.

5. If the length is already known circle, it is allowed to count square circle through it: S=L R/2. By the way, the length circle expressed through the radius by the formula L=2? R.

6. By constructing a central angle in a circle, it is possible to obtain a sector. A sector is a part of a circle bounded by an arc and two radii that connect the center of the circle with the ends of this arc. In order to discover square sectors, you need to know not only the radius, but also the angle?: S(sectors)=? R?/2. Here? – angle in radians. The length of the arc is determined by the relation L(arc)=? R.

7. In a comprehensive review, there is such an idiomatic representation as a unit circle - a circle of radius 1. Its square, accordingly, is equal to S=?.

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The cylinder is a spatial figure and consists of 2 equal bases, which are circles and a side surface connecting the lines limiting the bases. In order to calculate square cylinder, find the areas of all its surfaces and add them up.

You will need

• ruler;
• calculator;
• concept of area of ​​a circle and circumference.

Instructions

1. Define square reasons cylinder. To do this, measure the diameter of the base using a ruler, then divide it by 2. This will be the radius of the base cylinder. Calculate square one base. To do this, square the value of its radius and multiply by continuous?, Scr= ??R?, where R is the radius cylinder, huh??3.14.

2. Discover the universal square 2 reasons, based on the definition cylinder, which indicates that its bases are equal to each other. Multiply the area of ​​one circle of the base by 2, Sbasn=2?Scr=2???R?.

3. Calculate square lateral surface cylinder. To do this, find the circumference that limits one of the bases cylinder. If the radius is already known, then calculate it by multiplying the number 2 by? and base radius R, l= 2???R, where l is the circumference of the base.

4. Measure the length of the generatrix cylinder, which is equal to the length of the segment connecting the corresponding points of the base or their centers. In an ordinary straight cylinder, the generatrix L is numerically equal to its height H. Calculate square lateral surface cylinder, multiplying the length of its base by the generator Sside = 2???R?L.

5. Calculate square surfaces cylinder, summing up square bases and side surfaces. S=Smain+ Sside. Substituting the formula values ​​of the surfaces, you get S=2???R?+2???R?L, take out the universal factors S=2???R?(R+L). This will allow you to calculate the surface cylinder with a seamless formula.

6. Let's say the diameter of the base of a straight line cylinder is 8 cm and its height is 10 cm. Determine square its lateral surface. Calculate Radius cylinder. It is equal to R=8/2=4 cm. Generator of the straight line cylinder equal to its height, that is, L = 10 cm. For calculations, use the integral formula, it is more comfortable. Then S=2???R?(R+L), substitute the corresponding numerical values ​​S=2?3.14?4?(4+10)=351.68 cm?.

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If the cross-section of an object has a complex shape, to calculate its area it should be divided into sections of primitive shapes. Later, it will be possible to calculate the areas of these areas using the appropriate formulas, and then add them up.

Instructions

1. Divide the cross section of the object into areas having the shapes of triangles, rectangles, squares, sectors, circles, semicircles and quarter circles. If the distribution results in rhombuses, divide all of them into two triangles, and if parallelograms - into two triangles and one rectangle. Measure the dimensions of all of these areas: sides, radii. Make all measurements in identical units.

2. A right triangle can be represented as half a rectangle, divided in half diagonally. To calculate the area of ​​such a triangle, multiply by each other the lengths of those sides that are adjacent to the right angle (they are called legs), then divide the result of the multiplication by two. If the triangle is not rectangular, to calculate its area, first draw the height from each angle in it. It will be divided into two different triangles, each of which will be right-angled. Measure the lengths of the legs of all of them, and then, based on the results of the measurements, calculate their areas.

3. In order to calculate square rectangle, multiply the lengths of its 2 adjacent sides. For a square they are equal, therefore you can multiply the length of one side by itself, that is, build it into a square.

4. To determine the area of ​​a circle, divide, square its radius, and then multiply the total by the number ?. If the figure is not a circle, but a semicircle, divide square by two, and if it is a quarter of a circle - by four. For the sector, measure the angle between the center of the imaginary center and the ends of the arc, convert it from degrees to radians, multiply by the square of the radius, and then divide by two.

5. Add all the resulting areas together, and you get square, expressed in units of the same order as the initial data. Say, if you measured the lengths of the sides and radii in millimeters, square will be in square millimeters.

6. A device called a planimeter will help make measuring the area of ​​a difficult figure much easier. Set its scale to zero, then move the probe along the silhouette of the figure. Read the scale readings. The accuracy of such a measurement will be relatively small.

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Tip 14: How to calculate the area of ​​a figure bounded by a parabola

It is also known from the school course that in order to find the areas of figures on the coordinate plane, you need the ability to represent such as an integral. To use it to determine the areas of curvilinear trapezoids - this is what these figures are called - it is enough to know certain algorithms.

Instructions

1. To calculate the area of ​​a figure bounded by a parabola, draw it in a Cartesian coordinate system. To depict a parabola, you must know at least three points, one must be the vertex. In order to find the coordinate of the vertex along the X axis, substitute the known data into the formula x=-b/2a, and along the Y axis, substitute the resulting argument value into the function. Later, analyze the graph data included in the problem statement. If the vertex is below the X axis, then the branches will be directed upward, if higher - downward. The remaining 2 points are the coordinates of the intersection with the OX axis. Shade the resulting figure. This will make solving this problem much easier.

2. Later, determine the limits of integration. Usually they are indicated in the statement of the problem with the help of variables a and b. Place these values ​​at the top and bottom of the integral symbol, respectively. After the integral symbol, enter the value of the function in general form and multiply it by dx (say, (x²)dx in the case of a parabola). After this, calculate the antiderivative of the value of the function in general form, using the special table at the link given in the “Additional Sources” section, then substitute the limits of integration there and find the difference. The resulting difference will be the area.

3. There is also the possibility of calculating the integral using software. To do this, follow the link located in the “Additional Sources” section to a special mathematical site. In the text field that opens, enter integral of f(x), where f(x) is a record of a function whose graph limits the area of ​​the figure on the coordinate plane. After entering, click on the button in the form of the “equals” symbol. The page that opens will depict the resulting figure and also show the progress of calculating its area.

The question relates to analytical geometry. It is solved using equations of spatial lines and planes, the representation of a cube and its geometric properties, as well as using vector algebra. Methods for solving systems of linear equations may be required.

Instructions

1. Select these tasks so that they are comprehensive, but not redundant. Cutting plane? should be given by a general equation of the form Ax+By+Cz+D=0, which is in the best agreement with its arbitrary choice. To define a cube, the coordinates of any 3 of its vertices are absolutely enough. Take, say, points M1(x1,y1,z1), M2(x2,y2,z2), M3(x3,y3,z3), according to Figure 1. This figure illustrates a cross section of a cube. It intersects two side ribs and three base ribs.

2. Decide on a plan for subsequent work. We have to look for the coordinates of the points Q, L, N, W, R where the section intersects with the corresponding edges of the cube. To do this, you will have to find the equations of the lines containing these edges and look for the points of intersection of the edges with the plane?. Later this will be followed by partitioning the pentagon QLNWR into triangles (see Fig. 2) and calculating the area of ​​all of them using the properties of the vector product. The methodology is the same every time. Consequently, we can limit ourselves to the points Q and L and the area of ​​the triangle?QLN.

3. The direction vector h of the straight line, containing the edge M1M5 (and point Q), is found as the vector product M1M2=(x2-x1, y2-y1, z2-z1) and M2M3=(x3-x2, y3-y2, z3-z2), h=(m1, n1, p1)=. The resulting vector is a guide for all other side edges. Find the length of the edge of the cube as, say, ?=?((x2-x1)^2+(y2-y1)^2+(z2-z1)^2). If the magnitude of the vector h |h|??, then replace it with the corresponding collinear vector s=(m, n, p)=(h/|h|)?. Now write down the equation of the straight line containing M1M5 parametrically (see Fig. 3). After substituting the corresponding expressions into the equation of the cutting plane, you get A(x1+mt)+B(y1+nt)+C(z1+pt)+D=0. Determine t, substitute it into the equations for M1M5 and write down the coordinates of the point Q(qx, qy, qz) (Fig. 3).

4. Apparently, point M5 has coordinates M5(x1+m, y1+n, z1+p). The direction vector for the straight line containing the edge M5M8 coincides with M2M3=(x3-x2, y3-y2,z3-z2). After this, repeat the previous reasoning regarding the point L(lx, ly, lz) (see Fig. 4). Everything that follows for N(nx, ny, nz) is an exact copy of this step.

5. Write down the vectors QL=(lx-qx, ly-qy, lz-qz) and QN=(nx-qx, ny-qy, nz-qz). The geometric meaning of their vector product is that its modulus is equal to the area of ​​a parallelogram built on vectors. Consequently, the area?QLN S1=(1/2)||. Follow the suggested method and calculate the areas of the triangles ?QNW and ?QWR – S1 and S2. It is more convenient for everyone to find the cross product with the support of the determinant vector (see Fig. 5). Write down the final result S=S1+S2+S3.

Note!
Recalculate the resulting total twice: this way you will not make mistakes in the calculations.