The video course “Get an A” includes all the topics necessary for successful passing the Unified State Exam in mathematics for 60-65 points. Completely all problems 1-13 Profile Unified State Examination 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 necessary theory. Quick ways 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 big 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 spatial imagination. Trigonometry from scratch to problem 13. Understanding instead of cramming. Visual explanation complex concepts. Algebra. Roots, powers and logarithms, function and derivative. Basis for solution complex tasks 2 parts of the Unified State Exam.
Instructions
If you have the opportunity to use a protractor when constructing, start by choosing arbitrary point on a circle, which should become one of the vertices of the correct one. Label it, for example, with the letter A.
Draw an auxiliary segment connecting A to the center of the circle. Attach a protractor to this segment so that the zero division coincides with the center of the circle, and place an auxiliary point at the 120° mark. Through this point, draw another auxiliary segment with the beginning in the center of the circle at the intersection with circumference. Mark the intersection point with the letter B - this is the second vertex of the inscribed triangle.
Repeat the previous step, but apply the protractor to the second auxiliary segment, and the point of intersection with circumference designate it with the letter C. You will no longer need a protractor.
If there is no protractor, but there is a compass and , then start by calculating the length of the side triangle. You probably know that it can be expressed in terms of the radius of the circumscribed circle, multiplying it by triples to square root out of three, that is, by approximately 1.732050807568877. Round this to your desired precision and multiply by the radius of the circle.
Set aside the side length found in the fifth step on the compass. triangle and an auxiliary circle with a center at point A. Designate the intersection points of the two circles with the letters B and C - these are the other two vertices of the regular circle inscribed in the circle triangle.
Connect points A and B, B and C, C and A and the construction will be completed.
If a circle touches all three sides given triangle, and its center is inside the triangle, then it is called inscribed in the triangle.
You will need
- ruler, compass
Instructions
The point of intersection of the arcs along the ruler is connected to the vertex of the divisible angle;
The same is done with any other angle;
Sources:
- http://www.alleng.ru/d/math/math42.htm
Correct triangle- one in which all sides are the same length. Based on this definition, the construction of such a variety triangle but is not a difficult task.
You will need
- Ruler, sheet of lined paper, pencil
Instructions
note
In a regular (equilateral) triangle, all angles are equal to 60 degrees.
An equilateral triangle is also an isosceles triangle. If a triangle is isosceles, this means that 2 of its 3 sides are equal, and the third side is considered the base. Any regular triangle is isosceles, while the converse is not true.
Tip 4: How to find the area of a triangle inscribed in a circle
The area of a triangle can be calculated in several ways, depending on what value is known from the problem conditions. Given the base and height of a triangle, the area can be found by calculating the product of half the base and the height. In the second method, the area is calculated through the circumcircle of the triangle.
Instructions
In problems on planimetry, you have to find the area of a polygon inscribed in a circle or circumscribed around it. A polygon is considered circumscribed about a circle if it is outside and its sides touch the circle. A polygon located inside a circle is considered inscribed in it if its circles lie on it. If the problem is given , which is inscribed, all three of its vertices touch the circle. Depending on what kind of triangle is being considered, the method of the task is chosen.
The simplest case is when a regular triangle is inscribed in. Since such a triangle has everything, the radius of the circle equal to half its height. Therefore, of a triangle, you can find its area. Calculate this area in in this case can be done in any of the following ways, for example:
R=abc/4S, where S is the area of the triangle, a, b, c are the sides of the triangle
Another situation arises when the triangle is isosceles. If the base of the triangle coincides with the line of the diameter of the circle or the diameter is also the height of the triangle, the area can be calculated as follows:
S=1/2h*AC, where AC is the base of the triangle
If the radius of a circle, its angles, as well as the base coinciding with the diameter of the circle are known, the unknown height can be found using the Pythagorean theorem. The area of a triangle whose base coincides with the diameter of the circle is:
S=R*h
In another case, when the height is equal to the diameter of the circle circumscribed around isosceles triangle, its area is equal to:
S=R*AC
In a number of problems, a right triangle is inscribed in a circle. In this case, the center of the circle lies at the middle of the hypotenuse. Knowing the angles and base of a triangle, you can calculate the area using any of the methods described above.
In other cases, especially when the triangle is acute or obtuse, only the first of the above formulas is applicable.
The task is to fit into circle polygon can often confuse an adult. Her decision needs to be explained to a schoolchild, so parents go surfing the World Wide Web in search of a solution.
Instructions
Draw circle. Place the compass needle on the side of the circle, but do not change the radius. Draw two arcs crossing circle, turning the compass to the right and left.
Move the compass needle along the circle to the point where the arc intersects it. Turn the compass again and draw two more arcs, crossing the contour of the circle. This procedure repeat until you intersect the first point.
Draw circle. Draw the diameter through its center, the line should be horizontal. Construct a perpendicular to through the center of the circle, you get vertical line(SV, for example).
Divide the radius in half. Mark this point on the diameter line (label it A). Build circle with center at point A and radius AC. When crossing with horizontal line you will get another point (D, for example). As a result, the segment CD will be the side of the pentagon that needs to be inscribed.
Lay semicircles, the radius of which is equal to CD, along the contour of the circle. Thus, the original circle will be divided by five equal parts. Connect the dots with a ruler. The problem of inscribing a pentagon into circle also completed.
The following is described by fitting into circle square. Draw a diameter line. Take a protractor. Place it at the point where the diameter intersects the side of the circle. Open the compass to the length of the radius.
Draw two arcs until they intersect with circle yu, turning the compass in one direction or the other. Move the leg of the compass to the opposite point and draw two more arcs with the same solution. Connect the resulting dots.
Square the diameter, divide by two and take the root. As a result, you will get a side of a square that will easily fit into circle. Open the compass to this length. Put his needle on circle and draw an arc intersecting one side of the circle. Move the leg of the compass to the resulting point. Draw the arc again.
Repeat the procedure and draw two more points. Connect all four dots. This is an easier way to fit a square into circle.
Consider the task of fitting into circle. Draw circle. Take a point arbitrarily on the circle - it will be the vertex of the triangle. From this point, maintaining a compass, draw an arc until it intersects with circle Yu. This will be the second peak. Construct a third vertex from it in a similar way. Connect the dots with a ruler. The solution has been found.
Video on the topic
Being one of integral parts school curriculum, geometric problems to build regular polygons are quite trivial. As a rule, construction is carried out by inscribing a polygon into circle, which is drawn first. But what if circle given, but the figure is very complex?
You will need
- - ruler;
- - compass;
- - pencil;
- - paper.
Instructions
Construct a line segment perpendicular to AB and dividing it into two equal parts at the intersection point. Place the needle of the compass at point A. Place the leg with the lead at point B, or at any point on the segment that is closer to B than to A. Draw circle. Without changing the angle of the legs of the compass, set its needle to point B. Draw another circle.The drawn circles will intersect in two. Draw a straight line through them. Mark the point of intersection of this segment with segment AB as C. Mark the points of intersection of this segment with the original circle yu like D and E.
Construct a line segment DE dividing it in half. Carry out actions similar to those described in the previous step in relation to the segment DE. Let the drawn segment intersect DE at point O. This point will be the center of the circle. Also mark the points of intersection of the constructed perpendicular with the original one circle yu like F and G.
Set the opening of the compass legs so that the distance between their ends is the radius of the original circle. To do this, place the needle of the compass at one of points A, B, D, E, F or G. Place the end of the leg with the lead at point O.
Build regular hexagon. Place the compass needle at any point on the circle line. Label this point H. In a clockwise direction, make an arcuate notch with a compass so that it intersects the circle line. Label this point I. Move the compass needle to point I. Make a notch on the circle again and label the resulting point J. Similarly, construct points K, L, M. Consistently connect points H, I, J, K, L, M, H in pairs .Received
This publication contains another planimetry task for you. It relates to tasks increased complexity (profile level). But, as you will see, the solution process does not actually present any particular difficulty. Such a task can be considered a gift in the exam. So let's get started!
A circle is inscribed in a regular triangle with side “a”. A regular triangle is inscribed in this circle, into which a circle is inscribed, and so on.
a) Prove that the areas of circles form a geometric progression.
b) Find the sum of the areas of all circles.
*Reference! What is geometric progression? This is a sequence where each next member is equal to the previous one multiplied by the same number. A simple example: 3, 6, 12, 24, 48…. The previous term of the sequence is multiplied by 2 to obtain the next one. The number "2" is called the denominator geometric progression.
a) Let’s construct a regular triangle, inscribe a circle, inscribe a triangle into it and another circle into it (we’ll stop there):
Let's call the circles (from largest to smallest) simply “first” and “second”. Note that the radius of the first (larger) circle will be twice greater than radius second (in right triangle the leg lying opposite the angle of 30 degrees is equal to half the hypotenuse).
What happens to the areas of circles? We have:
That is, the area of the second circle is four times less area first. If we further consider the inscribed circles relative to each other, we will obtain the same relationship (dependence) of their areas relative to each other, that is, the area of each subsequent circle will be 4 times less than the area of the previous one. Let's write it down in more detail:
*The general formula for geometric progression is:
So we got a geometric progression. Its denominator is ¼. Proven!
b) The formula for an infinite geometric progression has the form:
This means that the sum of the areas of all circles will be equal to:
Now let’s express the radius of the first circle through the side of the triangle equal to “a”. We have (if the side is equal to “a”, then half the side is 0.5a):
Thus, we get: 
Second approach to the solution.
a) Since the radii of neighboring circles differ by a factor of two, it turns out that the similarity coefficient is 0.5 (circles are always similar). We can write:
This is a geometric progression.
b) Now let's calculate the sum of the areas of the circles. Let
It is known that in equilateral triangle The radius of the inscribed circle is equal to a third of its height, that is: 
So the area of the circle will be equal to: 