S of the lateral surface of the cone. How to find the generatrix of a cone

We know what a cone is, let's try to find its surface area. Why do you need to solve such a problem? For example, you need to understand how much dough will go into making a waffle cone? Or how many bricks does it take to make a brick castle roof?

Measuring the lateral surface area of ​​a cone simply cannot be done. But let’s imagine the same horn wrapped in fabric. To find the area of ​​a piece of fabric, you need to cut it and lay it out on the table. It will work out flat figure, we can find its area.

Rice. 1. Section of a cone along the generatrix

Let's do the same with the cone. Let’s “cut” its side surface along any generatrix, for example (see Fig. 1).

Now let’s “unwind” the side surface onto a plane. We get a sector. The center of this sector is the vertex of the cone, the radius of the sector equal to the generator cone, and the length of its arc coincides with the circumference of the base of the cone. This sector is called the development of the lateral surface of the cone (see Fig. 2).

Rice. 2. Development of the side surface

Rice. 3. Angle measurement in radians

Let's try to find the area of ​​the sector using the available data. First, let's introduce the notation: let the angle at the vertex of the sector be in radians (see Fig. 3).

We will often have to deal with the angle at the top of the sweep in problems. For now, let’s try to answer the question: can’t this angle turn out to be more than 360 degrees? That is, wouldn’t it turn out that the sweep would overlap itself? Of course not. Let's prove this mathematically. Let the scan “superpose” on itself. This means that the sweep arc length longer circle radius . But, as already mentioned, the length of the sweep arc is the length of the circle of radius . And the radius of the base of the cone, of course, is less than the generatrix, for example, because the leg of a right triangle is less than the hypotenuse

Then let’s remember two formulas from the planimetry course: arc length. Sector area: .

In our case, the role is played by the generator , and the length of the arc is equal to the circumference of the base of the cone, that is. We have:

Finally we get: .

Along with the area of ​​the lateral surface, one can also find the area full surface. To do this, the area of ​​the base must be added to the area of ​​the lateral surface. But the base is a circle of radius, whose area according to the formula is equal to .

Finally we have: , where is the radius of the base of the cylinder, is the generatrix.

Let's solve a couple of problems using the given formulas.

Rice. 4. Required angle

Example 1. The development of the lateral surface of the cone is a sector with an angle at the apex. Find this angle if the height of the cone is 4 cm and the radius of the base is 3 cm (see Fig. 4).

Rice. 5. Right Triangle Forming a Cone

By the first action, according to the Pythagorean theorem, we find the generator: 5 cm (see Fig. 5). Next, we know that .

Example 2. The axial cross-sectional area of ​​the cone is equal to , the height is equal to . Find the total surface area (see Fig. 6).

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Lesson type: a lesson in learning new material using elements of a problem-based developmental teaching method.

Lesson objectives:

• educational:
  • familiarization with new mathematical concept;
  • formation of new training centers;
  • formation of practical problem solving skills.
• developing:
  • development of independent thinking of students;
  • development of correct speech skills of schoolchildren.
• educational:
  • developing teamwork skills.

Lesson equipment: magnetic board, computer, screen, multimedia projector, cone model, lesson presentation, handouts.

Lesson objectives (for students):

• get acquainted with a new geometric concept - cone;
• derive a formula for calculating the surface area of ​​a cone;
• learn to apply the acquired knowledge when solving practical problems.

During the classes

Stage I. Organizational.

Returning notebooks from home test work on the topic covered.

Students are invited to find out the topic of the upcoming lesson by solving the puzzle (slide 1):

Picture 1.

Announcing the topic and objectives of the lesson to students (slide 2).

Stage II. Explanation of new material.

1) Teacher's lecture.

On the board there is a table with a picture of a cone. New material explained accompanied by program material"Stereometry". A three-dimensional image of a cone appears on the screen. The teacher gives the definition of a cone and talks about its elements. (slide 3). It is said that a cone is a body formed by the rotation of a right triangle relative to a leg. (slides 4, 5). An image of a scan of the side surface of the cone appears. (slide 6)

2) Practical work.

Updating basic knowledge: repeat the formulas for calculating the area of ​​a circle, the area of ​​a sector, the length of a circle, the length of an arc of a circle. (slides 7–10)

The class is divided into groups. Each group receives a scan of the lateral surface of the cone cut out of paper (a sector of a circle with an assigned number). Students take the necessary measurements and calculate the area of ​​the resulting sector. Instructions for performing work, questions - problem statements - appear on the screen (slides 11–14). A representative of each group writes down the results of the calculations in a table prepared on the board. Participants in each group glue together a model of a cone from the pattern they have. (slide 15)

3) Statement and solution of the problem.

How to calculate the lateral surface area of ​​a cone if only the radius of the base and the length of the generatrix of the cone are known? (slide 16)

Each group takes the necessary measurements and tries to derive a formula for calculating the required area using the available data. When doing this work, students should notice that the circumference of the base of the cone is equal to the length of the arc of the sector - the development of the lateral surface of this cone. (slides 17–21) Using the necessary formulas, the desired formula is derived. Students' arguments should look something like this:

The sector-sweep radius is equal to l, degree measure of arc – φ. The area of ​​the sector is calculated by the formula: the length of the arc bounding this sector is equal to the radius of the base of the cone R. The length of the circle lying at the base of the cone is C = 2πR. Note that since the area of ​​the lateral surface of the cone is equal to the development area of ​​its lateral surface, then

So, the area of ​​the lateral surface of the cone is calculated by the formula S BOD = πRl.

After calculating the area of ​​the lateral surface of the cone model using a formula derived independently, a representative of each group writes the result of the calculations in a table on the board in accordance with the model numbers. The calculation results in each line must be equal. Based on this, the teacher determines the correctness of each group’s conclusions. The results table should look like this:

Model parameters:

1. l=12 cm, φ =120°
2. l=10 cm, φ =150°
3. l=15 cm, φ =120°
4. l=10 cm, φ =170°
5. l=14 cm, φ =110°

The approximation of calculations is associated with measurement errors.

After checking the results, the output of the formulas for the areas of the lateral and total surfaces of the cone appears on the screen (slides 22–26), students keep notes in notebooks.

Stage III. Consolidation of the studied material.

1) Students are offered problems for oral solution on ready-made drawings.

Find the areas of the complete surfaces of the cones shown in the figures (slides 27–32).

2) Question: Are the areas of the surfaces of cones formed by rotating one right triangle about different legs equal? Students come up with a hypothesis and test it. The hypothesis is tested by solving problems and written by the student on the board.

Given:Δ ABC, ∠C=90°, AB=c, AC=b, BC=a;

ВАА", АВВ" – bodies of rotation.

Find: S PPK 1, S PPK 2.

Figure 5. (slide 33)

Solution:

1) R=BC = a; S PPK 1 = S BOD 1 + S main 1 = π a c + π a 2 = π a (a + c).

2) R=AC = b; S PPK 2 = S BOD 2 + S base 2 = π b c+π b 2 = π b (b + c).

If S PPK 1 = S PPK 2, then a 2 +ac = b 2 + bc, a 2 - b 2 + ac - bc = 0, (a-b)(a+b+c) = 0. Because a, b, c – positive numbers (the lengths of the sides of the triangle), the equality is true only if a =b.

Conclusion: The surface areas of two cones are equal only if the sides of the triangle are equal. (slide 34)

3) Solving the problem from the textbook: No. 565.

Stage IV. Summing up the lesson.

Homework: paragraphs 55, 56; No. 548, No. 561. (slide 35)

Announcement of assigned grades.

Conclusions during the lesson, repetition of the main information received during the lesson.

Literature (slide 36)

1. Geometry grades 10–11 – Atanasyan, V.F. Butuzov, S.B. Kadomtsev et al., M., “Prosveshchenie”, 2008.
2. “Mathematical puzzles and charades” - N.V. Udaltsova, library “First of September”, series “MATHEMATICS”, issue 35, M., Chistye Prudy, 2010.

The bodies of rotation studied in school are the cylinder, cone and ball.

If in a problem on the Unified State Exam in mathematics you need to calculate the volume of a cone or the area of ​​a sphere, consider yourself lucky.

Apply formulas for volume and surface area of ​​a cylinder, cone and sphere. All of them are in our table. Learn by heart. This is where knowledge of stereometry begins.

Sometimes it's good to draw the view from above. Or, as in this problem, from below.

2. How many times is the volume of a cone circumscribed about a regular quadrangular pyramid greater than the volume of a cone inscribed in this pyramid?

It's simple - draw the view from below. We see that the radius of the larger circle is times larger than the radius of the smaller one. The heights of both cones are the same. Therefore, the volume of the larger cone will be twice as large.

Another important point. Remember that in the problems of part B Unified State Exam options in mathematics, the answer is written as a whole number or a finite decimal fraction. Therefore, there should not be any or in your answer in part B. There is no need to substitute the approximate value of the number either! It must definitely shrink! It is for this purpose that in some problems the task is formulated, for example, as follows: “Find the area of ​​the lateral surface of the cylinder divided by.”

Where else are the formulas for volume and surface area of ​​bodies of revolution used? Of course, in problem C2 (16). We will also tell you about it.

Here are problems with cones, the condition is related to its surface area. In particular, in some problems there is a question of changing the area when increasing (decreasing) the height of the cone or the radius of its base. Theory for solving problems in . Let's consider the following tasks:

27135. The circumference of the base of the cone is 3, the generator is 2. Find the area of ​​the lateral surface of the cone.

The lateral surface area of ​​the cone is equal to:

Substituting the data:

75697. How many times will the area of ​​the lateral surface of the cone increase if its generatrix is ​​increased by 36 times, and the radius of the base remains the same?

Cone lateral surface area:

The generatrix increases 36 times. The radius remains the same, which means the circumference of the base has not changed.

This means that the lateral surface area of ​​the modified cone will have the form:

Thus, it will increase by 36 times.

*The relationship is straightforward, so this problem can be easily solved orally.

27137. How many times will the area of ​​the lateral surface of the cone decrease if the radius of its base is reduced by 1.5 times?

The lateral surface area of ​​the cone is equal to:

The radius decreases by 1.5 times, that is:

It was found that the lateral surface area decreased by 1.5 times.

27159. The height of the cone is 6, the generatrix is ​​10. Find the area of ​​its total surface divided by Pi.

Full cone surface:

You need to find the radius:

The height and generatrix are known, using the Pythagorean theorem we calculate the radius:

Thus:

Divide the result by Pi and write down the answer.

76299. The total surface area of ​​the cone is 108. A section is drawn parallel to the base of the cone, dividing the height in half. Find the total surface area of ​​the cut off cone.

The section passes through the middle of the height parallel to the base. This means that the radius of the base and the generatrix of the cut off cone will be 2 times less than the radius and generatrix of the original cone. Let us write down the surface area of ​​the cut off cone:

Got it to be 4 times less area surface of the original, that is, 108:4 = 27.

*Since the original and cut off cone are similar bodies, it was also possible to use the similarity property:

27167. The radius of the base of the cone is 3 and the height is 4. Find the total surface area of ​​the cone divided by Pi.

Formula for the total surface of a cone:

The radius is known, it is necessary to find the generatrix.

According to the Pythagorean theorem:

Thus:

Divide the result by Pi and write down the answer.

Task. The area of ​​the lateral surface of the cone is four times the area of ​​the base. Find what is the cosine of the angle between the generatrix of the cone and the plane of the base.

The area of ​​the base of the cone is: