How to find the total surface area of ​​a cone. Area of ​​the lateral and total surface of the cone

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 an integer or finite number decimal. 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 generator is 10. Find its area full 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 lateral surface area of ​​the cone is four times more area grounds. 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:

That is, the cosine will be equal to:

Answer: 0.25

Decide for yourself:

27136. How many times will the area of ​​the lateral surface of the cone increase if its generatrix is ​​increased by 3 times?

27160. The area of ​​the lateral surface of the cone is twice the area of ​​the base. Find the angle between the generatrix of the cone and the plane of the base. Give your answer in degrees. .

27161. The total surface area of ​​the cone is 12. 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.

That's all. Good luck to you!

Sincerely, Alexander.

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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;
  • skills development correct speech schoolchildren.
• educational:
  • developing teamwork skills.

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

Lesson objectives (for students):

• meet new people 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 is explained accompanied by the 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 rotation right triangle relative to the 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 necessary formulas, the required formula is displayed. Students' arguments should look something like this:

The sector-sweep radius is equal to l, degree measure arcs – 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 No.	I task	II task
	(125/3)π ~ 41.67 π
	(425/9)π ~ 47.22 π
	(539/9)π ~ 59.89 π

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 surface area of ​​a cone (or simply the surface of a cone) is equal to the sum of the areas of the base and the lateral surface.

The area of ​​the lateral surface of the cone is calculated by the formula: S = πR l, where R is the radius of the base of the cone, and l- forming a cone.

Since the area of ​​the base of the cone is equal to πR 2 (as the area of ​​a circle), the area of ​​the total surface of the cone will be equal to: πR 2 + πR l= πR(R+ l).

Obtaining the formula for the area of ​​the lateral surface of a cone can be explained by the following reasoning. Let the drawing show the development of the lateral surface of a cone. Let us divide the arc AB into possibly larger number equal parts and connect all the division points to the center of the arc, and the neighboring ones to each other by chords.

We get a series equal triangles. The area of ​​each triangle is ah / 2 where A- length of the base of the triangle, a h- his high.

The sum of the areas of all triangles will be: ah / 2 n = anh / 2 where n- number of triangles.

At large number divisions, the sum of the areas of the triangles becomes very close to the area of ​​the development, i.e., the area of ​​the lateral surface of the cone. The sum of the bases of the triangles, i.e. an, becomes very close to the length of the arc AB, i.e., to the circumference of the base of the cone. The height of each triangle becomes very close to the radius of the arc, i.e., to the generatrix of the cone.

Neglecting minor differences in the sizes of these quantities, we obtain the formula for the area of ​​the lateral surface of the cone (S):

S=C l / 2, where C is the circumference of the base of the cone, l- forming a cone.

Knowing that C = 2πR, where R is the radius of the circle of the base of the cone, we obtain: S = πR l.

Note. In the formula S = C l / 2 there is a sign of exact, not approximate equality, although based on the above reasoning we could consider this equality to be approximate. But in high school high school it is proved that the equality

S=C l / 2 is exact, not approximate.

Theorem. The lateral surface of the cone is equal to the product of the circumference of the base and half of the generatrix.

Let's write in the cone (Fig.) some correct pyramid and denote by letters R And l numbers expressing the lengths of the perimeter of the base and apothem of this pyramid.

Then side surface it will be expressed as the product 1/2 R l .

Let us now assume that the number of sides of the polygon inscribed in the base increases without limit. Then the perimeter R will tend to the limit taken as the length C of the base circumference, and the apothem l will have as a limit the generatrix of the cone (since ΔSAK it follows that SA - SK
1 / 2 R l, will tend to the limit of 1/2 C L. This limit is taken as the size of the lateral surface of the cone. Designating the lateral surface of the cone with the letter S, we can write:

S = 1/2 C L = C 1/2 L

Consequences.
1) Since C = 2 π R, then the lateral surface of the cone is expressed by the formula:

S = 1/2 2π R L= π R.L.

2) We obtain the full surface of the cone if we add the lateral surface to the area of ​​the base; therefore, denoting the complete surface by T, we will have:

T= π RL+ π R2= π R(L+R)

Theorem. Side surface truncated cone is equal to the product of half the sum of the lengths of the circles of the bases and the generator.

Let us write into the truncated cone (Fig.) some regular truncated pyramid and denote by letters r, r 1 and l numbers expressing in identical linear units the lengths of the perimeters of the lower and upper bases and apothem of this pyramid.

Then the lateral surface of the inscribed pyramid is equal to 1/2 ( p + p 1) l

With an unlimited increase in the number of lateral faces of the inscribed pyramid, the perimeters R And R 1 tend to the limits taken as the lengths C and C 1 of the base circles, and the apothem l has as a limit the generator L of a truncated cone. Consequently, the size of the lateral surface of the inscribed pyramid tends to a limit equal to (C + C 1) L. This limit is taken as the size of the lateral surface of the truncated cone. Denoting the lateral surface of the truncated cone with the letter S, we have:

S = 1 / 2 (C + C 1) L

Consequences.
1) If R and R 1 mean the radii of the circles of the lower and upper bases, then the lateral surface of the truncated cone will be:

S = 1 / 2 (2 π R+2 π R 1) L = π (R + R 1) L.

2) If in the trapezoid OO 1 A 1 A (Fig.), from the rotation of which a truncated cone is obtained, we draw midline BC, then we get:

BC = 1 / 2 (OA + O 1 A 1) = 1 / 2 (R + R 1),

R + R 1 = 2VS.

Hence,

S=2 π BC L,

i.e. the lateral surface of a truncated cone is equal to the product of the circumference of the middle section and the generatrix.

3) The total surface T of a truncated cone will be expressed as follows:

T= π (R 2 + R 1 2 + RL + R 1 L)