The lateral surface area of a sphere and a cone. Area of the lateral and total surface of the cone

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), then the area full surface 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 its lateral surface will be expressed by 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. Having designated lateral surface 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)

Back forward

Attention! Slide previews are for informational purposes only and may not represent all the features of the presentation. If you are interested this work, please download the full version.

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 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 area of the lateral surface 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 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 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:

l=12 cm, φ =120°
l=10 cm, φ =150°
l=15 cm, φ =120°
l=10 cm, φ =170°
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)

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

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 is equal to the generatrix of the cone, and the length of its arc coincides with the circumference of the base of the cone. Such a 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 length of the sweep arc is greater than the length of the circle of 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 lateral surface area, the total surface area can also be found. To do this, add the area of the base 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).