Find the total surface area of ​​the cone. The area of ​​the lateral and full surface of the cone

We know what a cone is, let's try to find its surface area. Why is it necessary to solve such a problem? For example, you need to understand how much the test will go to make a waffle cone? Or how many bricks would it take to lay down the brick roof of a castle?

It is not easy to measure the lateral surface area of ​​a cone. But imagine the same horn wrapped in cloth. To find the area of ​​a piece of fabric, you need to cut it and lay it out on the table. We get a flat figure, we can find its area.

Rice. 1. Section of the cone along the generatrix

Let's do the same with the cone. Let's "cut" its lateral surface along any generatrix, for example, (see Fig. 1).

Now we “unwind” the side surface onto a plane. We get a sector. The center of this sector is the top 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 a 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 according to the available data. First, let's introduce a notation: let the angle at the top of the sector be in radians (see Fig. 3).

We will often encounter the angle at the top of the sweep in tasks. In the meantime, let's try to answer the question: can't this angle turn out to be more than 360 degrees? That is, will it not turn out that the sweep will superimpose itself? Of course not. Let's prove it mathematically. Let the sweep "overlap" itself. This means that the length of the sweep arc is greater than the circumference of the radius . But, as already mentioned, the length of the sweep arc is the circumference of the 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 course of planimetry: arc length. Sector area: .

In our case, the role is played by the generatrix , 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, one can also find the area full surface. To do this, add the base area to the lateral surface area. But the base is a circle of radius , whose area, according to the formula, is .

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

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

Rice. 4. Desired 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 generatrix: 5 cm (see Fig. 5). Further, we know that .

Example 2. The area of ​​the axial section of the cone is , the height is . Find the total surface area (see Fig. 6).

The bodies of revolution studied at school are a cylinder, a cone and a ball.

If in a USE task 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 the knowledge of stereometry begins.

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

2. How many times the volume of a cone circumscribed near the correct quadrangular pyramid, greater than the volume of the cone inscribed in this pyramid?

Everything is simple - we draw a view from below. We see that the radius of the larger circle is several 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 tasks of part B USE options in mathematics, the answer is written as an integer or finite decimal fraction. Therefore, you should not have any or in your answer in part B. Substituting the approximate value of the number is also not necessary! It must be reduced! It is for this that in some tasks the task is formulated, for example, as follows: “Find the area of ​​the lateral surface of the cylinder divided by”.

And where else are the formulas for the 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 about changing the area with an increase (decrease) in the height of the cone or the radius of its base. Theory for problem solving in . Consider the following tasks:

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

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

Plugging in the data:

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

The area of ​​the lateral surface of the cone:

The generatrix is ​​increased by 36 times. The radius remains the same, which means the circumference of the base has not changed.

So the area of ​​the lateral surface of the modified cone will look like:

Thus, it will increase by 36 times.

*The dependence 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 area of ​​the lateral surface of the cone is:

The radius is reduced 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 surface of the cone:

Find the radius:

The height and generatrix are known, by 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 truncated cone.

The section passes through the mid-height parallel to the base. This means that the radius of the base and the generatrix of the truncated cone will be 2 times less than the radius and generatrix of the original cone. Let's write down what the surface area of ​​the cut-off cone is equal to:

Got her going 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, the height is 4. Find the total surface area of ​​the cone divided by pi.

The formula for the total surface of a cone is:

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 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 on your own:

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 truncated cone.

That's all. Good luck to you!

Sincerely, Alexander.

*Share information about the site with friends through social networks.

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 side 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- generatrix of a cone.

Since the area of ​​\u200b\u200bthe base of the cone is πR 2 (as the area of ​​\u200b\u200bthe circle), then the area of ​​\u200b\u200bthe full 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 such reasoning. Let the drawing show a development of the lateral surface of the cone. Divide the arc AB into possible more equal parts and connect all division points to the center of the arc, and adjacent ones to each other with 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 is: Ah / 2 n = anh / 2 , where n is the number of triangles.

With a large number of 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 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, that is, to the generatrix of the cone.

Neglecting slight 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- generatrix of a cone.

Knowing that C \u003d 2πR, where R is the radius of the circle of the base of the cone, we obtain: S \u003d πR l.

Note. In the formula S = C l / 2, the sign of exact, and not approximate, equality is given, although on the basis of 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 the generatrix.

Let us inscribe in a cone (Fig.) some correct pyramid and denote by letters R and l numbers expressing the lengths of the perimeter of the base and the 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 indefinitely. Then the perimeter R will tend to the limit taken as the length C of the circumference of the base, and the apothem l will have a cone generator as its limit (since ΔSAK implies that SA - SK
1 / 2 R l, will tend to the limit 1/2 C L. This limit is taken as the value of the lateral surface of the cone. Denoting 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 \u003d 2 π R, then the lateral surface of the cone is expressed by the formula:

S=1/2 2π R L= π RL

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

T= π RL+ π R2= π R(L+R)

Theorem. The lateral surface of a truncated cone is equal to the product of half the sum of the circumferences of the bases and the generatrix.

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

Then the lateral surface of the inscribed pyramid is 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 circles of the bases, and the apothem l has as its limit the generatrix L of the truncated cone. Consequently, the value of the lateral surface of the inscribed pyramid tends to the limit equal to (С + С 1) L. This limit is taken as the value of the lateral surface of the truncated cone. Denoting the side surface of the truncated cone with the letter S, we will have:

S \u003d 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+R1)L.

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

BC \u003d 1 / 2 (OA + O 1 A 1) \u003d 1 / 2 (R + R 1),

R + R 1 = 2BC.

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 average section and the generatrix.

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

T= π (R 2 + R 1 2 + RL + R 1 L)