A sphere whose volume is 8π is inscribed in a cube. Find the volume of the cube.

Solution

Let a be the side of the cube. Then the volume of the cube is V = a 3 .

Since the ball is inscribed in a cube, the radius of the ball is equal to half the edge of the cube, i.e. R = a/2 (see Fig.).

The volume of the ball is V w \u003d (4/3)πR 3 and is equal to 8π, therefore

(4/3)πR 3 = 8π,

And the volume of the cube is V = a 3 = (2R) 3 = 8R 3 = 8*6 = 48.

Task B9 (Case Study 2015)

The volume of the cone is 32. Through the middle of the height, a section is drawn parallel to the base of the cone, which is the base of a smaller cone with the same vertex. Find the volume of the smaller cone.

Solution

Consider the tasks:

72353. The volume of a cone is 10. A section is drawn through the middle of the height parallel to the base of the cone, which is the base of a smaller cone with the same vertex. Find the volume of the smaller cone.

Immediately, we note that the original and truncated cones are similar, and if we consider the truncated cone relative to the original, then we can say this: the smaller cone is similar to the larger one with a coefficient equal to one second or 0.5. We can write:

It could be written:

You could think so!

Consider the original cone with respect to the cut one. We can say that a larger cone is similar to a cut one with a factor of two, we write:

Now look at the solution without using similarity properties.

The volume of a cone is equal to one third of the product of the area of ​​its base and its height:

Consider a side projection (side view) with the specified section:

Let the radius of the larger cone be R, the height is H. The section (the base of the smaller cone) passes through the middle of the height, so its height will be equal to H / 2. And the radius of the base is R / 2, this follows from the similarity of triangles.

Let's write the volume of the original cone:

The volume of the cut off cone will be equal to:

Such detailed solutions are presented so that you can see how you can build reasoning. Act in any way - the main thing is that you understand the essence of the decision. Let the path you choose be not rational, the result is important (the correct result).

Answer: 1.25

318145. In a vessel shaped like a cone, the liquid level reaches half the height. The volume of liquid is 70 ml. How many milliliters of liquid must be added to completely fill the vessel?

This task is similar to the previous one. Although we are talking about a liquid here, the principle of the solution is the same.

We have two cones - this is the vessel itself and the "small" cone (filled with liquid), they are similar. It is known that the volumes of similar bodies are related as follows:

The original cone (vessel) is similar to a cone filled with liquid with a coefficient equal to 2, since it is said that the liquid level reaches half the height. You can write in more detail:

We calculate:

Thus, you need to add:

Other tasks with liquids.

74257. Find the volume V of a cone whose generatrix is ​​44 and is inclined to the plane of the base at an angle of 30 0 . Give your answer V/Pi.

Cone volume:

We find the height of the cone by the property of a right triangle.

The leg opposite the angle of 30° is equal to half of the hypotenuse. The hypotenuse, in this case, is the generatrix of the cone. Therefore, the height of the cone is 22.

We find the square of the radius of the base using the Pythagorean theorem:

*We need the square of the radius, not the radius itself.

Cone. Frustum

Tapered surface called the surface formed by all straight lines passing through each point of the given curve and a point outside the curve (Fig. 32).

This curve is called guide , direct - generating , dot - summit conical surface.

Straight circular tapered surface called the surface formed by all lines passing through each point of the given circle and a point on the line that is perpendicular to the plane of the circle and passes through its center. In what follows, this surface will be briefly referred to as conical surface (fig.33).

cone (straight circular cone ) is called a geometric body bounded by a conical surface and a plane that is parallel to the plane of the guide circle (Fig. 34).

Rice. 32 Fig. 33 Fig. 34

A cone can be considered as a body obtained by rotating a right triangle around an axis containing one of the legs of the triangle.

The circle that bounds the cone is called basis . The vertex of a conical surface is called summit cone. The line segment connecting the top of a cone with the center of its base is called height cone. The segments that form a conical surface are called generating cone. axis of a cone is a straight line passing through the vertex of the cone and the center of its base. Axial section called the section passing through the axis of the cone. Lateral surface development A cone is a sector whose radius is equal to the length of the generatrix of the cone, and the length of the arc of the sector is equal to the circumference of the base of the cone.

For a cone, the following formulas are true:

where R is the radius of the base;

H- height;

l- the length of the generatrix;

S main- base area;

S side

S full

V is the volume of the cone.

truncated cone called the part of the cone enclosed between the base and the cutting plane parallel to the base of the cone (Fig. 35).

A truncated cone can be considered as a body obtained by rotating a rectangular trapezoid about an axis containing the lateral side of the trapezoid, perpendicular to the bases.

The two circles that bound the cone are called its grounds . Height of a truncated cone is the distance between its bases. The segments that form the conical surface of a truncated cone are called generating . The straight line passing through the centers of the bases is called axis truncated cone. Axial section called the section passing through the axis of the truncated cone.

For a truncated cone, the following formulas are true:

(8)

where R is the radius of the lower base;

r is the radius of the upper base;

H is the height, l is the length of the generatrix;

S side is the lateral surface area;

S full is the total surface area;

V is the volume of the truncated cone.

Example 1 The section of the cone parallel to the base divides the height in a ratio of 1:3, counting from the top. Find the area of ​​the lateral surface of a truncated cone if the radius of the base and the height of the cone are 9 cm and 12 cm.

Solution. Let's make a drawing (Fig. 36).

To calculate the area of ​​the lateral surface of a truncated cone, we use formula (8). Find the radii of the bases About 1 A and About 1 V and generating AB.

Consider similar triangles SO 2 B and SO 1 A, coefficient of similarity , then

From here

Since then

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

Answer: .

Example2. A quarter circle of radius is folded into a conical surface. Find the radius of the base and the height of the cone.

Solution. The quadruple of a circle is a development of the lateral surface of the cone. Denote r is the radius of its base, H- height. The lateral surface area is calculated by the formula: . It is equal to the area of ​​a quarter of a circle: . We get an equation with two unknowns r and l(generator of a cone). In this case, the generatrix is ​​equal to the radius of a quarter of a circle R, so we get the following equation: , whence Knowing the radius of the base and generatrix, we find the height of the cone:

Answer: 2 cm, .

Example 3 A rectangular trapezoid with an acute angle of 45 O, a smaller base of 3 cm and an inclined side equal to , rotates around the side perpendicular to the bases. Find the volume of the obtained body of revolution.

Solution. Let's make a drawing (Fig. 37).

As a result of rotation, we get a truncated cone; in order to find its volume, we calculate the radius of the larger base and the height. in a trapeze O 1 O 2 AB we will spend AC^O 1 B. In we have: so this triangle is isosceles AC=BC\u003d 3 cm.

Answer:

Example 4 A triangle with sides 13 cm, 37 cm and 40 cm rotates around an external axis that is parallel to the larger side and is 3 cm away from it (the axis is located in the plane of the triangle). Find the surface area of ​​the resulting body of revolution.

Solution . Let's make a drawing (Fig. 38).

The surface of the resulting body of revolution consists of the side surfaces of two truncated cones and the side surface of the cylinder. In order to calculate these areas, it is necessary to know the radii of the bases of the cones and the cylinder ( BE and OC) forming cones ( BC and AC) and the height of the cylinder ( AB). The unknown is only CO. is the distance from the side of the triangle to the axis of rotation. Let's find DC. The area of ​​triangle ABC on one side is equal to the product of half of side AB and the height drawn to it DC, on the other hand, knowing all the sides of the triangle, we calculate its area using Heron's formula.

Cone volume. So we got to the cones and cylinders. In addition to those that have already been published, there will be about nine articles, we will consider all types of tasks. If new tasks are added to the open bank during the year, of course, they will also be posted on the blog. This article presents the theory, and examples in which it is used. It is not enough to know the formula for the volume of a cone, by the way, here it is:

We can write:

To solve some examples, you need to understand how the volumes of similar bodies correlate. It is to understand, and not just to learn the formula:

That is, if we increase (reduce) the linear dimensions of the body by k times, then the ratio of the volume of the resulting body to the volume of the original will be equal to k 3 .

NOTE! It doesn't matter how you define the volumes:

The fact is that in the process of solving problems when considering such bodies, some may get confused with the coefficient k. The question may arise - What is it equal to?

(depending on the value specified in the condition)

It all depends on which side you look at. It's important to understand this! Consider an example - a cube is given, the edge of the second cube is three times larger:

In this case, the similarity coefficient is equal to three (the edge is increased three times), which means that the ratio will look like this:

That is, the volume of the resulting (larger) cube will be 27 times larger.

You can look from the other side.

Given a cube, the edge of the second cube is three times smaller:

The similarity coefficient is equal to one third (reducing the edge by a factor of three), which means that the ratio will look like:

That is, the volume of the resulting cube will be 27 times smaller.

Conclusion! Indices are not important when designating volumes, it is important to understand how bodies are considered relative to each other.

It is clear that:

- if the original body increases, then the coefficient will be greater than one.

- if the original body decreases, then the coefficient will be less than one.

About the ratio of volumes, we can say the following:

- if in the problem we divide the volume of a larger body by a smaller one, then we get the cube of the similarity coefficient, and the coefficient itself will turn out to be greater than one.

- if we divide the volume of a smaller body by a larger one, then we get the cube of the similarity coefficient, and the coefficient itself will turn out to be less than one.

The most important thing to remember is that when it comes to VOLUME of similar bodies, the similarity coefficient has the THIRD degree, and not the second, as in the case of areas.

One more point regarding .

The condition contains such a thing as a generatrix of a cone. This is a segment connecting the top of the cone with the points of the circumference of the base (indicated by the letter L in the figure).

Here it is worth noting that we will analyze problems only with a direct cone (hereinafter simply a cone). The generators of a right cone are equal

Sincerely, Alexander Krutitskikh.

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