Today we will tell you about how to find the generatrix of a cone, which is often required in school geometry problems.
The concept of a generatrix of a cone
A right cone is a figure that results from the rotation of a right triangle around one of its legs. The base of the cone forms a circle. The vertical section of the cone is a triangle, the horizontal section is a circle. The height of a cone is the segment that connects the top of the cone to the center of the base. The generatrix of a cone is a segment that connects the vertex of the cone to any point on the line of the circumference of the base.
Since the cone is formed by the rotation of a right triangle, it turns out that the first leg of such a triangle is the height, the second is the radius of the circle lying at the base, and the generatrix of the cone will be the hypotenuse. It is easy to guess that the Pythagorean theorem is useful for calculating the length of the generatrix. And now more about how to find the length of the generatrix of the cone.
Finding a generatrix
The easiest way to understand how to find a generatrix is to specific example. Suppose the following conditions of the problem are given: the height is 9 cm, the diameter of the base circle is 18 cm. It is necessary to find the generatrix.
So, the height of the cone (9 cm) is one of the legs of the right triangle, with the help of which this cone was formed. The second leg will be the radius of the base circle. The radius is half the diameter. Thus, we divide the diameter given to us in half and get the length of the radius: 18:2 = 9. The radius is 9.
Now it is very easy to find the generatrix of the cone. Since it is the hypotenuse, the square of its length will be equal to the sum of the squares of the legs, that is, the sum of the squares of the radius and height. So, the square of the length of the generatrix = 64 (the square of the length of the radius) + 64 (the square of the length of the height) = 64x2 = 128. Now we extract Square root out of 128. As a result, we get eight roots out of two. This will be the generatrix of the cone.
As you can see, there is nothing complicated about this. For example, we took simple conditions tasks, but in the school course they can be more difficult. Remember that to calculate the length of the generatrix, you need to find out the radius of the circle and the height of the cone. Knowing these data, it is easy to find the length of the generatrix.
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 greater is the volume of a cone circumscribed near a regular quadrangular pyramid than the volume of a 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.
Geometry is a branch of mathematics that studies structures in space and the relationship between them. In turn, it also consists of sections, and one of them is stereometry. It provides for the study of the properties of volumetric figures located in space: a cube, a pyramid, a ball, a cone, a cylinder, etc.
A cone is a body in Euclidean space that bounds a conical surface and a plane on which the ends of its generators lie. Its formation occurs in the process of rotation of a right-angled triangle around any of its legs, therefore it belongs to the bodies of revolution.
Components of a cone
There are the following types of cones: oblique (or inclined) and straight. Oblique is the one whose axis intersects with the center of its base not at a right angle. For this reason, the height in such a cone does not coincide with the axis, since it is a segment that is lowered from the top of the body to the plane of its base at an angle of 90 °.
That cone, the axis of which is perpendicular to its base, is called a right cone. Axis and height in such geometric body coincide due to the fact that the top in it is located above the center of the diameter of the base.
The cone consists of the following elements:
- The circle that is its base.
- Lateral surface.
- A point not lying in the plane of the base, called the apex of the cone.
- Segments that connect the points of the circle of the base of the geometric body and its top.
All these segments are generators of the cone. They are inclined to the base of the geometric body, and in the case of a right cone their projections are equal, since the vertex is equidistant from the points of the base circle. Thus, we can conclude that in a regular (straight) cone, the generators are equal, that is, they have the same length and form the same angles with the axis (or height) and base.
Since in an oblique (or inclined) body of revolution the vertex is displaced with respect to the center of the base plane, the generators in such a body have different lengths and projections, since each of them is at a different distance from any two points of the base circle. In addition, the angles between them and the height of the cone will also differ.
The length of the generators in a right cone
As written earlier, the height in a straight geometric body of revolution is perpendicular to the plane of the base. Thus, the generatrix, the height and the radius of the base create a right triangle in the cone.
That is, knowing the radius of the base and the height, using the formula from the Pythagorean theorem, you can calculate the length of the generatrix, which will be equal to the sum of the squares of the base radius and height:
l 2 \u003d r 2 + h 2 or l \u003d √r 2 + h 2
where l - generatrix;
r - radius;
h - height.
Generator in an oblique cone
Based on the fact that in an oblique or oblique cone the generators do not have the same length, it will not work to calculate them without additional constructions and calculations.
First of all, you need to know the height, the length of the axis and the radius of the base.
r 1 \u003d √k 2 - h 2
where r 1 is the part of the radius between the axis and the height;
k - axis length;
h - height.
As a result of adding the radius (r) and its part lying between the axis and the height (r 1), you can find out the complete generatrix of the cone, its height and part of the diameter:
where R is the leg of a triangle formed by the height, generatrix and part of the diameter of the base;
r - base radius;
r 1 - part of the radius between the axis and the height.
Using the same formula from the Pythagorean theorem, you can find the length of the generatrix of the cone:
l \u003d √h 2 + R 2
or, without calculating R separately, combine the two formulas into one:
l = √h 2 + (r + r 1) 2 .
Regardless of whether the cone is straight or oblique and what kind of input, all methods for finding the length of the generatrix always come down to one result - the use of the Pythagorean theorem.
Cone section
An axial plane is a plane passing along its axis or height. In a right cone, such a section is an isosceles triangle, in which the height of the triangle is the height of the body, its sides are the generators, and the base is the diameter of the base. In an equilateral geometric body, the axial section is an equilateral triangle, since in this cone the diameter of the base and the generators are equal.
The plane of the axial section in a right cone is the plane of its symmetry. The reason for this is that its top is above the center of its base, that is, the plane of the axial section divides the cone into two identical parts.
Since in an oblique voluminous body height and axis do not match, the plane of the axial section may not include the height. If it is possible to build a set of axial sections in such a cone, since only one condition must be observed for this - it must pass only through the axis, then the axial section of the plane, which will belong to the height of this cone, can be carried out only one, because the number of conditions increases, and, as is known, two straight lines (together) can belong to only one plane.
Cross-sectional area
The axial section of the cone mentioned earlier is a triangle. Based on this, its area can be calculated using the formula for the area of a triangle:
S = 1/2 * d * h or S = 1/2 * 2r * h
where S is the cross-sectional area;
d - base diameter;
r - radius;
h - height.
In an oblique, or inclined cone, the cross section along the axis is also a triangle, so the cross-sectional area in it is calculated in a similar way.
Volume
Because the cone is voluminous figure in 3D space, you can calculate its volume. The volume of a cone is a number that characterizes this body in a unit of volume, that is, in m 3. The calculation does not depend on whether it is straight or oblique (oblique), since the formulas for these two types of bodies do not differ.
As stated earlier, the formation of a right cone occurs due to the rotation of a right triangle along one of its legs. An inclined or oblique cone is formed differently, since its height is shifted away from the center of the base plane of the body. Nevertheless, such differences in structure do not affect the method of calculating its volume.
Volume calculation
Any cone looks like this:
V = 1/3 * π * h * r2
where V is the volume of the cone;
h - height;
r - radius;
π is a constant equal to 3.14.
To calculate the height of a body, it is necessary to know the radius of the base and the length of its generatrix. Since the radius, height and generatrix are combined into a right triangle, the height can be calculated using the formula from the Pythagorean theorem (a 2 + b 2 \u003d c 2 or in our case h 2 + r 2 \u003d l 2, where l is the generatrix). In this case, the height will be calculated by extracting the square root of the difference between the squares of the hypotenuse and the other leg:
a \u003d √c 2 - b 2
That is, the height of the cone will be equal to the value obtained after extracting the square root from the difference between the square of the length of the generatrix and the square of the radius of the base:
h \u003d √l 2 - r 2
Having calculated the height by this method and knowing the radius of its base, it is possible to calculate the volume of the cone. In this case, the generatrix plays an important role, since it serves as an auxiliary element in the calculations.
Similarly, if you know the height of the body and the length of its generatrix, you can find the radius of its base by extracting the square root of the difference between the square of the generatrix and the square of the height:
r \u003d √l 2 - h 2
Then, using the same formula as above, calculate the volume of the cone.
Tilt cone volume
Since the formula for the volume of a cone is the same for all types of a body of revolution, the difference in its calculation is the search for height.
In order to find the height of an inclined cone, the input data must include the length of the generatrix, the radius of the base, and the distance between the center of the base and the intersection of the height of the body with the plane of its base. Knowing this, one can easily calculate that part of the diameter of the base, which will be the base of a right triangle (formed by the height, the generatrix and the plane of the base). Then, again using the Pythagorean theorem, calculate the height of the cone, and subsequently its volume.
Back forward
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Lesson type: a lesson in studying new material using elements of a problem-developing teaching method.
Lesson Objectives:
- cognitive:
- familiarization with a new mathematical concept;
- formation of new ZUN;
- the formation of practical skills for solving problems.
- developing:
- development of independent thinking of students;
- skills development correct speech schoolchildren.
- educational:
- development of teamwork skills.
Lesson equipment: magnetic board, computer, screen, multimedia projector, cone model, lesson presentation, handout.
Lesson objectives (for students):
- get acquainted with a new geometric concept - a cone;
- derive a formula for calculating the surface area of a cone;
- learn to apply the acquired knowledge in solving practical problems.
During the classes
I stage. Organizational.
Submission of notebooks with home test work on the topic covered.
Students are invited to find out the topic of the upcoming lesson by solving the rebus (slide 1):
Picture 1.
Announcement to students of the topic and objectives of the lesson (slide 2).
II stage. Explanation of new material.
1) Teacher's lecture.
On the board is a table with the image of a cone. new material explained in the accompanying program material "Stereometry". A three-dimensional image of a cone appears on the screen. The teacher gives a definition of a cone, 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 the leg. (slides 4, 5). An image of the development of the lateral surface of the cone appears. (slide 6)
2) Practical work.
Actualization of 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 circle sector with an assigned number). Students take the necessary measurements and calculate the area of the resulting sector. Instructions for doing work, questions - problem statements - appear on the screen (slides 11-14). The representative of each group writes the results of the calculations in a table prepared on the board. The participants of each group glue the model of the cone from the development 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 makes 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' reasoning should look something like this:
The radius of the sector - sweep is equal to l, the degree measure of the arc is 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 area of the development 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 lateral surface area of the cone model according to the 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 row must be equal. On this basis, the teacher determines the correctness of the conclusions of each group. The result table should look like this:
|
model no. |
I task |
II task |
(125/3)π ~ 41.67π |
||
(425/9)π ~ 47.22π |
||
(539/9)π ~ 59.89π |
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 full surfaces of the cone appears on the screen (slides 22-26) students keep notes in notebooks.
III stage. Consolidation of the studied material.
1) Students are offered tasks for oral solution on ready-made drawings.
Find the areas of the total surfaces of the cones shown in the figures (slides 27-32).
2) Question: Are the areas of the surfaces of cones formed by the rotation of one right triangle about different legs equal? Students make a hypothesis and test it. Hypothesis testing is carried out by solving problems and is written by the student on the blackboard.
Given:Δ ABC, ∠C=90°, AB=c, AC=b, BC=a;
BAA", ABV" - bodies of revolution.
Find: S PPC 1 , S PPC 2 .
Figure 5 (slide 33)
Solution:
1) R=BC = a; S PPC 1 = S BOD 1 + S main 1 = π a c + π a 2 \u003d π a (a + c).
2) R=AC = b; S PPC 2 = S BOD 2 + S main 2 = π b c + π b 2 \u003d π b (b + c).
If S PPC 1 = S PPC 2, then a 2 + ac \u003d b 2 + bc, a 2 - b 2 + ac - bc \u003d 0, (a-b) (a + b + c) \u003d 0. Because a, b, c positive numbers (the lengths of the sides of the triangle), the tore-equality is true only if a =b.
Conclusion: The areas of the surfaces of two cones are equal only if the legs of the triangle are equal. (slide 34)
3) Solution of the problem from the textbook: No. 565.
IV stage. Summing up the lesson.
Homework: p.55, 56; No. 548, No. 561. (slide 35)
Announcement of grades.
Conclusions during the lesson, repetition of the main information received in the lesson.
Literature (slide 36)
- Geometry grades 10–11 - Atanasyan, V. F. Butuzov, S. B. Kadomtsev et al., M., Enlightenment, 2008.
- "Mathematical puzzles and charades" - N.V. Udaltsov, library "First of September", series "MATHEMATICS", issue 35, M., Chistye Prudy, 2010.