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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 a new mathematical concept;
- formation of new training centers;
- formation of practical problem solving skills.
- developing:
- development of independent thinking of students;
- development of correct speech skills of schoolchildren.
- educational:
- developing teamwork skills.
Lesson equipment: magnetic board, computer, screen, multimedia projector, cone model, lesson presentation, handouts.
Lesson objectives (for students):
- get acquainted with a new 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.
Handing in notebooks with home test work on the covered topic.
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. The 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 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, schoolchildren 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' 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.
Today we will tell you how to find the generatrix of a cone, which is often required in school geometry problems.
The concept of a cone generatrix
A right cone is a figure that is obtained by rotating 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 connecting 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 with any point on the line of the base circle.
Since a cone is formed by rotating 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 hypotenuse is the generatrix of the cone. It is not difficult to guess that the Pythagorean theorem is useful for calculating the length of the generator. And now more about how to find the length of the generatrix of the cone.
Finding the generator
The easiest way to understand how to find a generator is with a 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 a 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 a 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 generator = 64 (the square of the length of the radius) + 64 (the square of the length of the height) = 64x2 = 128. Now we take the square root of 128. As a result, we get eight roots 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 of the problem, but in a school course they can be more complex. 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 this data, it is easy to find the length of the generatrix.
The bodies of rotation studied at 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. We remember that in the problems of part B of the Unified State Examination in mathematics, the answer is written as an integer or a final decimal fraction. 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.