Each student has heard of a round cone and imagines what this three-dimensional figure looks like. This article defines the development of a cone, provides formulas describing its characteristics, and describes how to construct it using a compass, protractor and ruler.
Round cone in geometry
We give a geometric definition of this figure. A round cone is a surface that is formed by straight segments connecting all points of a certain circle with a single point in space. This single point should not belong to the plane in which the circle lies. If instead of a circle we take a circle, then this method also leads to a cone.
The circle is called the base of the figure; its circle is the director. The segments connecting the point with the directrix are called generatrices or generators, and the point where they intersect is the vertex of the cone.
The round cone can be straight and inclined. Both figures are shown in the figure below.
The difference between them is as follows: if the perpendicular from the top of the cone falls exactly in the center of the circle, then the cone will be straight. For him, the perpendicular, which is called the height of the figure, is part of its axis. In the case of an inclined cone, the height and axis form an acute angle.
Due to the simplicity and symmetry of the figure, we will further consider the properties of only a straight cone with a round base.
Getting a shape using rotation
Before proceeding to consider the sweep of the surface of the cone, it is useful to find out how this spatial figure can be obtained by rotation.
Suppose we have a right triangle with sides a, b, c. The first two of them are legs, c is the hypotenuse. Place the triangle on the leg a and begin to rotate it around leg b. Hypotenuse c will describe the conical surface. This simple cone production technique is depicted in the diagram below.
Obviously, the leg a is the radius of the base of the figure, leg b is its height, and hypotenuse c corresponds to the generatrix of the round straight cone.
Cone Sweep Type
As you might guess, the cone is formed by two types of surfaces. One of them is a flat circle of the base. Suppose that it has a radius r. The second surface is lateral and is called conical. Let its generator be equal to g.
If we have a paper cone, then you can take the scissors and cut off the base from it. Then, the conical surface should be cut along any generatrix and deployed on a plane. In this way, we obtained a scan of the lateral surface of the cone. Two surfaces along with the initial cone are shown in the diagram below.
The bottom circle shows the base circle. In the center, a developed conical surface is shown. It turns out that it corresponds to some circular sector of a circle whose radius is equal to the length of the generator g.
Angle and sweep area
Now we obtain formulas that, according to the known parameters g and r, allow us to calculate the area and the sweep angle of the cone.
It is obvious that the arc of the circular sector shown in the figure above has a length equal to the circumference of the base, i.e.:
l = 2 * pi * r.
If the whole circle of radius g were built, then its length would be:
L = 2 * pi * g.
Since the length L corresponds to 2 * pi radians, then the angle on which the arc l rests can be determined from the corresponding proportion:
L ==> 2 * pi;
l ==> Ο.
Then the unknown angle Ο will be equal to:
Ο = 2 * pi * l / L.
Substituting the expressions for the lengths l and L, we arrive at the formula for the scan angle of the lateral surface of the cone:
Ο = 2 * pi * r / g.
The angle Ο is expressed here in radians.
To determine the area S b of the circular sector, we use the found value of Ο. We compose another proportion, only for areas. We have:
2 * pi ==> pi * g 2 ;
Ο ==> S b .
Where should S b be expressed, and then substitute the value of the angle Ο. We get:
S b = Ο * g 2 * pi / (2 * pi) = 2 * pi * r / g * g 2/2 = pi * r * g.
For the area of ββthe conical surface, we got a fairly compact formula. The value of S b is equal to the product of three factors: the number pi, the radius of the figure and its generatrix.
Then the area of ββthe entire surface of the figure will be equal to the sum of S b and S o (the area of ββthe round base). We get the formula:
S = S b + S o = pi * r * (g + r).
Building a cone sweep on paper
To complete this task, you will need a piece of paper, a pencil, a protractor, a ruler and a compass.
First of all, we draw a right-angled triangle with sides of 3 cm, 4 cm and 5 cm. Its rotation around the leg in 3 cm will give the desired cone. The figure has r = 3 cm, h = 4 cm, g = 5 cm.
We begin the development of the sweep by drawing with a compass a circle of radius r. Its length will be 6 * pi cm. Now next to it we draw another circle, but with a radius of g. Its length will correspond to 10 * pi cm. Now we need to cut off a circular sector from a large circle. Its angle Ο is equal to:
Ο = 2 * pi * r / g = 2 * pi * 3/5 = 216 o .
Now we postpone this angle with a protractor on a circle with radius g and draw two radii that will limit the circular sector.
Thus, we built a scan of the cone with the specified parameters of the radius, height and generatrix.
An example of solving a geometric problem
Dan round straight cone. It is known that the angle of its lateral sweep is 120 o . It is necessary to find the radius and generatrix of this figure, if it is known that the height h of the cone is 10 cm.
The task is not difficult if we recall that a round cone is a figure of rotation of a right triangle. From this triangle follows an unambiguous relationship between height, radius and generatrix. We write the corresponding formula:
g 2 = h 2 + r 2 .
The second expression that should be used in the solution is the formula for the angle Ο:
Ο = 2 * pi * r / g.
Thus, we have two equations relating two unknown quantities (r and g).
Express from the second formula g and substitute the result in the first, we get:
g = 2 * pi * r / Ο;
h 2 + r 2 = 4 * pi 2 * r 2 / Ο 2 =>
r = h / β (4 * pi 2 / Ο 2 - 1).
The angle Ο = 120 o in radians is 2 * pi / 3. Substituting this value, we obtain finite formulas for r and g:
r = h / β8;
g = 3 * h / β8.
It remains to substitute the height value and get an answer to the question of the problem: r β 3.54 cm, g β 10.61 cm.