The round cone in geometry is a symmetric spatial figure having an axis of rotation. One of its important characteristics is the axial cross-sectional area. In this article, we give the formula for the cross-sectional area of the axial cone of a straight line with a round base and truncated.
What figure will we talk about?
A round cone is a figure that can be obtained as follows. It is necessary to take a triangle with a straight angle and rotate it around one of the legs. Then we get the volumetric figure shown below.
The segment AC in the figure is called the radius of the base, which “draws” a circle when it is rotated centered at point A. The leg AB is the height of the cone. Obviously, the segment AB is perpendicular to the base and is part of the axis of rotation of the figure. Point B is the height of the figure in question. The segment BE is called the generatrix, or generatrix of the cone. The set of all generatrices forms the lateral surface of the cone. It is conical. The circle bounding the base is called the guide, or director of the cone.
Since the generatrix, radius and height are the hypotenuse and legs of the considered right triangle, then for them we can write the formula:
g 2 = r 2 + h 2
Here g is the generator, r is the radius, h is the height.
Axial section of the cone and its area
In order to write down the axial sectional area formula for a cone, you first need to get acquainted with the section itself. It turns out like this: you need to take a secant plane, place it parallel to the axis of the cone. Then it is necessary to cut the cone with a plane into two identical parts so that the top of the figure gets into the plane of the section.
It is easy to imagine that as a result of the described operation, an isosceles triangle will be obtained. The equal sides of the triangle will be the same as the length of the generatrix. And the third side will be equal to the diameter of the base.
The formula for the area of the axial section of the cone (see photo above) is not difficult. It corresponds to the formula for calculating this value for the described triangle. Since the area of the triangle is equal to the product of the base and the height, which should be divided in half, the desired equality for the axial section will take the form:
S = h * r
This formula suggests that S is twice as large as the area of a right-angled triangle, by rotation of which a cone was obtained.
Truncated cone and its axial section
A truncated cone is obtained from the usual using a secant plane, which is parallel to its base. The resulting figure under the plane will be a truncated cone. It is shown in the figure.
In addition to the lateral surface, this figure consists of two bases, which are large and small circles. Denote their radii as r 1 and r 2 . The distance between the bases is called the height, we denote it by the letter h.
The axial section of the cone under consideration will be a quadrangle, the two sides of which are generators. And the other two sides will be parallel to each other and equal to 2 * r 1 and 2 * r 2, respectively. This quadrilateral will be an isosceles trapezoid, which is shown in the figure below.
This fact allows us to use the expression for the trapezoid to write the formula for the cross-sectional area of the truncated axial cone. She will take the form:
S = (2 * r 1 + 2 * r 2 ) / 2 * h = h * (r 1 + r 2 )
That is, the area S is equal to the product of the sum of the radii of the bases of the truncated cone by its height.
To solve geometric problems, you may also need a formula for the connection between the generatrix of the figure and its parameters r 1 , r 2 and h. The corresponding expression takes the form:
g 2 = h 2 + (r 1 - r 2 ) 2
It is quite simple to get it yourself if you look at a right-angled triangle inside the cone, built on the sides g, h and (r 1 - r 2 ).
The task of determining the cross-sectional area of the axial cone of a truncated
We show how to find the axial section area by the example of a truncated cone.
It is known that the height of this figure is 10 cm. It is also known that for a cone of axial section, the area is equal to the difference in the area of the bases. Knowing that the diameters of the bases differ exactly twice, it is necessary to find the area of this section along the axis.
In accordance with the condition of the problem, two equations can be written:
r 1 = 2 * r 2 ;
h * (r 1 + r 2 ) = pi * (r 1 2 - r 2 2 )
The height value is known from the condition. Thus, we have 2 equalities and 2 unknown quantities. We solve this system:
h * (2 * r 2 + r 2 ) = pi * ((2 * r 2 ) 2 - r 2 2 ) =>
3 * pi * r 2 2 - 3 * h * r 2 = 0
We have obtained an incomplete quadratic equation, which should be solved with respect to the variable r 2 . The equation has 2 roots, but the solution r 2 = 0 is not physical, so we write only one single value for a small radius:
r 2 = h / pi
Then a large radius r 1 will be equal to:
r 1 = 2 * h / pi
Substituting these equalities into the formula for the area of the axial section of the cone, we obtain:
S = h * (r 1 + r 2 ) = 3 * h 2 / pi
We substitute the numerical value of h and write the answer: S ≈ 95.54 cm 2 .