The conditions of the problem when finding the diagonal axial section of the cylinder may be different. Carefully read the text of a task, select the known data.
The base radius and height of a cylinderIf your problem is known to such factors as the radius of the cylinder and its height, then based on this, find the diagonal. Since the axial section is a rectangle with sides that equal the height of the cylinder and the diameter at the base, then diagonal sections - is the hypotenuse of right triangles forming the axial section. The legs in this case are the base radius and height of the cylinder. By the Pythagorean theorem (c2 = a2 + b2) find the diagonal axis of the cross section:D = √〖(4R〗^2+H^2) where D is the diagonal axial section of the cylinder, R is the base radius, H – height of the cylinder.
The base diameter and the height of the cylinderIf the task of the diameter and height of cylinder are equal, then in front of you is an axial section in the form of a square, the only difference of this condition from the previous in that you will need to split into 2 base diameter. Continue to act in accordance with the Pythagorean theorem, as in the previous problem.
The height and the surface area of a cylinderRead carefully the conditions of the problem, with a known height and area must be the hidden data, for example, the caveat that the height is greater than the base radius 8 cm In this case, find the radius of this square, then use the radius calculate height, then the Pythagorean theorem – the diameter of the axial section:Sp = 2πRH+2nr^2 , where Sp is the surface area of the cylinder.Print out the formula for finding the height using the surface area of a cylinder, remember that the condition H = 8R.H = (Sp - 2nr^2) / 2nr.
Depicting a drawing on paper, try to use as much of the sheet size for the image cylinder. Than stronger and bigger the drawing, the will be presented the solution to the problem.
Advice 2: How to find a life-size section
Properties of figures in space involved in this section of geometry, as geometry of space. The basic method for solving problems in solid geometry is a method of cross-section polyhedra. It allows you to correctly build the cross-section of the polyhedra and to determine the species of these sections.
The definition of the cross-section of any shape, that is, the actual size of this cross-section, often assumed in formulating the tasks of building a sloping section. The oblique section should be called front-secant projective plane. And build it life-size enough to perform several actions.
With a ruler and pencil, draw in figure 3 the projections – front view, top view and side view. The main projection in the front view show the path that is front-projecting section plane, draw a sloping line.
On an incline direct select main points: points of entry section and exit section. If a figure is a rectangle, the points of entry and exit will be one. If a figure is a prism, then the number of points is doubled. Two points define the entry into and figure out. The other two identify points on the sides of the prism.
At an arbitrary distance guide line parallel to the front projecting section plane. Then, from the points located on the axis of the main view, swipe to the auxiliary line perpendicular to the inclined straight line until they intersect with the parallel axis. Thus you will get the projection of the points of the shape in the new coordinate system.
To determine the width of the figure, lower straight from the main point of view the figure of top view. Indicate the appropriate indices of the projection points at each intersection of a line and a shape. For example, if point a belongs to the mind of the figure the points A’ and A” belong to projective planes.
Put in the new coordinate system the distance that is formed between the vertical projections of the main points. The figure, which is obtained as the result of development, and is a natural value of the inclined section.