Instruction

1

The diagonal section of the cube has the shape of a rectangle,

**the area**(S) it is easy to calculate, knowing the length of any edge (a) three-dimensional figures. In this rectangle one of the sides is the height, which coincides with the length of the edge. The length of the other diagonal - calculate the Pythagorean theorem for the triangle in which it is the hypotenuse, and the two edges of the base - the legs. In General it can be written as: a*√2. The area of the diagonal**section**find by multiplying the two sides, the length of which you found out: S = a*a*√2 = a2*√2. For example, if the edge length of 20 cm**the area of the**diagonal**section**of the cube should be approximately equal 202*√2 ≈ 565,686 cm2.2

To compute the area of the diagonal

**cross section**of the parallelepiped (S) proceed the same way, but note that the Pythagorean theorem in this case involved the legs different lengths - length (l) and width (w) volumetric shapes. The length of the diagonal in this case is equal to √(l2+w2). Height (h) may also vary from the edge lengths of the bases, so in General the formula for the area**of the cross section**can be written as: S = h*√(l2+w2). For example, if the length, width and height of a cuboid are equal, respectively, 10, 20 and 30 cm**square**of its diagonal**section**approximately 30*√(102+202) = 30*√500 ≈ 670,82 cm2.3

The diagonal cross-section of the quadrangular pyramid has a triangular shape. If the height (H) of the polytope are known, and at its base lies a rectangle, the lengths of connected edges (a and b) which is also given in the conditions, the calculation of the area

**of the cross section**(S) begin with computing the length of a diagonal of the base. As in previous steps, use the triangle of the two edges of the base and diagonal, where the Pythagorean theorem the length of the hypotenuse is equal to √(a2+b2). The height of the pyramid in this polyhedron coincides with the height of the triangle diagonal**cross-section**, is lowered to the side whose length you just determined. So to find the area of a triangle, find half of the product height on the length of the diagonal: S = ½*H*√(a2+b2). For example, at a height of 30 cm and lengths of adjacent sides of the base 40 and 50 cm**the area of the**diagonal**section**should be approximately equal to ½ *30*√(402+502) = 15*√4100 ≈ 960,47 cm2.# Advice 2: How to find the area of a pyramid

Pyramid - complex geometrical body. It is formed of a flat polygon (the base of the pyramid), a point not lying in the plane of the polygon (the top of the pyramid) and all segments that connect the base of the pyramid to the vertex. How to find the area of the pyramid?

You will need

- ruler, pencil and paper

Instruction

1

The lateral surface area of any pyramid is equal to the sum of the areas of the lateral faces.

Because all the lateral faces of a pyramid are triangles, it is necessary to find the sum of the areas of all these triangles. The area of a triangle is calculated by multiplying the length of the base of the triangle to the length of its height.

Because all the lateral faces of a pyramid are triangles, it is necessary to find the sum of the areas of all these triangles. The area of a triangle is calculated by multiplying the length of the base of the triangle to the length of its height.

2

The base of the pyramid is a polygon. If this polygon be divided into triangles, the area of the polygon simple to compute as the sum of the squares poluchivshimsya when dividing triangles by the already known formula.

3

Finding the sum of the lateral surface area of a pyramid and the base of the pyramid, you can find the total surface area of the pyramid.

4

For calculations of area of regular pyramid using a special formula.

Example:

Before us is a pyramid. At the base is a regular n-gon with side a. The height of the side faces h (by the way, is the name apofema pyramid). The area of each lateral face is 1/2ah. The entire side surface of the pyramid has an area of n/2ha, calculated by summing the squares of the side faces. na is the perimeter of the base of the pyramid. The area of this pyramid we find: a work of apogamy of the pyramid and one half of the perimeter of its base equal to the area of lateral surface of regular pyramid.

Example:

Before us is a pyramid. At the base is a regular n-gon with side a. The height of the side faces h (by the way, is the name apofema pyramid). The area of each lateral face is 1/2ah. The entire side surface of the pyramid has an area of n/2ha, calculated by summing the squares of the side faces. na is the perimeter of the base of the pyramid. The area of this pyramid we find: a work of apogamy of the pyramid and one half of the perimeter of its base equal to the area of lateral surface of regular pyramid.

5

With regard to the area of the entire surface, then just add to the side area of the base, according to the principle discussed above.

# Advice 3: How to find the edge length of a pyramid

The pyramid is a figure that has a base in the form of a polygon and the lateral faces converging at the top with peaks. The boundaries of the lateral faces are called

**edges**. And how to find**the length of the**edges*of the pyramid*?Instruction

1

Find the boundary points of the ribs,

**the length**of which are looking for. Let it be points A and B.2

Set the coordinates of the points A and B. They need to ask three-dimensional, because the pyramid three – dimensional figure. Get A(x1, Y1, z1) and B(x2, y2, z2).

3

Calculate the needed

**length**, using the General formula: edge length*of a pyramid*is equal to square root of the sum of squares of differences of corresponding coordinates of boundary points. Substitute the numbers for your coordinates into the formula and find**the length of the**edges*of the pyramid*. In the same way, find**the length of the**ribs is not only the right*of the pyramid*, but rectangular, and truncated and arbitrary.4

Find

**the length of the**edges*of the pyramid*, in which all edges are equal, set the base figure and known height. Determine the location of the base height, i.e. the lower point. Since edges are equal, then it is possible to draw a circle whose center is the intersection point of the diagonals of the base.5

Draw straight lines connecting the opposite corners of the base

*of the pyramid*. Mark the point where they intersect. This point will be the lower bound of the height*of the pyramid*.6

Find

**the length**of diagonal of a rectangle using the Pythagorean theorem, where the sum of the squares of the legs of a right triangle is equal to the square of the hypotenuse. Get A2+b2=c2, where a and b are the legs and C is the hypotenuse. The hypotenuse will then be equal to the square root of the sum of the squares of the legs.7

Find

**the length of the**edges*of the pyramid*. First divide**the length**of the diagonal in half. All the data, substitute values in the formula of Pythagoras, are described above. Similar to the previous example, find the square root of the sum of the squares of the height*of the pyramid*and the half-diagonal.# Advice 4: How to find the area of the bases of the pyramid

Two grounds can only be a truncated

**pyramid**. In this case, the second base is formed with a cross-section parallel to the larger base**of the pyramid**. Find one**reason**in that case, if you know*the area*or the linear elements of the second.You will need

- properties of the pyramid;
- - trigonometric functions;
- the similarity of figures;
- - finding the areas of polygons.

Instruction

1

The area of the larger base

**of the pyramid**is*the area*of the polygon that represents it. If this is the right pyramid, its base lies a regular polygon. To know his*area*enough to know the only one of its sides.2

If the large base is a right triangle, find its

*area*by multiplying the square of side square root of 3 divided by 4. If the base is a square, erect his side in the second degree. In General, for any regular polygon, use the formula S=(n/4)•a2•ctg(180º/n), where n is the number of sides of a regular polygon, a is the length of its side.3

The side of the smaller base of the find, according to the formula b=2•(a/(2•tg(180º/n))-h/tg(α))•tg(180 ° /n). Here the a – side of the larger base, h – height of the truncated

**pyramid**, α is the dihedral angle at its base, n is the number of sides**of the bases**(it is the same). The area of the second base look similar to the first, using the formula the length of its sides S=(n/4)• b2•ctg(180º/n).4

If the reasons are other types of polygons, all known one

**reason**, and one of the sides of the other, the other hand, calculate as such. For example, the side of larger base 4, 6, 8 see the Big side of the smaller base of the wound is 4 cm. Calculate the coefficient of proportionality, 4/8=2 (take the large side in each of the**bases**), and calculate the other sides 6/2=3 cm, 4/2=2, see Receive side 2, 3, 4 cm smaller base side. Then, calculate their area, as the areas of triangles.5

If you know the ratio of the corresponding elements of the truncated pyramid, the ratio of the areas

**of the bases**is equal to the ratio of the squares of these elements. For example, if you know the corresponding sides**of the bases**a and A1, A2/A12=S/S1.# Advice 5: How to build a cross-section of a parallelepiped

In many textbooks there are jobs associated with the construction of sections of various geometric shapes including parallelepipeds. To cope with this task, you should be armed with some knowledge.

You will need

- paper;
- - handle;
- - the range.

Instruction

1

On a sheet of paper, draw a box. If your task says that the box needs to be rectangular, make it square corners. Remember that opposite edges must be parallel to each other. Name its vertices, for example, S1, T1, T, R, P, R1, P1 (as shown in the figure).

2

On the verge SS1TT1 put 2 points: A and C, let point a be on the segment S1T1 and a point on the segment S1S. If your task doesn't say where it should be these points, and do not specify the distance from the vertices, put them arbitrarily. Draw a straight line through points A and C. Continue this line to its intersection with the line segment ST. Label the point of intersection, let it be a point of M.

3

Put a point on the segment RT, label it as point B. draw a straight line through the points M and B. the intersection of this line with the edge of the label SP as point K.

4

Connect the points K and C. They must lie on the same face PP1SS1. Then through point B draw a straight line, parallel to the segment of the COP, will continue the line to the intersection with an edge of R1T1. The intersection of the label as point E.

5

Connect the points A and E. then select the resulting polygon ACKBE a different color – this will be a cross section of a given parallelepiped.

Note

Remember that when you build a cross-section of the parallelepiped is possible to connect only those points that lie in the same plane, if you had enough points for the build section, add them by extending the line segments to the intersection with the face on which the desired point.

Useful advice

Only the parallelepiped can be constructed of 4 sections: 2 diagonal and 2 transverse. For clarity, select the resulting polygon cross section, this can just stroke or stroke it in a different color.

# Advice 6: 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.Instruction

1

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.2

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.

3

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.4

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.

5

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.

6

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**.# Advice 7: How to find the area of a regular quadrangular pyramid

Pyramid - a polyhedron made up of a certain number having one common vertex flat lateral surfaces and one base. The Foundation, in turn, has on each side a common edge, and therefore its form determines the total number of faces of the shape. In a regular quadrangular pyramid such faces five, but to determine the total surface area is sufficient to calculate areas of only two of them.

Instruction

1

The total surface area of any polyhedron is the sum of the areas of its faces. In a regular quadrangular pyramid there are two forms of the polygons in the base has a square, the sides have a triangular configuration. Start calculations, e.g. calculation of the area of the quadrangular base of the pyramid (Sₒ). By definition, the right of the pyramid at its base must lie a regular polygon, in this case a square. If in terms of the length of an edge of (a), just erect it in the second degree: Sₒ = a2. If you know only the length of the diagonal of the base (l), to compute the area find the half of its square: Sₒ = l2/2.

2

Determine the area of the triangular lateral faces of the pyramid Sₐ. If you know the length of its shared base edge (a) and apofema (h), calculate half of the product of these two quantities: Sₐ = a*h/2. Under these conditions, the lengths of a side edge (b) and fin base (a) find half of a work the length of its base at the root of the difference between the squared length of a side edge and a quarter of square base length: Sₐ = ½*a*√(b2-a2/4). If in addition the length of the common base edge (a) given a plane angle at the vertex of the pyramid (α), calculate the ratio of the squared length of the edge to twice the cosine of half the plane angle: Sₐ = a2/(2*cos(α/2)).

3

Calculating the area of one side face (Sₐ), increase the value four times to calculate the lateral surface area of the regular quadrangular pyramid. At a certain apofema (h) and the perimeter of the base (P) is the action along with all the previous step can be replaced by calculating half the product of these two parameters: 4*Sₐ = ½*h*P. In any case, the lateral surface area will add up to calculated in the first step, with an area of square base figures - this will be the total surface area of a pyramid: S = Sₒ+4*Sₐ.