# Equivalent figures : Square equivalent [I]

Geometric figures can be compared with each other by reference for this comparison both its shape and its size.

These classifications are useful to facilitate understanding and handling, letting you group transformations are performed on them using criteria structured.

Based on the different combinations that can be found in these comparisons will classify in:

• Forms similar: Have the same shape but different size
• Forms equivalent: They have different but equal size (Volume of the area)
• Forms congruent: Have the same shape and size (equal)

In plane geometry two equivalent figures are those with equal area, so to get the equivalent of another given figure we meet the terms of their respective areas.

Area Figure 1 = Area Figure 2

This expression will be the basis for the study of this relationship. As they relate to us are quadratic forms of the utility theorems height and leg, and constructs derived from Power concept; These models solve we obtain proportional means.

Divide the study of the equivalence of geometric shapes in three different stages:

• Introduction to the concept
• Obtaining the square equivalent to a given form
• Getting a form equivalent to another given.

Overall, to obtain a form equivalent to another given, use an equivalent square as intermediate between two equivalent figures. Thus, first discuss how to obtain a square equivalent to a geometric figure.

## Introduction to the concept of equivalence between figures

The following figure shows a set of triangles equivalent. All share based (b), and have the same height (h) as two of its vertices are common (B to C) and the third is in all of them on a line parallel to the base, distance h, so that its area is in all cases b * h / 2 (based on the height between the).

Equivalent triangles

## Equivalent to a square triangle

To determine the equivalent area of ​​a triangle will make a construction that allows us to obtain a mean proportional, relating this area to the equivalent of a square. Thus we get the next “the” of a square having the same area as the triangle.

We may use any of the buildings that use quadratic forms, as those derived from the concept of power or theorems height and leg which are obtained from the geometry of the right triangle.

If we use Theorem hick, construction will be similar

It includes the construction finally power

## Equivalent square polygon

To determine the equivalent square polygon phase down to a triangle, removing vertices are replaced by others who keep the area but reduce the number of sides.

For example, will reduce the following quadrilateral to a triangle

We will use a diagonal set aside a single vertex. (in a ring worth any, in general not a polygon). For the vertex has been isolated from the rest (P4) will draw a parallel to the diagonal (P1-P3)

The idea is to replace the triangle P1-P3-P4 of equal area but has its apex in the extension of one side of the polygon. We will use the point P5 P4 to replace so that the new triangle shares the base with the previous (P1-P3) and has the same height as the apex is located in parallel to the base passing through P4.

The new polygon has a side less. Once reduced the number of sides three, resolve as we have seen in the previous case.

## Equivalent to a square rectangle

Let's look at how to determine the side of a square equivalent to a basic rectangle “b” and altitude “to”

The area of ​​the rectangle is obtained by multiplying the base times the height, and it must be equal to the square side “the” equivalent square.

In this case we will use Theorem height, but also could use the hick or model based on the concept of power, as in the previous cases.

To complete the construction we obtain by rotating the base of the square sought from the side that will be used as height.

## Equivalent to a circle squared

The equivalence relation can not be established accurately in all cases, such as from “squaring the circle“, but I can deal with sufficient approximation.

Squaring the circle is called the mathematical problem, Geometry insoluble, finding consistent with a rule-and-compass a square that has a area that is equal to that of a given circle. It can only be calculated by the method of successive iterations.
Solving this problem addressed repeatedly tried to, unsuccessful, from classical antiquity to the XIX century. Speaking figuratively, it says something that is “Squaring the circle” when rendering a very difficult or impossible to solve.(W)

### Method 1

An approximation of the number Pi is the sum the Following two and three root, 3.14626436994 that the error of a 0.0046

We can calculate these graphically segment from right triangles on the circumference.

These segments we turn to place them on a line that will be used to mean proportional building.

If we apply the theorem of root height between R and two more following three R we obtain the equivalent square penalty sought, with the precision that we discussed earlier.

### Method 2

Although many methods exist, with different approaches, discuss only one more to close this section, leaving the reader to discover other interesting task with varying approximation.

In this case the number Pi will approach as 22/7 = 3.14285714286 which gives us an error of 0.0012.

Take a long segment and a length R R * 22/7 to obtain the proportional side of the square as average between the two. A possible construction is as follows, in which shows how the radius is divided into 7 parts and how to construct segments are rotated by the average height theorem. The reader is left to the detailed analysis of construction.

Metric geometry