ConstituciÃ³n de USS

A Carré is a regular Polygone at four sides: it is a Quadrilatère which are at the same time a Rectangle (it has four right angles) and a Losange (its four sides have the same length).

The term Carré also indicates rise with the power 2 (undoubtedly in reference to the manner of calculating the surface starting from the side): “ has ²” can be read “ has squared”. The Curve représentatrice of the function ƒ ( X ) = X ² is a Parabole.

Properties

The four sides of a square are length equal The four angles of a square are right. The diagonals of a square are perpendicular and are cut in their medium. The sides opposite of a square are parallel two to two. That is to say " a" the length on a side of a square, then the diagonal measures has √2. The surface of the square is has ². The triangle is invariant by rotation of center O of π/4, π/2 (central symmetry) and 3π/4. The rectangle is invariant by axial symmetry according to the bisectrices on the sides and the diagonals. Any line passing by O divides the square into two superposable parts. It is as to note as the square has the properties of all the other quadrilaterals.

Construction

Construction with the compass alone

One wishes to build the square of tops $ABCD$ knowing only the points $A$ et $B$ . Let us pose $R$ the distance between $A$ et $B$ ; then, one proceeds as follows:

One traces $C_1$ le circle of center $A$ and ray $R$ (which contains $B$ then)

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one has a third point of the square on this curve.

One traces $C_2$ the circle of center $B$ et of ray $R$ (which contains $A$ then)

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the fourth point of the square is on this curve.

Let us pose $G$ a point of intersection of $C_1$ with $C_2$ ; one then builds $C_3$ centered in $G$ and of ray $R$ . This circle intersects $C_1$ in $B$ and another point $H$ .
$C_4$ , of center $H$ and ray $R$ , intersects $C_1$ in $G$ and a new point $I$ .
Let us pose $S$ the distance between $G$ and $I$ ; one then builds $C_5$ of center $I$ and ray $S$ (it contains $G$ inevitably).
$C_6$ is obtained while taking for center $B$ and ray $S$ (it contains $H$ inevitably). One notes $J$ the point of intersection between $C_6$ and $C_5$ which is same side as $G$ compared to the line $AB$ .
If $T$ is the distance between $A$ and $J$ , one builds $C_7$ the circle of center $A$ and ray $T$ (it contains $J$ inevitably).

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the point

C

is obtained by intersection between

C_7

and

C_2

One then builds $C_8$ of center $C$ and $R$ .

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the intersection of

C_8

and

C_1

is the point

D

See too

Simple: Public garden (geometry)

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