DE910847C - Educational game - Google Patents

Description

Educational game according to the Pythagorean theorem is right-angled Triangle the hypotenuse square equal to the sum of the cathetus squares and in extension of this theorem, in each triangle the square over one side is equal to the sum of the Squares over the other two sides decreased or doubled Rectangle from the second side and the projection of the third side onto the second, depending on whether the first side is at an acute or obtuse angle.

The invention now shows an educational game, which the development and Extension of the above mentioned mathematical principles from a parallelogram in the form that pieces held to certain dimensions (rectangles, parallelograms, Squares, triangles) can be put together to form mathematical figures and that these pieces of colored, transparent or opaque materials so manufactured and optionally painted in color or limited that at one Merge the figure below with the parts that cover each other Can be seen through areas.

Drawings I to IV provide evidence for the hereby set up Claims: r. In each parallelogram, the squares are above the four sides equal to the squares above the two diagonals; z. in each triangle are the squares over two sides equal to half the square from the third side plus double that Square the center line to the third side and continue to show the shape and the dimensional relationships between the individual parts: of the educational game need to have.

Fig. I shows the rectangle ÄB CD, which congruent twice from two right-angled triangles (ACB-DCB, BAD-CAD) is composed have -whose hypotenuse constituting the diagonals of the rectangle. Hence: In every rectangle the squares over the sides are equal to the squares over the two diagonals.

If the sides of the rectangle are shifted to form a parallelogram (Fig. II), the doubles of the rectangles EFBG and HDI h complement each other in Fig. III (increased for an obtuse angle, less the projection for an acute angle). Hence: In every parallelogram, the squares over the four sides are equal to the squares over the two diagonals.

A parallelogram halved by its diagonal is a triangle (ABD, Fig IV). Hence: In every triangle the squares are the same over two sides half the square from the third page plus double the square of the message line on the third page.

In order to use the evidence of the To represent the new theorem by means of a teaching game, it requires laying pieces in twelve various designs, as shown in Figs. V and XIV and by the numbers i to 1a are marked. In detail, Fig. V around a basic rectangle i, in Fig. VI around four congruent ones. right triangles 2 from hypotenuse = diagonal of the basic rectangle i, cathetus = large and small side of the basic rectangle i, in Fig. VII around a parallelogram 3 of two sides = small ones Side of the basic rectangle i, two sides = large side of the basic rectangle i, at Fig. VIII around two congruent triangles from one side a = large side of the basic rectangle i, one side h> the small side of the basic rectangle i, including an angle v on the sides a and b = large angle of the parallelogram 3, in Fig. IX by two congruent triangles 5 from one side a = large side of the basic rectangle i, one Side b = small side of the basic rectangle i. an angle enclosed by the Sides a and b = small angle of the parallelogram 3.

in Fig. X by three squares 6 from one side = large side of the basic rectangle r, in Fig. XI by two squares 7 from one side = small side of the basic rectangle i, in Fig. XII uni a square 8 from one side = diagonal of the basic rectangle i, in Fig. XIII by two rectangles g from two sides = diagonal of the basic rectangle i, two sides = half a diagonal of the basic rectangle i, in Fig. XIV by two right-angled ones Triangles io from hypotenuse = small side of rectangle i, an angle = smaller Angle of the parallelogram 3.

in Fig. XV by four rectangles i i from two sides = large side of the Basic rectangle i, two sides = small cathetus of the triangle io, in Fig. XVI around two squares 12 from one side = half the middle side of the triangle 5.

Claims (1)

PATENT CLAIM: Means to illustrate the theorem, that in every triangle the sum of the squares over two sides is equal to half the square from the third side plus your double square of the center line on the third Page is characterized by flat structures that can be laid next to one another, that when these structures are formed as triangles, rectangles, parallelograms or Squares have the following configuration: i. a basic rectangle i, 2. four congruent, Right triangles 2 from hypotenuse = diagonal of the basic rectangle i, cathetus = large and small side of the basic rectangle i, 3. a parallelogram 3 of two Sides = small side of the basic rectangle i, two sides = large side of the basic rectangle i, q .. two congruent triangles .4 from one side a = large side of the basic rectangle i, one side b = small side of the basic rectangle i, including an angle from the sides a and b = large angle of the parallelogram 3, 5. two congruent Triangles 5 from one side a = large side of the basic rectangle i, one side b = small side of the basic rectangle i, an angle enclosed by the sides a and b = small angle of the parallelogram 3, 6. three squares 6 from one side = large side of the basic rectangle i, 7. two squares 7 from one side = small Side of the basic rectangle r, B. a square 8 from one side = diagonal of the basic rectangle i, g. two rectangles g from two sides = diagonal of the basic rectangle i, two sides = half diagonal of the basic rectangle i, io. two right triangles io from hypotenuse = small side: of the rectangle i, an angle = small angle of the parallelogram 3, i i. four rectangles i i from two sides = large side of the basic rectangle i, two Sides = small legs of the triangle io, 12. two squares 12 from one side = half middle side of the triangle 5.