Page:Popular Science Monthly Volume 55.djvu/828

806 demonstrations that lie does not understand. Taking the parallelogram, for example, let us suppose a figure made like Fig. 4, and we saw through it along the lines A A' and B C. It does not need a very great effort of attention to recognize, experimentally if need be, that the two triangles A A' D and B B' C may be placed one upon the other and are identical. If, from the figure thus formed, we take away the right-hand triangle the parallelogram will remain; if we take away the other triangle a rectangle will be left, or a peculiar parallelogram, of which also we give the idea to the child as a figure in which the angles are formed by straight lines perpendicular to one another. Here, then, the child gains the notion of the equivalence of a parallelogram and a rectangle of the same base and height; and this notion, obtained by cutting up a piece of board or pasteboard, he will carry so seriously and firmly in his head that he will never lose it. By cutting the same parallelogram in two, along a diagonal A C, it may be easily shown that the two triangles can be placed exactly one upon the other, and that, consequently, they have equal areas. These lessons constitute a series of classical theorems in geometry which the child can try with his fingers and learn without even giving them the form of theorems. I might show the same as to the area of the trapeze and with many other theorems, but my purpose is only to present as many examples as will make my idea understood, without going into details.

Yet I can not leave this subject without showing how we can make a very child understand some of the geometrical theorems that have acquired a bad reputation in the world of candidates for degrees, including even such as the pons asinorum of Pythagoras; the demonstration, that is, that if we construct the triangles B and C on the sides of a right-angled triangle, their sum will be equal to the square A constructed on the hypotenuse. The usual demonstration of this theorem is not very complicated, but there is something tiresome, artificial, and hard in it. The demonstration I propose is almost intuitive, and the reasoning of it is both simple and rigorous.

Suppose we take two equal squares, and, making equal lengths on the four sides of one of them, join the points so obtained as indicated in the first of the two figures (Figs. 5 and 6) so as to form four right-angled triangles, and then place four other squares in the corners of the original square. These right-angled triangles are of such sort that the sum of their sides is equal to the side of the square.