[Illustration: Fig. 22.] Thus the square ACDE (fig. 21) contains one point, and has four points at the four corners. Since one-fourth of each of these four belongs to the square, the four together count as one point, and the point value of the square is two points—the one inside and the four at the corner make two points belonging to it exclusively. Now the area of this square is two unit squares, as can be seen by drawing two diagonals in fig. 22. We also notice that the square in question is equal to the sum of the squares on the sides AB, BC, of the right-angled triangle ABC. Thus we recognise the proposition that the square on the hypothenuse is equal to the sum of the squares on the two sides of a right-angled triangle. Now suppose we set ourselves the question of determining the whereabouts in the ordered system of points, the end of a line would come when it turned about a point keeping one extremity fixed at the point. We can solve this problem in a particular case. If we can find a square lying slantwise amongst the dots which is equal to one which goes regularly, we shall know that the two sides are equal, and that the slanting side is equal to the straight-way side. Thus the volume and shape of a figure remaining unchanged will be the test of its having rotated about the point, so that we can say that its side in its first position would turn into its side in the second position. Now, such a square can be found in the one whose side is five units in length. [Illustration: Fig. 23.] In fig. 23, in the square on AB, there are— 9 points interior 9 4 at the corners 1 4 sides with 3 on each side, considered as 1½ on each side, because belonging equally to two squares 6 The total is 16. There are 9 points in the square on BC. In the square on AC there are— 24 points inside 24 4 at the corners 1 or 25 altogether. Hence we see again that the square on the hypothenuse is equal to the squares on the sides. Now take the square AFHG, which is larger than the square on AB. It contains 25 points. 16 inside 16 16 on the sides, counting as 8 4 on the corners 1 making 25 altogether. If two squares are equal we conclude the sides are equal. Hence, the line AF turning round A would move so that it would after a certain turning coincide with AC. This is preliminary, but it involves all the mathematical difficulties that will present themselves. There are two alterations of a body by which its volume is not changed. One is the one we have just considered, rotation, the other is what is called shear. Consider a book, or heap of loose pages. They can be slid so that each one slips over the preceding one, and the whole assumes the shape _b_ in fig. 24. [Illustration: Fig. 24.] This deformation is not shear alone, but shear accompanied by rotation.