108 



Let us take two equal squares A, and B, Figure 16, and 

 four equal right-angled trangles F, G, II, and I, and arrange 

 them as shown so as to leave a square C inscribed in A. The 

 square C and the four triangles occupy the whole of square 

 A. Now let us take the same four equal right-angled triangles 

 and arrange them in the two parallelograms upon square B, 

 leaving two squares D and E. The two squares D and E and 

 the four triangles occupy the whole of square B. If from the 

 equal squares A and B we take the equal right-angled triangles 

 the remainders are equal, therefore the square C must be 

 equal to the sum of squares D and E. 



But the square C is the square of the hypotenuse of each 

 of the equal right-angled triangles F, G, H, and I, and the 

 square D is the square 'of one of the sides containing the right 

 angle, and E is the square of the other side containing the 

 right angle. Wherefore the square of the hypotenuse of these 

 right-angled triangles is equal to the sum of the squares of the 

 sides containing the right angle. And as it is evident that 

 the same result would follow whatever shaped right-angled 

 triangles we took to arrange as we have shown — and in 

 practice I would make students use other shaped ones — it 

 follows that in all right-angled triangles the square of the 

 hypotenuse is equal to the sum of the square of the other sides. 



The truth thus demonstrated is the basis of all trigonometry, 

 and by it we can not only exactly measure the inaccessible, 

 but we can also sufficiently exactly measure the incommensur- 

 able. It is beyond the scope of this paper to show how this is 

 done. But to illustrate how concrete demonstrations may be 

 used to simplify the proofs of measurement of the incommensur- 



able, let us take the rule for ascertaining the area of 



3 k 



by multiplying half the circumference by the radius 



If the circle (Figure 17) be divided into 24 equal parts, as in 

 this model, half these parts may be arranged with their points 

 one way and half with their points the other (one of these parts 

 being again divided into two). The circle has now assumed a 

 rectangular shape (Figure 18), the length of which is half of the 

 circumference, and the height of which is the radius. The 

 bases of the triangles are not perfectly straight lines, but had 

 the circle been divided into hundreds'or thousands of triangles 

 instead of twenty-four, the eye could not have detected the 

 little curves. And as the calculated length of the circumference 

 is taken in the rule it may be held to be exact. 



SOLID GEOMETRY. 



However useful concrete demonstrations may be in Plane 

 Geometry they are greatly more so in Solid Geometry, as it 

 is always difficult to explain — to beginners at least — the 



