MR EDWARD SANG ON THE THEORY OF COMMENSURABLES. 



737 



Section 4. — On Trigonal Areas. 



40. The only regular figures which can be used to cover surface are the 

 trigon, the tetragon, and the hexagon ; and any one of these may be used in the 

 measurement of surface. Thus we may as well say, that the area of a figure is 

 so many trigonal inches, as that it is so many square inches. If we were accus- 

 tomed to it, the one mode of expression would be as intelligible as the other ; and 

 by confining om'selves to the rectangular or tessular method of denoting areas, 

 we may fail to discover all the relations of geometrical magnitudes ; or even to 

 apprehend the cause of the superior convenience of the actual system. 



Since the regular hexagon contains exactly six regular trigons, there is no 

 need for considering the trigonal and hexagonal systems separately ; it will be 

 enough for us to assume the surface of the equilateral trigon, constructed on the 

 linear unit as the unit of surface. For the sake of brevity, we shall use the 

 expression trigonal area^ as an equivalent for the area measured in equilateral 

 trigons constructed on the linear unit. 



41. If one angle of a trigon be 60° or 120°, that is ^tt or l^r, its trigonal area 

 is expressed by the product of the numbers representing the containing sides. 



42. The trigonal area of any equilateral trigon is represented by the second 

 power of the number of units in its side. 



The truth of these two theorems is apparent; theyare merely quoted for reference. 



43. If a trigon have one angle 120°, the equilateral trigon, constructed on the 

 subtense, is equivalent to the sum of the equilateral trigons, constructed on the 

 two containing sides, together with the original trigon. 



If the angle ACB be the third part of a revolution, the equilateral trigon 

 ADB, constructed on the subtense AB, 

 is equivalent to AEC and CFB, con- 

 structed on the containing sides AC, 

 CB, together with the trigon ACB. 



For ACF and ECB are straight 

 lines; join DC. Then it is easy to 

 show that the trigon DAC is equal to 

 BAE, and DBC to ABF, wherefore 

 the tetragon DACB is equivalent to 

 the pentagon AECFB, together with the 

 trigon ACB. That is, the trigon ABD 

 is equivalent to the pentagon AECFB. 



If a, h, c denote the three sides of 

 such a trigon, c being the subtense of 

 120°, we have,— 



a2+afc + 52 = c^ 



VOL. III. PART III. 



9 M 



