Ill 



for the four pyramids. These different solids make up the 

 entire heap, whose volume is therefore 96 + 48 + 24 + 16 = 

 184 cubic feet. 



But the error in the two systems usually adopted could have 

 been shown in the concrete without any calculations. To take 

 the first method : Mean length X mean breadth x height. 

 The top length is the length of a long prism, and the bottom 

 length is the length of the long prism and that of the bases of 

 two pyramids, therefore the mean length is that of the long 

 prism and of the base of one pyramid. In like manner it is 

 clear that the mean breadth is the length of a short prism and 

 of the base of a pyramid. If, therefore, this method be right, 

 the heap is equal to a parallelepiped of this mean length and 

 breadth, with a height equal to that of the heap. The models 

 arranged as in figure 25 show such a parallelepiped, and it is 

 made up of the central parallelepiped of the heap, the two long 

 and the two short prisms arranged as in figure 19, and three 

 pyramids arranged as in figure 23. But the heap had four 

 pyramids ; the parallelepiped therefore is not equal to the 

 heap, and the method is consequently incorrect. 



So also is the second method — Volume equal to mean base 

 multiplied by height. The top base is that of the central 

 parallelepiped of the heap. The bottom base is that of the 

 same solid, together with those of two long and two short 

 prisms and four pyramids. The sum of these is twice the base 

 of the central solid, of the long prism, of the short prism, and 

 four times that of a pyramid, so the mean base will be that of 

 the central solid, a long prism, a short prism, and two 

 pyramids, and this should be the base of the parallepiped of 

 the height of the heap, whose volume should (if this method 

 be right) equal that of the heap. That is, it would be neces- 

 sary to add to figure 25 the base of another pyramid and com- 

 plete the parallelepiped upon that base up to the height of the 

 rest. To do this two more pyramids would be required, but 

 all the constituent parts of the heap are already used, and 

 consequently this method is wrong also. 



The first method was wrong by the omission of the volume 

 of one pyramid. This volume is equal to its base multiplied 

 by a third of its height. Its base is equal to the difference 

 between the extreme and mean lengths and breadths of the 

 heap ; therefore, if these differences be multiplied together, and 

 their product by a third of the height, and this be added to the 

 product of the first method, the true contents of the heap will 

 be found. 



These are a few illustrations of a method I venture to recom- 

 mend for the preliminary teaching of mathematics. You will 

 observe that very often it strikingly shows the intimate relations 



