ON IMMERSED RIGHT ANGLED PARALLELOGRAMS. 23 



udc is right angled at d; therefore, by the property of the right- 

 angled triangle, we have 



aczz -v/ ad* 

 or by employing the appropriate symbols, we have 



Dizrv/r-M 2 . 



But, according to the construction, and by the nature of the centre 

 of gravity, we have 



am"=z -i-ac, and an='%ac, 



or symbolically, we obtain 



am -ly Z 2 -f 6 2 , and an zr f^/^-j-fe 2 . 



Now, by reason of the parallel lines em, fn, and be, the triangles 

 a em, afn, and a be, are similar among themselves ; consequently, by 

 the property of similar triangles, we have 



ac : be : : an :fn : : am : em; 



therefore, by separating the analogies, and employing the symbols, 

 it is 



D:Z::|V~F+& 2 :/ w , 

 and again, we have 



D : I :: i\/^ + & : em '> 



from these analogies, therefore, we obtain fn n: |7, and em~\l\ 

 which is otherwise manifest by drawing the dotted lines mt and nu. 



Now, in the right angled triangles erm and fsn, there are given 

 the hypothenuses em and fn, and the equal angles mer and nfs, to 

 find rm and sn, the perpendicular depths of the centres of gravity; 

 consequently, by Plane Trigonometry, we have, from the triangle 

 mer, 



rad. : sin. $ : : $1 : d, 

 and from the triangle nfs, it is 



rad. : sin. <j> : : %l : , 

 and since radius is equal to unity, these analogies become 



d-=.\l sin. 0, and zz f I sin, 0. 



But according to Inf. 2, Proposition (A), the pressure sustained by 

 each triangle, in a direction perpendicular to its surface, 



Is expressed by the product of its area, drawn into the 

 perpendicular depth of the centre of gravity. 



Now, the area of each triangle is manifestly equal to half the area 

 of the given parallelogram, and by the principles of mensuration, the 

 area of the rectangular parallelogram is equal to the product of its 

 two dimensions ; that is, of the length drawn into the breadth ; there- 

 fore, we have for the pressure on the triangle bd, 



