534 



THE FORMS OF TISSUES 



[CH. 



the solid AECB = AeCO; whence it appears that the solid content of the 

 whole cell will be the same, wherever the point E be taken in Bb, and will 

 in fact be identical with the content of the hexagonal prism. We have -then 

 to enquire where E is to be taken in Bb, in order that the combined surfaces 

 of the rhomb and of the two trapezia may be a minimum. 



Because Ee is perpendicular to ^C in P, the area of the rhombus = P^.^C'; 

 and the area of the two trapezia = {Aa + Eb) x BC. The total area in question, 

 then, is PE .AC + 2BC .Bb-BC. BE. But BC . Bb is constant ; so the question 

 remains. When is PE.AC — BC.BE a minimum? 



Let a point L be so taken in Bb that BL : PL :: BC : AC. From the 

 centre P, in the plane PBE, describe the circular arc ER, meeting PL in R; 

 and on PL let fall the perpendiculars ES, BT. 



Fig. 206. 



The triangles LES, LBT, LPB are all similar. Therefore 

 LS:LE::LT:LB::LB:LP, 

 and (by hypothesis) : : BC : AC. 



Hence {LT -LS) : (LB-LE) :: BC : AC, 



i.e. ST: BE:: BC: AC. 



Therefore ST.AC = BE.BC 



and consequently, PE .AC -BE .BC = PE .AC -ST .AC 



= AC{PE-8T). 

 But PE^PR; therefore AC {PR - ST) ^ AC {PT + RS). 



