36 SECTIONS OF RECTANGULAR PARALLELOGRAMS 



therefore, by the principles of mensuration, the area of the triangle 

 efd is 



bx bx* 



. 4* X T = 2T 



Now, according to the construction, dn is equal to one third of ed, 

 and an is equal to ad minus dn; but sn is obviously equal to an\ 



hence we have 



sninl \x, 



and by the principles of Plane Trigonometry, it is 



rm zz c?~ (I ^ x) sin. <f> ; 

 consequently, the pressure on the triangle edf becomes 



_ bx*s (31 x} sin. <p 

 P ~ ~6T 



and this, by the conditions of the problem, is equal to half the pres- 

 sure on the entire parallelogram ; therefore, and by equation (7), we 

 have 



bx* s (3 1 x") sin. < _ b l*s sin. p e 



61 ~T~ 



hence, by expunging the common quantities, we get 



, 



and furthermore, by separating and transposing the terms, it 'IB 



2x 3 6lx*= 3l 3 , 

 and dividing all the terms by 2, we obtain 



x 8 3lx*=i 1.51 s . (18). 



It is somewhat remarkable, that the solution of a problem appa- 

 rently so simple, should require the reduction of a cubic equation ; 

 but so it happens, and it may be proper to observe, that in the present 

 instance, it cannot be resolved by means of an equation of a lower 

 degree. 



Now, in order to determine the value of x from the above equation, 

 we have only to substitute the numerical value of / as given in the 

 question, and then to resolve the equation by the rules given for that 

 purpose. 



53. EXAMPLE 8. Suppose the immersed length of the rectangle, or 

 that tending downwards, to be 20 feet ; how far below the surface of 

 the fluid must the point be situated, through which a line drawn 

 parallel to the diagonal, will divide the parallelogram into two parts 

 sustaining equal pressures ? 



