32 OF RECTANGULAR PARALLELOGRAMS DIVIDED INTO 



consequently, the absolute pressures as referred to the respective 

 portions, are 



but according to the conditions of the problem, these pressures are 

 equal to one another ; hence by comparison, we have 



and this by a little farther reduction, becomes 



2* 2 = Z 2 . (15). 



47. The equation in its present form, suggests a very simple geometri- 

 cal construction ; for since Z 2 is equal to twice z 2 , it is manifest, that Z 

 is the diagonal of a square of svhich the side is x ; hence 

 the following process. 



Draw the straight line AB of the same length as the 

 side of the given parallelogram, and bisect AB per- 

 pendicularly in c by the straight line CD; on AB as a 

 diameter, and about the centre c describe the semi- 

 circle ADB, cutting the straight line CD in the point 

 D; join AD, and about the point A as a centre, with 

 the distance A D, describe the arc DE meeting AB in E ; then is F the 

 point of division sought. 



Upon A B and with the given horizontal breadth, describe the paral- 

 lelogram ABHG, and through the point E, draw the straight line EF 

 parallel to AG or BH ; then will EF divide the parallelogram, exactly 

 after the manner required in the problem. The truth of the above 

 construction is manifest ; for by the property of the right angled 

 triangle, we have 



A D 2 A c 2 -J- c D ? ; 



but AC is equal toco, these being radii of the same circle, hence 

 we get 



AD 2 =2 AC 2 ; 



but by the construction, we have 



AE m AD ; 

 consequently, by substitution, it is 



AE 2 Z= 2 AC 2 , 



and doubling both sides of the equation, we get 



now AC is equal to one half of AB, and it is demonstrated by the 

 writers on geometry, that the square of any quantity is equal to four 

 times the square of its half; consequently, we have 



4AC 2 =IAB 2 ; 



therefore, by substitution, we obtain 



2AE 2 ZZZ AB 2 , 



