50 CENTRE OF GRAVITY OF PARALLELOGRAMS 



co-ordinate sr or sm, can very easily be found; for through the 

 point b the bisection of FC, draw bn parallel to BC and meeting AB 

 perpendicularly in n ; then, the triangles ban and mar are similar to 

 one another, and the sides bn, mr and an, are given to find ar, ajid 

 from thence the rectangular co-ordinate sr or sm\ consequently, 

 we have 



b a : na : : mr : r a; 



therefore, by subtraction, we get 



BrorsrazzaB r a. 



Now, CZZBW., is obviously equal to half the difference between 

 D c, the breadth of the parallelogram, and DF, the base of the triangle 

 ADF ; therefore, we have 



B = j(6.-0); 

 but an zr a B BW ; that is, a n zz J/3, and 



3(24-0)- 



therefore, by reducing the analogy, we get 

 _0(36 20). 

 ' 6(26-0)' 

 hence, by subtraction, we obtain 



36(6-0)4-0* 



3(26 0) (26). 



After the same manner that equation (25) is deducible from equa- 

 tion (20), by putting Z'zzZ; so also, is equation (26) deducible from 

 equation (21), by means of the same equality ; we might therefore have 

 dispensed with the preceding investigation, and derived the expression 

 from principles already established ; we however preferred obtaining 

 it as above, for the purpose of exhibiting that agreeable variety which 

 gives additional embellishment to scientific investigations. The method 

 of establishing the formulee, on the supposition that the side BC is 

 horizontal, is sufficiently obvious from what has been done in the 

 sixth problem preceding, and therefore, it need not be repeated here. 



COROL. By substituting the numerical values of b and 0, as given 

 in the preceding example, we shall have from equation (26) 



Therefore, from the point B, set off BS and sr respectively equal to 

 16.8 and 8.4 feet; and through the points s and r, draw sm and rm 

 parallel to AB and BC, the perpendicular sides of the given trapezoid, 



