33-] 



DETERMINATION OF CENTROIDS. 



32. It has been shown in Art. 28 (b) how the centroid of a 

 homogeneous triangular area ABC can be found. 



Dividing the area into linear elements by drawing lines 

 parallel to one of the sides, say AB (Fig. 3, p. 13), it appears 

 that the centroid of each element, such as PQ, lies at its middle 

 point. The locus of these middle points is 

 the median CO of the triangle ; on this line, 

 then, the centroid G of the triangle must be 

 situated. Resolving the triangle into linear 

 elements parallel to the side BC, or to CA, 

 it follows in the same way that G must lie on 

 each of the other two medians of the triangle. 

 The intersection of these medians is there- 

 fore the centroid G. 



The point G trisects each median so that CG/GC' = 2. For 

 if A A' (Fig. 5) is another median, the triangles AGC and A'GC' 

 are similar, and A'C' = %AC' t hence CG=\CG. 



It follows from Art. 25, that the centroid of the homogeneous 

 triangular area coincides with that of three particles of equal 

 mass placed at the vertices. 



33. Homogeneous Quadrilateral. The centroid is found graphi- 

 cally by resolving the quadrilateral into triangles, finding their 

 centroids, and deducing from them the centroid of the quadri- 

 lateral. This process applies generally to any polygon and can 

 be carried out in various ways. 



Thus for the quadrilateral A BCD (Fig. 6) drawing the 

 diagonal AC and determining the centroids of the triangles 



ABC and ADC, we obtain by join- 

 ing these centroids one line on 

 which the required centroid of the 

 quadrilateral must lie. Repeating 

 the same construction for the tri- 

 angles obtained by drawing the 

 other diagonal BD, we find a second line on which the centroid 



