THE DERIVED FIGURES 261 



If y is the perpendicular distance of the centroid from the 

 axis OX, 



Then y 



area 



_ h x area of the first derived figure 

 area of the figure 



Let RS be the projection of A. 1 B 1 on LN and the lines PR and 

 PS drawn cutting AjBj in A 2 and B 2 respectively. 



., RS A ,1) , 



By similar triangles -=- = - - 

 n y 



Then h A 2 B 2 = y RS 



= y A^, since RS = AjBj 



1/2 



= ^- AB, since y AB = h AjBj 



and 7t 2 A 2 B 2 = y z AB 



Hence h* A 2 B 2 8t/ = AB y z y 



But AB y 2 $y is the moment of inertia of the strip about 

 the axis OX. The moment of inertia of the irregular area about 

 OX would be obtained by taking the sum of the moments of 

 inertia of all these strips, 



and 



Now ZA 2 B 2 Sy is the area of the figure obtained by joining 

 all the points derived in the same manner as A 2 and B 2 for 

 different positions of the horizontal line AB between the limits 

 y = and y = h. This figure is spoken of as the " second de- 

 rived figure." 



Then I ox = h 2 x area of the second derived figure 



If K is the point where the axis OY touches the boundary of 

 the irregular figure and vertical lines are drawn across the 

 figure, then by working with the projections of these lines on NM, 

 the other pair of derived figures can be obtained, 



k x area of the first derived figure 



and x = 7 r- ^ 



area of the figure 



I OY = k* x area of the second derived figure 

 when k = OM, the length of the rectangle. 



141. Working with the irregular figure given in the previous 

 example, and let this figure be divided into 10 horizontal strips 

 of equal breadth and the mid-ordinate of each strip drawn. 



