80 
ARCHIMEDES 
area. Lastly (Prop. 7), if there be two parabolic segments, 
their centres of gravity divide their diameters in the same 
ratio (Archimedes enunciates this of similar segments only, 
but it is true of any two segments and is required of any two 
segments in Prop. 8). Prop. 8 now finds the centre of gravity 
of any segment by using the last proposition. It is the 
geometrical equivalent of the solution of a simple equation in 
the ratio (m, say) of AG to AO, where G is' the centre of 
gravity of the segment. 
Since the segment = f (A ABB'), the sum of the two seg 
ments AQB, AQ'B' = i(A ABB'). 
Further, if QD, Q'D' are the diameters of these segments, 
QD, Q'D' are equal, and, since the centres 
of gravity H, H' of the segments divide 
QD, Q'D' proportionally, HR' is parallel 
to QQ', and the centre of gravity of the 
two segments together is at K, the point 
where HR' meets AO. 
Now AO — 4AV (Lemma 3 to Prop. 
2), and QD = ^AO — AV = AV. But 
R divides QD in the same ratio as G 
divides AO (Prop. 7); therefore 
VK = QH = m.QD = m.AV 
Taking moments about A of the segment, the triangle ABB' 
and the sum of the small segments, we have (dividing out by 
A V and A ABB') 
i (1 + m) + |. 4 = |. 4m, 
or 15m =9, 
and m = f. 
That is, AG = f AO, or AG: GO = 3:2. 
The final proposition (10) finds the Centre of gravity of the 
portion of a parabola cut off* between two parallel chords PP', 
BB'. If PP' is the shorter of the chords and the diameter 
bisecting PP', BB' meets them in A r , 0 respectively, Archi 
medes proves that, if N0 be divided into five equal parts of 
which LM is the middle one (L being nearer to JV* than M is), 
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1523.2