90 
ARCHIMEDES 
Taking the limit, we have, if A denote the area of the 
triangle EqQ, so that A = nX, 
'A 
area of segment = 
X 2 dX 
èA• 
II. The purely geometrical method simply exhausts the 
parabolic segment by inscribing successive figures ‘ in the 
recognized manner’ (see p. 79, above). For this purpose 
it is necessary to find, in terms of the triangle with the same 
base and height, the area added to the 
inscribed figure by doubling the number of 
sides other than the base of the segment. 
Let QPq be the triangle inscribed ‘ in the 
recognized manner ’, P being the point of 
contact of the tangent parallel to Qq, and 
PV the diameter bisecting Qq. If QV, Vq 
be bisected in XI, m, and PM, rm be drawn 
parallel to PT meeting the curve in P, r, 
the latter points are vertices of the next 
figure inscribed c in the recognized manner’, 
for PY, ry are diameters bisecting PQ, Pq 
respectively. 
Now QV 2 = 4PW 2 , so that PV = 4PW, or PM = 3PW. 
But YM = |PF = 2PIT, so that YM = 2 RY. 
Therefore A PPQ = \ A PQM = A PQV. 
Similarly 
APrq = ±APVq; whence {APPQ + APrq)= %PQq. (Prop. 21.) 
In like manner it can be proved that the next addition 
to the inscribed figure, adds \ of the sum of As PPQ, Prq, 
and so on. 
Therefore the area of the inscribed figure 
= {1 +\ + (^) 2 +...}. APQq. (Prop. 22.) 
Further, each addition to the inscribed figure is greater 
than half the segments of the parabola left over before the 
addition is made. For, if we draw the tangent at P and 
complete the parallelogram EQqe with side EQ parallel to PV,