A history of Greek mathematics - Wilbourhall.org

90 ARCHIMEDES
Taking the limit, wc have, if A denote the area of the
triangle EqQ, so that A = nX,
area of segment
1
'"J 2
= IA.
X 2 dX
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
'
being the point of
recognized manner ', P
contact of the tangent parallel to Qq, and
PV the diameter bisecting Qq. If QV, Vq
be bisected in M, m, and RM, rm be drawn
parallel to PV meeting the curve in R, r,
the latter points are vertices of the next
Now QV 2
But
Therefore
Similarly
figure inscribed ' in the recognized manner ',
for RY, ry are diameters bisecting PQ, Pq
respectively.
4RW 2 , so that PV = 4PW, or RM = 3PJT.
YM=±PV 2PW, so that YM =2RY.
APRQ = ±APQM=±APQV.
APrq = ±APVq; whence (APRQ + APrq)= ±PQq. (Prop. 21.)
In like manner it can be proved that the next addition
to the inscribed figure adds J of the sum of AsPRQ, Prq,
and so on.
Therefore the area of the inscribed figure
= [l+i+a) 2 + ...}.APQg. (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,