The foregoing considerations carry us a long way in our investigations of many of the simpler forms of cell-division. Strictly speaking they are limited to the case of flattened cells, in which we can treat the problem as though we were simply partitioning a plane surface. But it is obvious that, though they do not teach us the whole conformation of the partition which divides a more complicated solid into two halves, yet they do, even in such a case, enlighten us so far, that they tell us the appearance presented in one plane of the actual solid. And as this is all that we see in a microscopic section, it follows that the results we have arrived at will greatly help us in the interpretation of microscopic appearances, even in comparatively complex cases of cell-division. [Illustration: Fig. 151.] Let us now return to our quadrant cell (_OAPB_), which we have found to be divided into a triangular and a quadrilateral portion, as in Fig. 147 or Fig. 151; and let us now suppose the whole system to grow, in a uniform fashion, as a prelude to further subdivision. The whole quadrant, growing uniformly (or with equal radial increments), will still remain a quadrant, and it is obvious, therefore, that for every new increment of size, more will be added to the margin of its triangular portion than to the {368} narrower margin of its quadrilateral portion; and these increments will be in proportion to the angles of arc, viz. 55° 22′ : 34° 38′, or as ·96 : ·60, i.e. as 8 : 5. And accordingly, if we may assume (and the assumption is a very plausible one), that, just as the quadrant itself divided into two halves after it got to a certain size, so each of its two halves will reach the same size before again dividing, it is obvious that the triangular portion will be doubled in size, and therefore ready to divide, a considerable time before the quadrilateral part. To work out the problem in detail would lead us into troublesome mathematics; but if we simply assume that the increments are proportional to the increasing radii of the circle, we have the following equations:― Let us call the triangular cell _T_, and the quadrilateral, _Q_ (Fig. 151); let the radius, _OA_, of the original quadrantal cell = _a_ = 1; and let the increment which is required to add on a portion equal to _T_ (such as _PP′A′A_) be called _x_, and let that required, similarly, for the doubling of _Q_ be called _x′_. Then we see that the area of the original quadrant = _T_ + _Q_ = ¼π_a_^2 = ·7854_a_^2, while the area of _T_ = _Q_ = ·3927_a_^2. The area of the enlarged sector, _p′OA′_, = (_a_ + _x_)^2 × (55° 22′) ÷ 2 = ·4831(_a_ + _x_)^2, and the area _OPA_ = _a_^2 × (55° 22′) ÷ 2  = ·4831_a_^2. Therefore the area of the added portion, _T′_, = ·4831 ((_a_ + _x_)^2 − _a_^2). And this, by hypothesis, = _T_ = ·3927_a_^2. We get, accordingly, since _a_ = 1, _x_^2 + 2_x_ = ·3927/·4831 = ·810, and, solving, _x_ + 1 = √1·81 = 1·345, or _x_ = 0·345.