Mathematics in Ancient and Medieval India c6.s9.83

c6.s9.83

In this problem (Fig. 14b), AOB is the vertical diameter of the circle of centre O and EDC, the horizontal chord, denotes the water level. OA is the lotus stalk, AD lotus flower, and O the root of the lotus, and E and C the points where the lotus flowers just dips into the water. Here AD=6 angulas=¼ cubits DC=2 cubits and DB=(DC²)/AD = (2²)/(¼) = 2² × 4 = 16 cubits. Like that of previous problem, OA = height of the plant &= \textstyle1/2\ (DB+AD) / &= \textstyle1/2\ (16 + \textstyle1/4) / &= 65/8 = 8\textstyle1/8 cubits OD &= depth of the water / &= OA-AD = \textstyle65/8 - 1/4 = 7 7/8 cubits.