Mathematics in Ancient and Medieval India c6.s9.81

c6.s9.81

Let AOB be the horizontal diameter of the circle of centre O. CDE, the vertical chord intersects AOB at D. Imagining that the hawk is sitting at C and the rat at B, they see each other. The rat then runs to his hole at D, but is killed by the hawk at O. Here CO = BO, Since speed of the hawk and rat are same. Given CD = 18 cubits, DB = 81 cubits, since CD² = AD × DB, AD = 4 cubits. \therefore DO &= DB - OB / &= DB - OA / &= DB - (DO + AD) / or DO &= 1/2(DB - AD) / &= 1/2(81 - 4) = 1/2 \ . \ 77 / &= 38 1/2 cubits / Similarly BO &= 1/2(DB + AD) / &= 1/2(81 + 4) = 1/2.85 = 42 1/2 cubits Hence the rat has crossed a distance of 42½ cubits, and the horizontal movement of the hawk along CO is DO = 38½ cubits.