Label one endpoint of the chord A, the other end B, and the center of circle O. 1). AB=10(cm), and OA=OB=√50(cm). So, AB^2=100, OA^2+OB^2 = (√50)^2+(√50)^2 = 100. Thus, AB^2 = OA^2+OB^2. This indicates that the sides of ΔOAB satisfies the Pythagorean Triple: c^2 = a^2+b^2. So, the central angle ∠AOB=90°, and ΔOAB is a 45-45-90 triangle. 2). Since angles around the center O total 360° and central angle ∠AOB=90°=1/4x360°, the area of sector AOB = 1/4x(area of circle O). 3). The area of circle segment AB, S, is expressed as follows: S = (area of sector AOB)－(area of ΔOAB) = 1/4x(area of circle O)－(area of ΔOAB) = 1/4xπxOA^2－1/2xOAxOB = 1/4xπx√50x√50－1/2x√50x√50 = (50π－2x50)/4, If we take the value of π = 3.14, S = (3.14x50 －100)/4 = 57/4 = 14.25(cm^2). The area of circle segment AB is 14.25 cm^2.