如图,由两个同心圆组成的圆环,O为圆心,A、B为内圆切线与外圆的交点,C为切点.则圆环面积为πCB2.证明连接OC与OB,如图.圆环的面积S=S大圆-S小圆=π(OB2-OC2).∵AB与内圆相切,∴OC⊥AB,△OCB为直角三角形.OB2-OC2=CB2.∴圆环的面积S=πCB2.综上得出结论,两个同心圆组成的圆环面积为与内圆相切的外圆的弦的长度的一半的 As shown in the figure, the circle formed by two concentric circles, O is the center of the circle, A, B is the intersection of the tangent of the inner circle and the outer circle, C is the tangent point, then the ring area is πCB2. . The area of ​​the ring S = S circle - S circle = π (OB2 - OC2). ∵ AB and the inner circle tangent, ∴ OC ⊥ AB, △ OCB for the right triangle .OB2-OC2 = Area S = πCB2 .To sum up, the area of ​​the two concentric rings is half of the length of the outer circular tangent to the inner circle