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(2)AC+AD=BC 理由如下:
连接 OD .
∵ AC 切小圆 O 于点 A , BC 切小圆 O 于点 E
∴ CE = CA
∵ 在 Rt△OAD 与 Rt△OEB 中, OA = OE,OD = OB,∠OAD = ∠OEB = 90
∴ Rt△OAD ≌ Rt△OEB (HL)
∴ EB = AD
∵ BC = CE + EB
∴ BC = AC + AD
(3)∵ ∠BAC = 90 , AB = 8,BC = 10
∴ AC = 6
∵ BC = AC + AD
∴ AD = BC AC = 4
∵ 圆环的面积 S = πODx2 πOAx2 = π (OD x2 +OAx2 )
又∵ OD +OA = AD , S = 4 π = 16πcm
连接 OD .
∵ AC 切小圆 O 于点 A , BC 切小圆 O 于点 E
∴ CE = CA
∵ 在 Rt△OAD 与 Rt△OEB 中, OA = OE,OD = OB,∠OAD = ∠OEB = 90
∴ Rt△OAD ≌ Rt△OEB (HL)
∴ EB = AD
∵ BC = CE + EB
∴ BC = AC + AD
(3)∵ ∠BAC = 90 , AB = 8,BC = 10
∴ AC = 6
∵ BC = AC + AD
∴ AD = BC AC = 4
∵ 圆环的面积 S = πODx2 πOAx2 = π (OD x2 +OAx2 )
又∵ OD +OA = AD , S = 4 π = 16πcm
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