岩侧 幼苗
共回答了20个问题采纳率:80% 举报
(1)见图:
因为折叠的关系,△DFC ≌ △EFC;
所以CD=CE;用圆规,以C为圆心,CD为半径,交AB于点E,可得CD=CE;
所以∠FEC = ∠FDC = 90°;作EF垂直于CE交AD于F.
(2) FC2 = DC2 + DF2
∵ sin∠CEB = CB/CE = 1/2
∴ ∠CEB = 30° ,又∠FEC = 90°
∴ ∠FEA = 180° - ∠FEC- ∠CEB = 60°
设DF为 x ,则FA为 1 - x ,又DF = FE,得
sin∠FEA = FA/DF = ( 1 - x ) / x = √3/2
解得 x = 4 - 2√3
故FC2 = 22 + (4 - 2√3)2 = 32 - 8√3
1年前