题目 #20298

如图①，当∠CDM = 90°时，△DCM是直角三角形，在Rt △ABC中，∠B = 90°，∠A = 60°，AC = 2√3 + 4，∴∠C = 30°，AB = 1\(1 _{2}\)AC = √3 + 2，由折叠可得，∠MDN = ∠A = 60°，∴∠BDN = 30°，∴BN = _{3}AB = √3+\(2 _{3}\)，∴AN = 2BN = 3 2√3+4，∵∠DNB = 60°，∴∠ANM = ∠DNM = 60°，∴∠AMN = 60°，∴AN = MN = 2√3+4.如图②，当∠CDM = 90°时，△DCM是直角三角形，由题可得，∠CDM = 60°，∠A = ∠MDN = 60°， 3 ∴∠BDN = 60°，∠BND = 30°，∴BD = 1\(1 _{2}\)DN = _{2}AN，BN = √3BD，又∵AB = √3 + 2，∴BD = 1，AN = 1 2，BN = √3，过点N作NH ⊥AM于点H，则∠ANH = 30°，∴AH = _{2}AN = 1，HN = √3，∴∠AMN = 60°， ∴AN = MN = 2√3+4，由折叠可得，∠AMN = ∠DMN = 45°，∴△MNH是等腰直角三角形，∴HM = HN = 3 √3，∴MN = √6.综上，MN的长为2√3+4或√6. 3 15.7或1 当点A, 点C′, 点D′共线时, 根据题意知: DG = DG′= AB = 4, C′D′=CD= AB = 4, ∠AD′G = ∠CD′G = ∠D = 90∘, ∴AG = AD −DG = 5.在Rt △AD′G中, AD′=\(\sqrt{AG^{2}}\)+ D′\(G^{2}\)= 3, ∴AC′= AD′+ D′C′=3+4 = 7; 当点A, 点C′, 点D′共线时, 根据题意可知DG = D′G =AB= 4, C′D′= CD = AB = 4, ∠AD′G = ∠D = 90∘, 所以AG = AD −DG= 5. 在Rt △AD′G中, AD′=\(\sqrt{AG^{2}}\)−D′\(G^{2}\)= 3, ∴AC′= C′D′−AD’= 4 −3 = 1. 故答案为: 7或1.