题目 #20650

因为CF = 4cm, FB′= 1cm, ∴CB′= CF + FB′= 5cm. 由翻折, 菱形的性质得:质得: CB = CD = CB′= 5cm, CB ∥AD, ∠B = ∠D, ∵CB′⊥AD, ∴CB′⊥BC,∴∠BCB′= 90∘, ∴∠BCE = ∠B′CE = 45∘, ∵CD = 5cm, CF = 4cm,∠CFD= 90∘, ∴FD =\(\sqrt{CD^{2}}\)−\(CF^{2}\)= 3cm, 如图所示, 过点E作EG ⊥BC,设CG = xcm, 则EG = xcm, BG = (5 −x)cm,因为∠B = ∠D, ∠BGE =∠DFC,所以\(\Delta{}EGB\)∼\(\Delta{}CFD\)所以 EG EB GB x EB 5−\(x _{CF}\)= _{DC}= _{DF}, 所以 _{4}= _{5}= _{3},解得:x = 20\(25 _{7}\), ∴BE = _{7}, 故答案为: _{7}.^{25} 10.2: 5 ∵AD ∥BE, ∴∠DAF = ∠BEF, ∠ADF = ∠EBF, ∴\(\Delta{}BEF\)∼\(\Delta{}DAF,∵BE:\)EC = 2: 3, ∴BE: BC = 2: (2 + 3) = 2: 5. 又∵AD = BC,∴BE: AD = 2: 5, ∴\(\Delta{}BEF\)和\(\Delta{}DAF\)的周长比为2: 5. 11.主；左 主视图: 两层, 底层两个小正方形, 上层左边一个小正方形; 左视图:两层, 底层两个小正方形, 上层左边一个小正方形; 俯视图: 两行, 第一行两个小正方形, 第二行左边一个小正方形； 12.2√2 如图,过点C作PC ⊥AC交AF的延长线于点P, ∵∠BAC = 90∘, AE ⊥BD,所以∠CAP + ∠BAP = 90∘, ∠DBA + ∠BAP = 90∘. ∴∠CAP = ∠DBA.因为PC ⊥AC, ∴∠PCA = ∠BAD = 90∘. 又因为AB = AC, 所以\(\Delta{}PCA\)≅△DAB(ASA). ∴PC = AD.因为 D 是 AC 边的中点, ∴AD =\(1 _{2}\times\)AC= 3. ∴PC = 3. 因为PC ⊥ AC, ∠BAC = 90∘, ∴PC ∥AB. ∴\(\Delta{}PCF\)∼△ABF∴ CF PC 3 1 1\(1 _{FB}\)= _{AB}= _{6}= _{2}. ∴CF = _{2}FB = _{3}BC. ∵BC =\(\sqrt{AC^{2}}\)+\(AB^{2}\)等于\(\sqrt{6^{2}}\)+\(6^{2}\)= 6√2, ∴CF = 2√2. 故答案是2√2. 13.2: 5 ∵在 △ABC 中,点 D, E 分别在边 AB, AC 上, DE ∥BC, ∴∠ADE = ∠B,∵∠A = ∠A, ∴△ADE ∼△ABC, ∴\(AD _{AB}\)= DE AD 2 DE AD AD\(2 _{BC}\), ∵ _{DB}= _{3},∴_{BC}= _{AB}= _{AD+DB}= _{5}, ∴△ADE与△ABC的周长比为2: 5. 14.孝 与"文"相对的是"孝", 与"明"相对的是"强", 与"百"相对的是"义". 6√5