题目 #20470

如图①所示, 令旋转1秒后点D的对应点为点E, 分别过点D和点E作y轴和x轴的垂线, 垂足分别为M和N, 由旋转可知, OD = OE, ∠DOE=90°∴∠DON + ∠NOE = 90∘, 又∵∠MON = 90∘, ∴ ∠DMO = ∠ENO ∠MOD + ∠DON=90°,∴∠MOD = ∠NOE,在 △DOM 和 △EON 中， {∠MOD = ∠NOE,所以△DOM ≅△ DO = EO EON(AAS), ∴EN = DM, ON = MO, ∵点D坐标为D(1,2),所以NE = DM = 1, ON = OM = 2, ∴点E的坐标为 (2, −1). 如图②: 同理可得, 旋转2秒后点D的对应点坐标为(−1, −2), 旋转3秒后点D的对应点坐标为 (−2,1), 旋转4秒后点D的对应点坐标为(1,2), 旋转5秒后点D的对应点坐标为(2, −1), … 由此可见, 点D的对应点按(2, −1), (−1, −2), (−2,1), (1,2)循环出现, 又∵2025 =\(4 \times 506 + 1\)∴旋转 2025秒后点D的对应点D′的坐标为(2, −1), 故选D． 2.D AD BC BC AD 5 ∵AB ∥CD ∥EF, ∴ _{DF}= _{CE}, ∵BC = 5, CE = 8, ∴ _{CE}= _{DF}= _{8}, 故选D. 3.B ∵两个相似三角形的最长边分别为10cm和6cm, ∴相似比为10: 6 = 5: 3,∴较大三角形与较小三角形的周长比为5: 3, ∵它们的周长之和为48cm, ∴较小三角形的周长为 3 :\(48 \times _{5+3}= 18cm\)故选B. 4.D 由题意得C(2,0), AC = 4, ∴将△ABC先绕点C逆时针旋转90°, 得到A的对应点的坐标为(2,4), ∴ 向左平移5个单位长度, 变换后点A的对应点的坐标为(−3,4), 故选D. 5.C ∵四边形 ABCD 为平行四边形, ∴AB ∥CD, AB = CD,∴GF AG EG\(BG _{FC}\)= _{CD}, _{EC}= _{CD}, 设GF为x, ∵EF = 1, EC = 3, ∴EG = 1 + x,BG= AG + CD, ∴ x AG 1+x AG 1+x\(x _{4}\)= _{CD}, _{3}=\(\frac{AG+CD}{CD} = 1 + _{CD}\)∴ _{3}= 1 + _{4},即8 −x = 0, 得x = 8, ∴GF = 8. 故选C． 6.\(A _{3}\)）^{2}=2 延长AG 到 BC 于点 N, ∵ 点 G 是 △ABC 的重心, GD ∥BC, ∴ AG 2 AG\(2 _{GN}\)= _{1},∴_{AN}= _{3}, ∴S△ADG: S△ANC= （\(4 _{9}\), ∵点G是△ABC的重心,则AN是三角形中线, ∴S△ANC= S△ABN, ∴S△ADG: S△ABC= 4: 18=2: 9. 故选A． 19 7. 3 过点A作AG ∥CD, 交EF于点H, 交BC于点G, ∵AD ∥EF ∥BC,所以四边形AHFD, AGCD均为平行四边形, ∴ CG = HF = AD = 3,因为EF = 5, ∴EH = EF −HF = 5 −3 = 2, ∵EF ∥BC, ∴△AEH ∼△ABG,所以BG\(AB _{EH}\)= _{AE}, ∵ AE: EB = 3: 2, ∴ _{EH}= _{AE}= _{3}, ∴BG = _{3}EH = _{3},所以BC = BG + CG = _{3}+ 3 = _{3}, 故答案为 _{3}.^{19} BG AB 5 5 10 10 19 8.见解析. ∵∠BCE = ∠ACD, ∴∠BCE + ∠ACE = ∠ACD + ∠ACE, ∴∠DCE =∠ACB, 又 CD ⋅BC = CE ⋅AC,则 CD\(CE _{CA}\)= _{CB}, ∴△ ABC ∼△DEC. 9.(1)见解析; (2) _{2}.^{9} (1)在Rt △ABC中, ∠ACB = 90∘, ∵CD是△ABC的中线,∴AD = BD = CD, ∴∠DCB = ∠DBC, ∵AE ⊥ CD,∴∠AEC = ∠ACB = 90∘, ∴∠CAE = 90∘−∠ACE = ∠DCB,∴∠CAE = ∠DBC, ∴△ACE ∼△BAC; (2)∵AC = 3, CE = 1,由(1)知 △ACE ∼△BAC, ∴ CE AC 1\(3 _{AC}\)= _{AB}, ∴ _{3}=_{AB}, ∴AB = 9, ∵CD是△ABC的中线, ∴CD =\(1 _{2}\)AB = _{2}.^{9} 10.2: 3 因为DE ∥BC, ∴△ADE ∼△ABC, ∴ DE AD AD\(2 _{BC}\)= _{AB}, ∵AD = 2BD, ∴ _{AB}= _{3},所以DE: BC = 2: 3, 故答案为2: 3. 11.(1)见详解; (2)CD = 4. (1)∵CA = CB, ∴∠A = ∠B, ∵∠CDB = 2∠B, ∴∠CDB = 2∠A, 又∵∠CDB= ∠A + ∠ACD, ∴∠ACD = ∠B, ∵∠A = ∠A, ∴△ABC ∼△ACD. (2)∵AC = 6, ∴BC = AC = 6, ∵△ABC ∼△ACD, ∴ AB BC 9\(6 _{AC}\)= _{CD}, ∴ _{6}= _{CD},解得CD = 4. 12.1: 2 ∵四边形 ABCD 是平行四边形, ∴BC = AD, AD ∥BC, ∵ 点 E 是 AD 的中点,∴DE =\(1 _{2}\)AD, ∴DE: BC = 1: 2, ∵ AD ∥BC, ∴△DEF ∼△BCF,∴C△DEF: C△BCF= 1: 2, 故答案为1: 2． 13.8 DE AD 1 4 1 ∵DE ∥BC, ∴△ADE ∼△ABC, ∴ _{BC}= _{AB}= _{2}, 即 _{BC}= _{2}, ∴BC = 8cm, 故答案为8. 14.见解析. ∵△ACE ≅△ABF, ∴∠CAE = ∠QAF, AE = AF, ∵ AE AC AE\(AC _{AQ}\)= _{AE},∴_{AQ}= _{AF}, ∴△ACE ∼△AFQ. 5 15. 2 ∵四边形 ABCD 是平行四边形, ∴AB = CD, AB ∥CD, ∴∠AEF=∠CDF, ∠EAF = ∠DCF, ∴△EAF ∼△DCF, ∴\(DF _{EF}\)= CD AB AE 2 AB 5 S△ADF DF AB\(5 _{AE}\)=_{AE}, ∵ _{EB}= _{3}, ∴ _{AE}= _{2}, ∴ _{S}△AEF= _{EF}= _{AE}= _{2},故答案为： _{2}.^{5}