课时训练·基础达标
1 |
D |
2 |
A |
3 |
C |
4 |
B |
5 |
D |
6 |
D |
7 |
C |
8、3.42
9、4.5
10、7 cm
11、5
12、解:设另两根的长度分别为x cm和y cm.
∵ 两三角形的对应边不确定,故分三种情况讨论:
① 当边长30 cm与边长60 cm是对应边时,有30/60 = 40/x = 50/y.
∴ x = 80 , y = 100,
故另两边的长度为80 cm和100 cm,
② 当边长40 cm与边长60 cm是对应边时,有30/x = 50/y = 40/60.
∴ x = 45.y = 75,
故另两边的长度为45 cm和75 cm.
③ 当边长50 cm与边长60 cm是对应边时,有30/x = 40/y = 50/60.
∴ x = 36 ,y = 48,
故另两边的长度为36 cm和48 cm.
综上所知,当另两边长为80 cm和100 cm或45 cm和75 cm或36 cm和48 cm时,
两个三角形相似.
13、解:∵ AB/CD = 6/7.5 = 4/5 ,BC/AC = 5/4.
∴ AB/CD = BC/AC.
又∵∠B = ∠ACD.
∴ △ABC ∽ △DCA.
∴ AC/AD = AB/CD,
即 5/AD = 4/5,
∴ AD = 25/4.
14、(1)证明:∵ BE平分∠DBC,
∴ ∠CBE = ∠DBC.
∵ ∠CBE = ∠CDF,
∴ ∠DBG = ∠CDF.
∵ ∠BGD = ∠DGE,
∴ △BDG ∽ △DEG.
(2)解:∵ △BDG ∽ △DEG
∴ DG/EG = BG/DG,
∴ DG² = BG·EG = 4,∴ DG = 2.
∵ ∠EBC + ∠BEC = 90°,∠BEC = ∠DEG,
∠EBC = ∠EDG,
∴ ∠BGD = 90°.∵ ∠DBG = ∠FBG,BG = BG,
∴ △BDG ≌ △BFG.∴ FG = DG = 2 ,∴ DF = 4.
∵ BE = DF,
∴ BE = DF = 4.