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請以英文作答為佳.並寫出步驟.In a figure,D and E are the mid-points of AB and AC respectively.F and G are points on BC such that BF=FG=GC.DF and EG are produced to meet at H. (a)Prove that 2DE=3FG (c)Find the ratios of HG:GE and HF:FD (d)Prove that TriangleCEG~TriangleHBG (e)Hence prove that ABHC... 顯示更多 請以英文作答為佳.並寫出步驟. In a figure,D and E are the mid-points of AB and AC respectively.F and G are points on BC such that BF=FG=GC.DF and EG are produced to meet at H. (a)Prove that 2DE=3FG (c)Find the ratios of HG:GE and HF:FD (d)Prove that TriangleCEG~TriangleHBG (e)Hence prove that ABHC is a parallelogram The length of the diagonal BD of the base of the square pyramib is 12√2 cm and its height is 8 cm.Find the total surface area of the pyramid.

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Because AD=DB, AE=CE, by mid-pt theorem, (1) (a) BC=2DE, but FG= 1/3BC=2/3DE so 2DE=3FG (b) By mid-pt theorem, DE//BC Angle FHG= Angle DHE (common) Angle HFG= Angle HDE (alt Angle DE//BC) Angle HGF= Angle HED (alt Angle DE//BC) So Triangle HFG is similar to triangle HDE (AAA) HG/HE = FG/DE=2/3 , so HG:GE=2:1 (corr side of similar triangle) Similarly HF:FD=2:1 (intercept theorem) (c) Angle CGE= Angle HGB (vert opp angle) EG:HG= 1:2 by (b) Because CG:CB=1:3, so CG:BG=1:2 so TriangleCEG~TriangleBHG (ratio of 2 sides, inc angle) (d) Angle ECG = Angle GBH (corr angle of similar triangle) so EC//BH ( alt angle equal) EG/HG=CE/BH=1/2 (corr side of similar triangle) so BH=2CE, but AC=2CE , so BH=AC so ABHC is a //gram (2 sides equal and //) (2) Let the side of the square base be x cm (√2x^2) =12√2 , x=12 Height of the 4 triangles= (√6^2+8^2)=10cm Total surface area of pyramid= 1/2(12)(10)(4)+12^2= 384cm^2

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