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20 14 Changning mathematics single-mode answer
A model test paper in Changning District of Grade Three! I'm doing it, too

In fact, it is the easiest to put the numbers together. The benefits of filling in the blanks

Solution:

Δ ABC is equilateral, g is the midpoint of Δ ABC, connecting AG/BG/CG, and AG=BG=CG.

Let AB=a, AG = BG = CG = √ 3/3 AD = AE = 2/3 BD = CE =1/3.

To make △BDM∽△CEM.

Known ∠DBC=∠BCE=60? There is DB/BM=CM/CE.

BM+CM=BC=a, let BM=x, CM =1-X.

Enumerable formula:1/3: x = (1-x):1/3.

X= 1/2+√5/6 or x= 1/2-√5/6.

S△BDM= 1/2×BD×BM×sin60?

S△CEM= 1/2×CE×CM×sin60?

S△BDM=√3/24+√ 15/72 or √3/24-√ 15/72.

S△CEM=√3/24-√ 15/72 or √3/24+√ 15/72.