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平面几何中有这样一道题:在△ABC中,AD是BC边上的中线。BC=8cm,AB=4~(1/3)cm∠BAD=30°,求△ADC的面积。一部分同学是这样计算的: 作出图(右)。作BH垂直AD的延长线于点H。易知AH=ABcos30°=4~(1/3)·3~(1/2)/2=6。又BH=(1/2)AB=2~(1/3)。从而求得BH=2~(1/3)。DH=(4~2-(2~(1/3)~2)~(1/2)~2)=2,AD=6-2=4。故S_(△ABD)=1/2AD·BH=(1/2)·4·2~(1/3)=4~(1/3) 又S_(△ADC)=S_(△ABD), 故△ADC的面积为4~(1/3)cm~2。上述解法为基础方法,思路清晰,似乎无懈可
There is such a question in plane geometry: In △ABC, AD is the midline on the BC side. BC=8cm, AB=4~(1/3)cm∠BAD=30°, find the area of ΔADC. Some classmates calculated this: Making a map (right). Make an extension of the BH vertical AD at point H. It is easy to know AH=ABcos30°=4~(1/3)·3~(1/2)/2=6. BH=(1/2)AB=2~(1/3). In order to obtain BH = 2 ~ (1/3). DH=(4~2-(2~(1/3)~2)~(1/2)~2)=2, AD=6-2=4. Therefore, S_(ΔABD)=1/2AD·BH=(1/2)·4·2~(1/3)=4~(1/3) and S_(ΔADC)=S_(△ABD), so The ADC area is 4~(1/3)cm~2. The above solution is the basic method and the thinking is clear.