[Toán 9]Hình học

D

dragonkshb

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V

vy000

Ta có:AM;ANAM;AN là phân giác 2 góc ngoài của BAC^\widehat {BAC}

ADMN\Rightarrow AD \bot MN

AD//NC//MB\Rightarrow AD//NC//MB

Kẻ CKADCK=ANCK \bot AD \Rightarrow CK=AN

SACD=CK.AD2=AN2AD\Rightarrow S_{ACD}=\dfrac{CK.AD}2=\dfrac{AN}2AD

CMTT:SADB=AM2AD S_{ADB}=\dfrac{AM}2AD

SABC=SACD+SADB=AM+AN2AD=NM.AD2=22.72=77(cm2)\Rightarrow S_{ABC}=S_{ACD}+ S_{ADB}=\dfrac{AM+AN}2AD=\dfrac{NM.AD}2=\dfrac{22.7}2=77(cm^2)
 
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