Toán hình 9

H

hien_vuthithanh

Coi $\widehat{AOB}=\widehat{COD}=120^o$

Có : $S_{AOB}=\dfrac{1}{2}AO.BO.sin\widehat{AOB}=\dfrac{\sqrt{3}}{4}AO.BO$

$\rightarrow S_{AOB}=S_{COD}=\dfrac{\sqrt{3}}{4}AO.BO$ (*)

$S_{AOD}=\dfrac{1}{2}AO.DO.sin\widehat{AOD}=\dfrac{\sqrt{3}}{4}AO.DO$

$\rightarrow S_{AOD}=S_{BOC}=\dfrac{\sqrt{3}}{4}AO.DO=\dfrac{ \sqrt{3}}{4}AO.BO$(*)(*)

Cộng theo vế $\rightarrow S_{ABCD}=4.\dfrac{ \sqrt{3}}{4}AO.BO=\dfrac{\sqrt{3}}{4}AC.BD$
 
You must log in or register to reply here.
Top Bottom