hình học lớp 6

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$AE = 2EB <=> EB = \dfrac{AB}{3} <=> S_{EBG} = \dfrac{S_{ABG}}{3} <=> S_{ABG} = 30$
$AD = \dfrac{DC}{2} <=> AD = \dfrac{AC}{3} <=> S_{ADG} = \dfrac{S_{AGC}}{3} và S_{ADB} = \dfrac{S_{ABC}}{3}$
Suy ra $S_{ABD} - S_{AGD} = \dfrac{S_{ABC} - S_{AGC}}{3} = \dfrac{S_{AGB} + S_{BGC}}{3} = \dfrac{30 + S_{BGC}}{3} = 10 + \dfrac{S_{BGC}}{3}$
Mặt khác, ta có $S_{ABD} - S_{AGD} = S_{ABG} = 30$
Suy ra $30 = 10 + \dfrac{S_{BGC}}{3} => \dfrac{S_{BGC}}{3} = 20 ==> S_{BGC} = 60$
 
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