Giải
Ta có \(\dfrac{S_{BMN}}{S_{ABN}}=\dfrac{BM}{BA}\) (chung đường cao từ N)
mà \(\dfrac{AM}{AB}=\dfrac{1}{3}\)
Do đó: \(\dfrac{AB-AM}{AB}=\dfrac{3-1}{3}\) hay \(\dfrac{BM}{AB}=\dfrac{2}{3}\)
Nên \(\dfrac{S_{BMN}}{S_{ABN}}=\dfrac{2}{3}\)
Tương tự: \(\dfrac{S_{ABN}}{S_{ABC}}=\dfrac{BN}{BC}=\dfrac{1}{3}\) (chung đường cao từ A)
\(\Rightarrow\) \(\dfrac{S_{BMN}}{S_{ABN}}.\dfrac{S_{ABN}}{S_{ABC}}=\dfrac{2}{3}.\dfrac{1}{3}\)
\(\Rightarrow\) \(\dfrac{S_{BMN}}{S_{ABC}}=\dfrac{2}{9}\)
Tương tự: \(\dfrac{S_{DNC}}{S_{ABC}}=\dfrac{2}{9}\); \(\dfrac{S_{ADM}}{S_{ABC}}=\dfrac{2}{9}\)
Vậy SMND = SABC - SADM - SBMN - SDNC
= SABC - 3 . \(\dfrac{2}{9}\)SABC = \(\dfrac{1}{3}\)SABC = \(\dfrac{1}{3}\) . 30
= 10 (cm2)
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