Kẻ trung tuyến AM
\(\Rightarrow S_{ABM}=S_{ACM}=\dfrac{1}{2}S_{ABC}=15\left(cm^2\right)\)
Lại có \(\dfrac{MG}{AG}=\dfrac{1}{3}\Rightarrow\dfrac{S_{BGM}}{S_{ABM}}=\dfrac{S_{CGM}}{S_{ACM}}=\dfrac{MG}{AG}=\dfrac{1}{3}\)
\(\Rightarrow S_{BGM}=S_{CGM}=\dfrac{1}{3}\cdot S_{ABM}=5\left(cm^2\right)\\ \Rightarrow S_{BGC}=S_{BGM}+S_{CGM}=10\left(cm^2\right)\)