Phương pháp giải:

Sử dụng tỉ số diện tích giữa hai tam giác.

Lời giải chi tiết:

Ta có: \(\left\{ \begin{array}{l}{S_{\Delta DFG}} = \frac{1}{2}d\left( {D;\,\,FG} \right).FG\\{S_{\Delta DEB}} = \frac{1}{2}d\left( {D;\,\,FG} \right).BE\end{array} \right. \Rightarrow \frac{{{S_{\Delta DFG}}}}{{{S_{\Delta DEB}}}} = \frac{{FG}}{{BE}} = \frac{1}{6}\) \( \Rightarrow {S_{\Delta DFG}} = \frac{1}{6}{S_{\Delta DEB}}.\)

\(\left\{ \begin{array}{l}{S_{\Delta DEB}} = \frac{1}{2}d\left( {D;\,\,BE} \right).BE\\{S_{\Delta BEC}} = \frac{1}{2}d\left( {C;\,\,BE} \right).BE\end{array} \right.\)\( \Rightarrow \frac{{{S_{\Delta DEB}}}}{{{S_{\Delta BEC}}}} = \frac{{d\left( {D;\,\,BE} \right)}}{{d\left( {C;\,\,BE} \right)}} = \frac{{BD}}{{BC}} = \frac{4}{5}\)\( \Rightarrow {S_{\Delta DEB}} = \frac{4}{5}{S_{\Delta BEC}}.\)

\(\left\{ \begin{array}{l}{S_{\Delta BEC}} = \frac{1}{2}d\left( {B;\,\,EC} \right).EC\\{S_{\Delta ABC}} = \frac{1}{2}d\left( {B;\,\,AC} \right).AC\end{array} \right.\) \( \Rightarrow \frac{{{S_{\Delta BEC}}}}{{{S_{\Delta ABC}}}} = \frac{{EC}}{{AC}} = \frac{3}{4}\)\( \Rightarrow {S_{\Delta BEC}} = \frac{3}{4}{S_{\Delta ABC}}.\)

 \( \Rightarrow {S_{\Delta DFG}} = \frac{1}{6}.\frac{4}{5}.\frac{3}{4}.{S_{\Delta ABC}} = \frac{1}{{10}}.900 = 90\,c{m^2}.\)

Chọn B.