Bài giải:

${{S}_{AM\text{D}}}={{S}_{ABN}}={{S}_{BPC}}={{S}_{Q\text{D}C}}=\frac{1}{6}{{S}_{ABC\text{D}}}$

${{S}_{MQ\text{D}}}=\frac{1}{18}{{S}_{ABC\text{D}}};{{S}_{M\text{D}C}}=\frac{1}{2}{{S}_{ABC\text{D}}};\frac{{{S}_{MQ\text{D}}}}{{{S}_{M\text{D}C}}}=\frac{1}{18}{{S}_{ABC\text{D}}}:\frac{1}{2}{{S}_{ABC\text{D}}}=\frac{1}{9}$

$\Rightarrow {{S}_{Q\text{D}I}}=\frac{1}{9}{{S}_{I\text{D}C}}=\frac{1}{10}{{S}_{Q\text{D}C}}=\frac{1}{60}{{S}_{ABC\text{D}}}$

$\Rightarrow {{S}_{DIHP}}={{S}_{Q\text{D}C}}-\left( {{S}_{Q\text{D}I}}+{{S}_{HPC}} \right)=\frac{1}{6}{{S}_{ABC\text{D}}}-\left( \frac{1}{60}{{S}_{ABC\text{D}}}+\frac{1}{60}{{S}_{ABC\text{D}}} \right)=\frac{2}{15}{{S}_{ABC\text{D}}}$

Chứng minh tương tự ta được ${{S}_{MKGB}}=\frac{2}{15}{{S}_{ABC\text{D}}}$

$\Rightarrow {{S}_{GHIK}}={{S}_{ABC\text{D}}}-\left( {{S}_{AM\text{D}}}+{{S}_{BPC}}+{{S}_{DIHP}}+{{S}_{MKGB}} \right)={{S}_{ABC\text{D}}}-\left( \frac{1}{6}{{S}_{ABC\text{D}}}+\frac{1}{6}{{S}_{ABC\text{D}}}+\frac{2}{15}{{S}_{ABC\text{D}}}+\frac{2}{15}{{S}_{ABC\text{D}}} \right)=\frac{2}{5}{{S}_{ABC\text{D}}}$ Vậy ${{S}_{GHIK}}=900\times \frac{2}{5}=360c{{m}^{2}}$