Bài giải:

Gọi K là trung điểm cạnh CD ta có:

${{S}_{A\text{D}I}}+{{S}_{BCK}}=\frac{1}{2}{{S}_{A\text{D}B}}+\frac{1}{2}{{S}_{BC\text{D}}}=\frac{1}{2}{{S}_{ABC\text{D}}}$

${{S}_{BCI}}+{{S}_{CAK}}=\frac{1}{2}{{S}_{BCA}}+\frac{1}{2}{{S}_{DCA}}=\frac{1}{2}{{S}_{ABC\text{D}}}$

Từ đó:

${{S}_{A\text{D}I}}+{{S}_{BCI}}+\left( {{S}_{BKC}}+{{S}_{DAK}} \right)={{S}_{ABC\text{D}}}$

Nhưng ${{S}_{A\text{D}I}}+{{S}_{BCI}}+{{S}_{IC\text{D}}}={{S}_{ABC\text{D}}}$

Vậy

${{S}_{IC\text{D}}}={{S}_{BKC}}+{{S}_{DK\text{A}}}=\frac{1}{2}{{S}_{B\text{D}C}}+\frac{1}{2}{{S}_{A\text{D}C}}$ = 18 : 2 + 12 : 2 = 15($c{{m}^{2}}$)

Đáp số: 15cm2