1. $144 \mathrm{~cm}^{2}$

Solution As shown in the figure, in $\triangle A B C$, let $A B=A C, A D \perp B C, A M=M C$, then $A D=18 \mathrm{~cm}, B M=15 \mathrm{~cm}$.

Also, let $A D$ intersect $B M$ at $G$, then $G$ is the centroid of $\triangle A B C$.
$\therefore \quad G D=\frac{1}{3} A D=6 \mathrm{~cm} . \quad B G=\frac{2}{3} B M=10 \mathrm{~cm}$.
$\therefore B D=8 \mathrm{~cm}$. Therefore,
$S_{\triangle A B C}=\frac{1}{2} B C \cdot A D=144 \mathrm{~cm}^{2}$
1. If the height on the base of an isosceles triangle equals $18 \mathrm{~cm}$, and the median to the leg equals $15 \mathrm{~cm}$, then the area of this isosceles triangle is $\qquad$ .