4. In $\triangle A B C$, $D$ is a point on side $B C$, it is known that $A C$ $=5, A D=6, B D=10, C D=5$. Then, the area of $\triangle A B C$ is ( ).
(A) 30
(B) 36
(C) 72
(D) 125
4. (13).

As shown in the figure, draw $CH \perp AD$ at $H$. Since $\triangle ACD$ is an isosceles right triangle, in the right triangle $\triangle ACH$, $AC=5, AH=3$. Therefore, $CH=4$. Thus,
$$
\begin{array}{l}
S_{\text {AUT }}=\frac{1}{2} A \cdot CH=\frac{1}{2} \times 6 \times 4=12, \\
S_{\triangle ABC}=\frac{BC}{DC} S_{\triangle AC1}=\frac{15}{5} \times 12=36 .
\end{array}
$$