4. Given the three sides of $\triangle A B C$ are $A B=$ $2 \sqrt{a^{2}+576}, B C=\sqrt{a^{2}+14 a+625}, A C=$ $\sqrt{a^{2}-14 a+625}$, where $a>7$. Then the area of $\triangle A B C$ is $\qquad$
4.168.

Notice
$$
\begin{array}{l}
A B^{2}=(2 a)^{2}+48^{2}, \\
B C^{2}=(a+7)^{2}+24^{2}, \\
A C^{2}=(a-7)^{2}+24^{2} .
\end{array}
$$

As shown in Figure 6, with $A B$ as the hypotenuse, construct a right triangle $\triangle A B D$ on one side of $\triangle A B C$, such that
$$
\begin{array}{l}
B D=2 a, A D=48, \\
\angle A D B=90^{\circ} .
\end{array}
$$

Take a point $E$ on $B D$ such that $B E=a+7, E D=a-7$, and take the midpoint $F$ of $A D$, construct rectangle $E D F C_{1}$.
Since $B C_{1}^{2}=B E^{2}+E C_{1}^{2}=(a+7)^{2}+24^{2}=B C^{2}$,
$$
A C_{1}^{2}=C_{1} F^{2}+A F^{2}=(a-7)^{2}+24^{2}=A C^{2} \text {, }
$$

thus point $C$ coincides with point $C_{1}$.
And $S_{\triangle A B D}=48 a, S_{\triangle C B D}=24 a, S_{\triangle A C D}=24(a-7)$,
then $S_{\triangle B B C}=S_{\triangle A B D}-S_{\triangle C B D}-S_{\triangle A C D}=168$.