6. In $\triangle A B C$, if $A B=B C=29, A C=$ 42, then the area of $\triangle A B C$ is ( ).
(A) 100
(B) 420
(C) 500
(D) 609
(E) 701
6. B.

Draw $B D \perp A C$ at point $D$. Then
$$
\begin{array}{l}
S_{\triangle A B C}=2 S_{\triangle A B D}=2 \times \frac{1}{2} A D \cdot B D \\
=\sqrt{29^{2}-\left(\frac{42}{2}\right)^{2}} \times \frac{42}{2}=420 .
\end{array}
$$