  We give $q$-analogues of the following two Ramanujan-type formulas for
$1/\pi$: \begin{align*} \sum_{k=0}^\infty (6k+1)\frac{(\frac{1}{2})_k^3}{k!^3
4^k} =\frac{4}{\pi} \quad\text{and}\quad \sum_{k=0}^\infty
(-1)^k(6k+1)\frac{(\frac{1}{2})_k^3}{k!^3 8^k } =\frac{2\sqrt{2}}{\pi}.
\end{align*} Our proof is based on two $q$-WZ pairs found by the first author
in his earlier work.
