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