A rod having a rest length L travels at a relativistic speed. Let...

The length of a moving rod depends on its orientation with respect to the direction of motion.

Let the rod be oriented at an angle $\theta$ with the direction of motion in its rest frame.

The component of the length parallel to the velocity is $L_{\parallel} = L \cos\theta$.

The component of the length perpendicular to the velocity is $L_{\perp} = L \sin\theta$.

Due to length contraction, the parallel component in the moving frame becomes $L'_{\parallel} = \dfrac{L \cos\theta}{\gamma}$.

The perpendicular component remains unchanged, so $L'_{\perp} = L \sin\theta$.

The total length of the rod in the moving frame is $L_{moving} = \sqrt{(L'_{\parallel})^2 + (L'_{\perp})^2} = L \sqrt{\dfrac{\cos^2\theta}{\gamma^2} + \sin^2\theta}$.

Given $L' = \dfrac{L}{\gamma}$.

If \theta = 0^{}^{{}^{\circ}}, $L_{moving} = \dfrac{L}{\gamma} = L'$.

If \theta = 90^{}^{{}^{\circ}}, $L_{moving} = L$. Thus, the length may be equal to $L$.

If 0^{}^{{}^{\circ}} < \theta < 90^{}^{{}^{\circ}}, $L' < L_{moving} < L$. Thus, the length may be more than $L'$ but less than $L$.