Your reasoning is correct. It is indeed possible and correct to calculate the are of some shape by using the contracted lengths.
For example, for a square with $A=x\cdot y$ where in the square's rest frame $x=y$, we would calculate the new area (assuming the square moves in $x$ direction with velocity $v$) by
$$A'=y \cdot x' = x^2\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=A\gamma$$$$A'=y \cdot x' = x^2{\sqrt{1-\frac{v^2}{c^2}}}=\frac{A}{\gamma}$$
(To avoid cunfusion: primed' indicates a measurement made on an object with relative velocity $v$ while non-primed is a measurement made in the object's rest frame)
In fact, it seems to me that the relationship $A'=\gamma A$$A'=\frac{A}{\gamma}$ also applies to volumes, for example for a cube with volume $V=x\cdot y\cdot z$ where in the cube's rest frame $x=y=z$, we have
$$V'=y\cdot z\cdot x'=yzx\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=x^3\gamma=V\gamma$$$$V'=y\cdot z\cdot x'=yzx\sqrt{1-\frac{v^2}{c^2}}=\frac{x^3}{\gamma}=\frac{V}{\gamma}$$
The same would also apply to a sphere with $V=\frac 4 3\pi r^3$:
$$V'=\frac 4 3\pi r^2r'=\frac 4 3\pi r^2r\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}=V\gamma$$$$V'=\frac 4 3\pi r^2r'=\frac 4 3\pi r^2r\sqrt{1-\frac{v^2}{c^2}}=\frac{V}{\gamma}$$
It can thus be said, assuming constant velocity in one direction that
$$A'=\gamma A\tag{1}$$$$A'=\frac{A}{\gamma}\tag{1}$$
$$V'=\gamma V\tag{2}$$$$V'=\frac{V}{\gamma}\tag{2}$$
where $\gamma$ is the Lorentz Factor $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
I am aware that the volume part isn't exactly related to the question, but I nevertheless wanted to expand a little in my answer, since the first part is basically just "yes"