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Glancing at your other questions, you've already had some accurate answers on this.

You do understand what Hardy-Weinberg is supposed to do? It predicts the frequency of the genotypes if the alleles are evenly distributed.

If you have the relative frequencies of the two alleles, these are A and a (in this case) and the homozygous types will have frequency of A^2 and a^2 if equilibrium is manifest.

The simplest thing to do is to square root the actual AA and aa frequencies and see if the answers add up to 1. If they do, you've got equilibrium.

As has been said, aa square rooted gives you 0.1, so you'd expect the homozygous frequency to be (0.9)^2 =0.81. It's higher than you'd expect.