Can you derive density from a sample that includes different quadrant sizes? I wish to estimate the percentage covering of vegetation in an area 1 km long and 600 m wide. Within this area I have percentage cover of vegetation for 10 circles with 10 m diameter and percentage cover for 12 rectangles that are each 2 m by 50 m. Can I combine these data to provide an estimate of percentage cover for the entire area with error estimates or confidence intervals?
 A: @whuber Below is the initial data treatment I envisaged put somewhat more explicitly. Thank you for your, as usual deep, insights. The one outstanding comment I did not understand was "Such a normalization would lose information about relative precision of the estimates." Does the example treatment below address that relative precision comment or did you have something else in mind, please? 
At present there is not enough information to answer the question. Let us assume that there is no overlap between any of the regions (unstated at present). Let us further assume that the distribution of vegetation coverage for each of the two area methods is (separately) normally distributed enough to allow the mean value of each different area geometry to be a good estimator of location of the vegetation coverage for that geometry (unstated at present).
Then we take the mean value of each geometry's vegetation coverage and construct "residual" vectors thus 
$X_{round,residual}=X_{round}-\bar{X}_{round}$
$X_{rectangular,residual}=X_{rectangular}-\bar{X}_{rectangular}$
We then area normalize everything yielding
$Y_{round+rectangular}=\frac{X_{round}}{Area_{round}}+\frac{X_{rectangular}}{Area_{rectangular}}$
$Y_{round+rectangular,residual}=\frac{X_{round,residual}}{Area_{round}}+\frac{X_{rectangular,residual}}{Area_{rectangular}}$
We then examine (e.g., histogram) the vectors $Y_{round+rectangular}$ and $Y_{round+rectangular,residual}$ and determine what (if any) measure of location can be used for both vectors. We then measure those locations.
Now since there are assumptions unstated at present 1) This is only one possible answer and 2) There are other approaches, some completely different, to this same problem.
