\(\text { Let } \mathrm{ABCD} \text { be the parallelogram having adjacent sides } \mathrm{AB}=10 \text { and } \mathrm{BC}=8\)




In triangle $\(\mathrm{ABD}, \mathrm{AO}=\mathrm{OD}\)$, so BO is a median, using Apollonius theorem in triangle ABD , we get $\(\mathrm{AB}^2+\mathrm{BD}^2=2\left(\mathrm{BO}^2+\mathrm{AO}^2\right)\)$ i.e. $\(2\left(\mathrm{BO}^2+\mathrm{AO}^2\right)=10^2+8^2=164 \Rightarrow \mathrm{BO}^2+\mathrm{AO}^2=82\)$ which is sum of the squares of half of the diagonals of the parallelogram.

The expression $\(\mathrm{BO}^2+\mathrm{AO}^2=82\)$ can be true for many different integer/non-integer values of BO and AO out of which one pair is $\((\mathrm{BO}, \mathrm{AO}) /(\mathrm{AO}, \mathrm{BO})=(\sqrt{75}, \sqrt{7})\)$

So, the diagonals of the parallelogram are $\(2 \times \sqrt{75}=10 \sqrt{3} \& 2 \times \sqrt{7}=2 \sqrt{7}\)$ out of which $\(10 \sqrt{3}\)$ is greater than 8 but $\(2 \sqrt{7}\)$ is less than 8 .

Hence a unique comparison cannot be formed between column $\(A \&$\) column $\(B\)$ quantities, so the answer is (D).