Identify compound `(A)` How many isomers including geometrical isomers are possible for `(A)` ?

Explanation

`D.U` in `(A) = ((2nC + ) 0 n_(H))/(2) = ((2xx10+2)-12)/(2) = 5^(@)`
Since on oxidation phalic acid is obtained. `(A)` contains benzene ring `(4.D.U.)` One `D.U` must be due to cyclic ring fused with benzene containing four `C` atom. There is no psoobiley of side chain with me `(c=C)` bond to count `1. D.U` becuase compound `(A)` does not give not test for unsaaturation (i.e. does not decolourise `Br_(2)` solution) (Note: Benzene gives negative `Br_(2)` test.)
The possible structures for `A`:

(All the compound `(I)` to `(VI)` on oxidation give phtalic acid.)