Here electronegativity difference between 'A' and 'B' has negligible value. So there is no consideration of pulling electron pair and reducing bond length. Internuclear distance between 'A' atoms is 10A'. So atomic size is half of it. So, '5A'. Similarly in'B', atomic size is '5A' Similarly in 'B' , atomic size is '3A'. If they both combine, then the resultant internuclear distance in AB is 5A+3A =8A.