Respuesta :
Solid A has five(5) rectangular blocks that are co-joined. The dimension of each, is 45miles by 90miles.
Thus,
[tex]\begin{gathered} Total\text{ surface area=5}\times Area\text{ of one rectangular block} \\ \text{Total Surface Area=5}\times(45\times90) \\ T\mathrm{}S\mathrm{}A=5\times4050 \\ T\mathrm{}S\mathrm{}A=20250mi^2 \end{gathered}[/tex]Solid B has five(5) rectangular blocks that are co-joined. The dimension of each, is 30mi by 60mi.
Thus,
[tex]\begin{gathered} \text{Total Surface Area= 5}\times Area\text{ of one rectangular block} \\ T\mathrm{}S\mathrm{}A=5\times(30\times60) \\ T\mathrm{}S\mathrm{}A=5\times1800 \\ T\mathrm{}S\mathrm{}A=9000mi^2 \end{gathered}[/tex]The ratio of the T.S.A of the similar solids is given below:
[tex]\begin{gathered} T\mathrm{}S\mathrm{}A_{solid\text{ A}}\colon T.S.A_{solid\text{ B}} \\ 20250\colon9000 \\ \text{Divide both by 2250, we have:} \\ 9\colon4 \end{gathered}[/tex]Hence, the correct option is Option A
