If two hexagons are similar, then their areas are related by:
[tex]\frac{A_1}{A_2}=k^2...(1)[/tex]
Where k is a constant of proportionality. From the problem, we identify:
[tex]\begin{gathered} k=\frac{12}{8}=\frac{3}{2} \\ \\ A_1=60\text{ units}^2 \end{gathered}[/tex]
Finally, using (1), we can find the area of the smaller hexagon:
[tex]\begin{gathered} \frac{60}{A_2}=(\frac{3}{2})^2 \\ \\ \frac{A_2}{60}=\frac{4}{9} \\ \\ \therefore A_2=\frac{80}{3}\text{ units}^2 \end{gathered}[/tex]
