Let's call the sides of the figure on the left [tex] a,b,c \text{ and } d [/tex], and [tex] a',b',c' \text{ and } d' [/tex] the sides of the figure on the right. Since the figure on the right is similar to the one on the left, each of its side is obtained by multiplying the correspondant original side by a same amount, say [tex] r [/tex].
So, we have
[tex] a' = ar,\quad b' = br,\quad c' = cr,\quad d' = dr[/tex]
So, the two perimeters are
[tex] p=a+b+c+d,\quad p'=a'+b'+c'+d' = ar+br+cr+dr = r(a+b+c+d)=pr [/tex]
So, we can find the scale ratio by computing
[tex] \cfrac{p'}{p} = \cfrac{pr}{p} = r = \cfrac{28}{20} = \cfrac{7}{5} [/tex]
Now, the two areas are not in this same proportion, i.e. it is not true that [tex] A' = Ar [/tex], but almost: you have to square the scale ratio. So, it is true that
[tex] A' = Ar^2 = 18.6 \times \cfrac{49}{25} = 36.456 [/tex]
