Suppose that a triangle with sides $\,a\,$, $\,b\,$, and $\,c\,$
has been scaled by a positive number $\,s\,$
to get a similar triangle with corresponding sides $\,A\,$, $\,B\,$, and $\,C\,$.
Thus, $\,A = sa\,$ and $\,B = sb\,$ and $\,C = sc\,$.
Let's investigate the relationship between the perimeters of these two triangles:
$\text{perimeter of original triangle} = a + b + c\,$
Thus, the perimeter ends up being scaled by the same factor that scales the sides!
This can be rephrased in terms of ratios.
Notice that:
$$
\cssId{s17}{\frac{\text{scaled perimeter}}{\text{original perimeter}}}
\cssId{s18}{= \frac{A+B+C}{a+b+c}}
\cssId{s19}{= \frac{sa+sb+sc}{a+b+c}}
\cssId{s20}{= \frac{s(a+b+c)}{a+b+c}}
\cssId{s21}{= s}
\cssId{s22}{= \text{scaling factor}}
$$
and
$$
\cssId{s24}{\frac{\text{scaled side}}{\text{original side}}}
\cssId{s25}{= \frac{A}{a}}
\cssId{s26}{= \frac{sa}{a}}
\cssId{s27}{= s}
$$
$$
\cssId{s28}{\frac{\text{scaled side}}{\text{original side}} = \frac{B}{b} = \frac{sb}{b} = s}
$$
$$
\cssId{s29}{\frac{\text{scaled side}}{\text{original side}} = \frac{C}{c} = \frac{sc}{c} = s}
$$
So, the ratio of the perimeters is the same as the ratio of corresponding sides.
A similar calculation shows that this result is indeed true for polygons in general:
Area, on the other hand, behaves quite differently.
Suppose you have a square of side $\,a\,$
that has been scaled by $\,s\,$
to get a square of side $\,A\,$. Let's investigate the relationship between the areas of these two squares: $ \cssId{s45}{\text{area of original square}} \cssId{s46}{= a\cdot a} \cssId{s47}{= a^2} $
$\text{area of scaled square}$
$= A\cdot A$
$= (sa)\cdot(sa)$
$= s^2\cdot a^2$
$= s^2(\text{area of original square})$
Thus, the area ends up being scaled by the square of the factor that scales the sides! 
Again, this can be rephrased in terms of ratios.
Notice that:
$$
\cssId{s56}{\frac{\text{scaled square area}}{\text{original square area}}}
\cssId{s57}{= \frac{A\cdot A}{a \cdot a}}
\cssId{s58}{= \frac{(sa)\cdot (sa)}{a\cdot a}}
\cssId{s59}{= \frac{s^2a^2}{a^2}}
\cssId{s60}{= s^2}
\cssId{s61}{= \text{the square of the ratio of corresponding sides}}
$$
This idea leads to the following theorem:
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
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