Suppose that a triangle with sides
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$\,a\,$,
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$\,b\,$, and
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$\,c\,$ has
been scaled by a positive number
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$\,s\,$
to get a similar triangle
with corresponding sides
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$\,A\,$,
$\,B\,$, and
$\,C\,$.
Thus,
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$\,A = sa\,$ and
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$\,B = sb\,$ and
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$\,C = sc\,$.
Let's investigate the relationship between the perimeters of these two triangles:
[beautiful math coming... please be patient] $\text{perimeter of original triangle} = a + b + c\,$
Thus, the perimeter ends up being scaled by the same factor that scales the sides!
This can be re-phrased in terms of ratios.
Notice that:
$$
\frac{\text{scaled perimeter}}{\text{original perimeter}}
= \frac{A+B+C}{a+b+c}
= \frac{sa+sb+sc}{a+b+c}
= \frac{s(a+b+c)}{a+b+c}
= s = \text{scaling factor}
$$
and
$$\frac{\text{scaled side}}{\text{original side}} = \frac{A}{a} = \frac{sa}{a} = s$$
$$\frac{\text{scaled side}}{\text{original side}} = \frac{B}{b} = \frac{sb}{b} = s$$
$$\frac{\text{scaled side}}{\text{original side}} = \frac{C}{c} = \frac{sc}{c} = s$$
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:
That is: $$ \frac{\text{perimeter of scaled polygon}}{\text{perimeter of original polygon}} = \text{the scaling factor} = \frac{\text{scaled side}}{\text{original side}} $$
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Area, on the other hand, behaves quite differently.
Suppose you have a square of side
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$\,a\,$
that has been scaled by
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$\,s\,$
to get a square of side
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$\,A\,$. Let's investigate the relationship between the areas of these two squares: [beautiful math coming... please be patient] $\text{area of original square} = a\cdot a = a^2$
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$\text{area of scaled square}$
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$= A\cdot A$
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$= (sa)\cdot(sa)$
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$= s^2\cdot a^2$
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$= s^2(\text{area of original square})$
Thus, the area ends up being scaled by the square of the factor that scales the sides! |
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Again, this can be re-phrased in terms of ratios.
Notice that:
$$
\frac{\text{scaled square area}}{\text{original square area}}
= \frac{A\cdot A}{a \cdot a}
= \frac{(sa)\cdot (sa)}{a\cdot a}
= \frac{s^2a^2}{a^2}
= s^2 = \text{the square of the ratio of corresponding sides}
$$
This idea leads to the following theorem:
Suppose that a triangle has sides of lengths $\,x\,$, $\,y\,$, and $\,z\,$.
The sides of the triangle are scaled by a factor of $\,5\,$ to get a similar triangle.
What is a formula for the perimeter of the new triangle?
The perimeter is scaled by the same factor as the sides.
The original perimeter is $\,x+y+z\,$.
The new (scaled) perimeter is $\,5(x+y+z)\,$.
Suppose that the sides of a polygon are scaled by a factor of $\,3\,$ to get a similar polygon.
This new polygon has area $\,198\,$.
What is the area of the original polygon?
The area is scaled by the square of the factor that scales the sides.
Let $\,A\,$ denote the (unknown) original area.
We are told that the new (scaled) area is $\,198\,$, and the scaling factor is $\,3\,$.
So, $\,A\,$ gets scaled by $\,3^2\,$ to give $\,198\,$; i.e., $\,3^2A = 198\,$.
Solving for $\,A\,$ gives $\,A = \frac{198}{3^2} = 22\,$, so the area of the original polygon is $\,22\,$.
by practicing the exercise at the bottom of this page.
When you're done practicing, move on to:
Two Special Triangles