Perimeters and Areas of Similar Polygons

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:

$\text{perimeter of original triangle} = a + b + c\,$

$\text{perimeter of scaled triangle}$

$= A + B + C$

$= sa + sb + sc$

$= s(a + b + c)$

$= s(\text{perimeter of original triangle})$

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:

THEOREM Perimeters of Similar Polygons

The ratio of the perimeters of two similar polygons is equal to the ratio of the corresponding sides.

That is: $$ \frac{\text{perimeter of scaled polygon}}{\text{perimeter of original polygon}} = \text{the scaling factor} = \frac{\text{scaled side}}{\text{original side}} $$

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 $\,A\,$.
Thus, $\,A = sa\,$.

Let's investigate the relationship between the areas of these two squares:

$\text{area of original square} = a\cdot a = a^2$

$\text{area of scaled square}$

$= A\cdot A$

$= (sa)\cdot(sa)$

$= s^2\cdot a^2$

$= s^2(\text{area of original square})$

Thus, the area ends up being scaled by the square of the factor that scales the sides!

Again, this can be 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:

THEOREM Areas of Similar Polygons

The ratio of the areas of two similar polygons is equal to the square of the ratio of the corresponding sides.

EXAMPLES:

Question:
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?

Solution:
The perimeter is scaled by the same factor as the sides.
The original perimeter is $\,x+y+z\,$.
The new (scaled) perimeter is $\,5(x+y+z)\,$.

Question:
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?

Solution:
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 $\,22\,$.

Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Two Special Triangles

(MAX is 11; there are 11 different problem types.)