PERIMETERS and AREAS of SIMILAR POLYGONS

- Jump right to the exercises!
- You may want to review: Similarity, Ratios, and Proportions

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

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$\text{perimeter of scaled triangle}$

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$= A + B + C$

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$= sa + sb + sc$

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$= s(a + b + c)$

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

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

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
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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 |

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?

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 $\,5(x+y+z)\,$.

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?

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 $\,22\,$.

The area is scaled by the

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

by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:

Two Special Triangles

On this exercise, you will not key in your answer.

However, you can check to see if your answer is correct.

However, you can check to see if your answer is correct.