Do I have to check both angles and sides, or is one enough?

For general polygons you must check both. A square and a long rectangle have matching angles but not matching side ratios; a square and a slanted rhombus have matching side ratios but not matching angles. Neither pair is similar. Triangles are the exception: for triangles, congruent angles force proportional sides and vice versa, which is why the AA, SSS, and SAS similarity shortcuts exist.

Why does area use k^2 instead of k?

Area depends on two dimensions. If you scale a base by k and a height by k, the product scales by k · k = k^2. Picture a 1-by-1 square scaled by 3: the new square is 3 by 3, so its perimeter went from 4 to 12 (times 3) but its area went from 1 to 9 (times 9). Any polygon can be cut into triangles, so the same factor applies to all of them.

Does it matter which figure I put on top of the ratio?

Not for correctness, only for consistency. The scale factor from small to large is the reciprocal of the factor from large to small. Pick a direction, write it down, and use that same direction in every proportion in the problem. Then check that your answer is bigger or smaller in the way the picture suggests.

Are all congruent figures similar?

Yes. Congruent polygons satisfy both similarity conditions with scale factor k = 1: corresponding angles are congruent and every side ratio equals 1. The reverse is not true — similar figures are congruent only when k = 1.

GEOM-6.2

Similar Polygons

Learn how to test two polygons for similarity, use the scale factor k to find missing sides, and apply the k and k² ratios for perimeter and area.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Similar Polygons, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

Two shapes can look alike without being mathematically similar, and two shapes can be similar even when one is rotated, flipped, or many times larger. In this lesson you will learn the exact test: corresponding angles congruent and corresponding sides proportional. Both conditions matter, and skipping either one is the most common mistake students make with polygons that are not triangles.

Once you know two polygons are similar, one number — the scale factor

k

— unlocks everything else. Multiply a side by

k

to get the matching side in the other figure. Multiply the perimeter by

k

. Multiply the area by

k^2

. You will practice writing similarity statements in the correct order, solving proportions for unknown lengths, and explaining why area grows faster than length. These ideas carry directly into the triangle similarity shortcuts and proportionality theorems in the rest of this unit.

The Two-Part Definition of Similar Polygons

Two polygons are similar if their corresponding angles are congruent and their corresponding sides are proportional. Write it as

ABCD \sim EFGH

, where the order of the letters tells you which vertices match:

A

with

E

B

with

F

, and so on. That ordering is not decoration — it is how you know which sides to pair when you set up ratios.

Both conditions are required for general polygons. A square and a non-square rectangle have four right angles each, so every pair of corresponding angles is congruent, but a 2-by-2 square and a 2-by-6 rectangle are not similar because

\frac{2}{2} \neq \frac{2}{6}

. A square and a rhombus with 60-degree angles have all four sides in a constant ratio, but their angles do not match, so again they are not similar. Only for triangles does one condition force the other, which is why triangles get the special AA, SSS, and SAS shortcuts.

To check similarity in practice, first match up the vertices using the angle measures or the given similarity statement. Then compute the ratio

\frac{\text{side of first figure}}{\text{corresponding side of second figure}}

for every pair. If all of those ratios reduce to the same number, the sides are proportional.

Condition met	Angles congruent	Sides proportional	Similar?
Square and larger square	Yes	Yes	Yes
Square and 2-by-6 rectangle	Yes	No	No
Square and 60-degree rhombus	No	Yes	No
Any two rectangles	Yes	Only sometimes	Only if ratios match

Congruent polygons are a special case of similar polygons where every ratio equals 1.

Scale Factor and Finding Unknown Sides

The common ratio of corresponding sides is the scale factor, usually called

k

. If

ABCD \sim EFGH

, then

k = \frac{AB}{EF} = \frac{BC}{FG} = \frac{CD}{GH} = \frac{DA}{HE}.

Direction matters. The scale factor from the small figure to the large figure is greater than 1; the scale factor from large to small is between 0 and 1, and it is the reciprocal. Neither is wrong, but you must stay consistent inside a single problem. Writing a sentence such as "

k = 3

going from

ABCD

EFGH

" before you compute prevents most sign-free arithmetic errors.

To find an unknown side, set up a proportion using one pair of sides whose lengths you know both of, plus the pair containing the unknown. Suppose

ABCD \sim EFGH

with

AB = 6

EF = 9

, and

BC = 10

. Then

\frac{AB}{EF} = \frac{BC}{FG} \quad \Rightarrow \quad \frac{6}{9} = \frac{10}{FG}.

Cross multiply:

6 \cdot FG = 90

, so

FG = 15

. Equivalently, going from

ABCD

EFGH

multiplies every length by

\frac{9}{6} = 1.5

, and

10 \cdot 1.5 = 15

.

Where students go wrong: mixing figures inside one ratio. Writing

\frac{AB}{BC} = \frac{FG}{EF}

scrambles the correspondence. A safe habit is to keep the first figure on top in every fraction. Another frequent slip is using the scale factor upside down, which produces an answer that is obviously too small when the second figure is clearly larger — always sanity-check the size of your result against the picture.

Perimeter Scales by k, Area Scales by k Squared

If two polygons are similar with scale factor

k

, then

\frac{\text{perimeter}_1}{\text{perimeter}_2} = k \qquad \text{and} \qquad \frac{\text{area}_1}{\text{area}_2} = k^2.

The perimeter rule follows directly from the definition: every side of the first figure is

k

times the matching side of the second, so the sum of the sides is also

k

times as large. Factoring

k

out of the sum is the whole proof.

Area behaves differently because area involves two dimensions. A rectangle with base

b

and height

h

has area

bh

; scaling both by

k

gives

(kb)(kh) = k^2bh

. The same reasoning works for a triangle,

\frac{1}{2}(kb)(kh) = k^2 \cdot \frac{1}{2}bh

, and for any polygon you can decompose into triangles.

Concretely, if

k = 3

, the larger polygon has 3 times the perimeter but 9 times the area. Tripling the side lengths of a tile means you need nine of the original tiles to cover it. This is the single most-missed idea in the lesson: students multiply area by 3 instead of 9, or divide area by 2 when the sides were halved instead of dividing by 4.

Working backwards requires a square root. If you know the ratio of the areas is

\frac{49}{16}

, then

k = \sqrt{\frac{49}{16}} = \frac{7}{4}

, and that

\frac{7}{4}

is the ratio you use for sides and perimeter. Take the square root of the area ratio before comparing lengths, never after.

Reading Problems and Avoiding Common Traps

Most similar-polygon problems come in one of four shapes, and recognizing the type tells you what to do first.

Given	First move	Then
Two figures with all sides labeled	Check every ratio	Conclude similar or not
Similarity statement plus some sides	Find $k$ from a complete pair	Multiply or divide to get unknowns
Perimeters of similar figures	Ratio of perimeters is $k$	Use $k$ for individual sides
Areas of similar figures	Take $\sqrt{\ }$ of the area ratio to get $k$	Use $k$ for sides and perimeter

A few traps worth naming. First, "proportional" does not mean "differ by the same amount." Sides of 3, 4, 5 and 5, 6, 7 each grew by 2, but

\frac{3}{5} \neq \frac{4}{6}

, so those triangles are not similar. Additive thinking is a genuinely common wrong answer.

Second, a figure that is rotated or reflected can still be similar. Orientation never affects similarity — only angle measures and side ratios do. Match vertices by the angles, not by which side happens to be drawn on the bottom.

Third, when a problem gives the ratio of perimeters as something like 5 to 2, that ratio is

k

; do not square it or take a root. Squaring is only for moving from lengths to area.

Finally, keep units straight. Lengths are in centimeters or inches, areas in square units. If an answer for area comes out in plain centimeters, you likely used

k

where

k^2

belonged.

Key terms

Similar polygons.: Two polygons whose corresponding angles are all congruent and whose corresponding sides are all proportional; written with the symbol $\sim$ .
Corresponding parts.: The angles or sides that match up between two figures, determined by the order of vertices in the similarity statement.
Scale factor ( $k$ ).: The constant ratio of a side length in one figure to the matching side length in the similar figure.
Proportional.: Related by a constant multiplier, so that every pair of corresponding sides has the same ratio — not the same difference.
Similarity statement.: A notation such as $ABCD \sim EFGH$ in which vertex order encodes exactly which parts correspond.
Ratio of perimeters.: For similar polygons, this equals the scale factor $k$ .
Ratio of areas.: For similar polygons, this equals $k^2$ , the square of the scale factor.
Congruent polygons.: Similar polygons with scale factor $k = 1$ ; same shape and same size.

Worked example

Quadrilateral

ABCD \sim

quadrilateral

PQRS

. In

ABCD

AB = 8

cm,

BC = 12

cm,

CD = 10

cm, and

DA = 6

cm. In

PQRS

PQ = 12

cm. Find

QR

, the perimeter of

PQRS

, and the area of

PQRS

if the area of

ABCD

is 60 square centimeters.

Step 1: Find the scale factor. The similarity statement pairs

A

with

P

and

B

with

Q

, so

AB

corresponds to

PQ

. Going from

ABCD

PQRS

k = \frac{PQ}{AB} = \frac{12}{8} = \frac{3}{2} = 1.5.

Since

k > 1

PQRS

is the larger figure. Every answer should come out bigger than its counterpart in

ABCD

— a quick check to run at the end.

Step 2: Find $QR$ .

BC

corresponds to

QR

, so

QR = k \cdot BC = 1.5 \cdot 12 = 18 \text{ cm}.

Step 3: Find the perimeter of $PQRS$ . First get the perimeter of

ABCD

8 + 12 + 10 + 6 = 36

cm. Because perimeter scales by

k

itself,

\text{perimeter of } PQRS = 1.5 \cdot 36 = 54 \text{ cm}.

You could also scale each side separately (

12 + 18 + 15 + 9 = 54

) and get the same total, which confirms the rule.

Step 4: Find the area of $PQRS$ . Area scales by

k^2

, not by

k

k^2 = (1.5)^2 = 2.25,

\text{area of } PQRS = 2.25 \cdot 60 = 135 \text{ square centimeters}.

Step 5: Check. All three results are larger than the originals, and the area grew by a bigger factor (2.25) than the lengths did (1.5), exactly as expected. Multiplying 60 by 1.5 to get 90 would have been the classic error.

Practice questions

Two similar pentagons have areas of 45 square inches and 80 square inches. What is the ratio of their corresponding side lengths, smaller to larger?

9 to 16
3 to 4
45 to 80
$\sqrt{45}$ to 80

Answer: 3 to 4

The ratio of the areas is

\frac{45}{80} = \frac{9}{16}

. Since the area ratio equals

k^2

, take the square root:

k = \sqrt{\frac{9}{16}} = \frac{3}{4}

. So the sides are in a 3 to 4 ratio. The choice 9 to 16 is the reduced area ratio, not the length ratio, and it is the most common wrong answer here — it skips the square root step entirely.

A rectangle measures 4 by 9. A second rectangle measures 6 by 11. Are the two rectangles similar? Justify your answer.

Answer: No. The angles all match (four right angles each), but the sides are not proportional: $\frac{4}{6} = \frac{2}{3}$ while $\frac{9}{11}$ does not reduce to $\frac{2}{3}$ , so the two ratios are unequal.

Similarity requires both conditions, and rectangles automatically satisfy the angle condition, so the side ratios decide the question. Comparing

\frac{4}{6}

and

\frac{9}{11}

by cross multiplication gives

4 \cdot 11 = 44

and

6 \cdot 9 = 54

; since

44 \neq 54

, the ratios differ. Notice that each side grew by 2, which tempts students to say yes. Adding the same amount to every side is not the same as multiplying by the same amount, and only multiplying produces similar figures. For these to be similar, the second rectangle would need dimensions 6 by 13.5.

Triangle

JKL \sim

triangle

MNO

with a scale factor of

\frac{2}{5}

from

JKL

MNO

. If the perimeter of

MNO

is 60 cm, find the perimeter of

JKL

and the ratio of the area of

JKL

to the area of

MNO

Answer: Perimeter of $JKL$ is 24 cm; the area ratio is $\frac{4}{25}$ .

Perimeter scales by the same factor as the sides, so

\frac{\text{perimeter of } JKL}{\text{perimeter of } MNO} = \frac{2}{5}

. That gives perimeter of

JKL = \frac{2}{5} \cdot 60 = 24

cm. For area, square the scale factor:

\left(\frac{2}{5}\right)^2 = \frac{4}{25}

. Triangle

JKL

has less than one-fifth the area of

MNO

even though its sides are two-fifths as long, which shows how quickly area shrinks when lengths shrink.

FAQ

Do I have to check both angles and sides, or is one enough?: For general polygons you must check both. A square and a long rectangle have matching angles but not matching side ratios; a square and a slanted rhombus have matching side ratios but not matching angles. Neither pair is similar. Triangles are the exception: for triangles, congruent angles force proportional sides and vice versa, which is why the AA, SSS, and SAS similarity shortcuts exist.
Why does area use $k^2$ instead of $k$ ?: Area depends on two dimensions. If you scale a base by $k$ and a height by $k$ , the product scales by $k \cdot k = k^2$ . Picture a 1-by-1 square scaled by 3: the new square is 3 by 3, so its perimeter went from 4 to 12 (times 3) but its area went from 1 to 9 (times 9). Any polygon can be cut into triangles, so the same factor applies to all of them.
Does it matter which figure I put on top of the ratio?: Not for correctness, only for consistency. The scale factor from small to large is the reciprocal of the factor from large to small. Pick a direction, write it down, and use that same direction in every proportion in the problem. Then check that your answer is bigger or smaller in the way the picture suggests.
Are all congruent figures similar?: Yes. Congruent polygons satisfy both similarity conditions with scale factor $k = 1$ : corresponding angles are congruent and every side ratio equals 1. The reverse is not true — similar figures are congruent only when $k = 1$ .

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