7.2 Similar polygons and triangles

Similar Polygons and Triangles

Two polygons are similar if they have the exact same shape but not necessarily the same size. Their corresponding angles are congruent, and their corresponding sides are proportional. This concept connects proportional reasoning to geometry, giving you tools to find unknown measurements in figures that share the same shape.

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Definition of Similar Polygons

For two polygons to be similar, two conditions must both be true:

1. All pairs of corresponding angles are congruent.
2. All pairs of corresponding sides are proportional (their ratios are equal).

Similarity is written with the symbol $\sim$. For example, $\triangle ABC \sim \triangle DEF$ means triangle ABC is similar to triangle DEF. The order of the letters matters: it tells you which vertices correspond. Here, $A$ corresponds to $D$, $B$ to $E$, and $C$ to $F$. If you mix up the order, you'll set up the wrong proportions and get wrong answers, so pay close attention to the correspondence every time.

Some polygon families are always similar to others of their kind. Any two equilateral triangles are similar, any two squares are similar, and any two regular pentagons (or regular hexagons, etc.) are similar. Regular polygons of the same type always have congruent angles and proportional sides (since all sides are equal), so both similarity conditions are automatically satisfied.

Similarity Theorems for Triangles

Triangles are special because you don't need to check every angle and every side to prove similarity. Three shortcut theorems let you establish similarity with less information:

• AA (Angle-Angle) Similarity: If two angles of one triangle are congruent to two angles of another triangle, the triangles are similar. Since triangle angles always sum to $180°$, matching two angles guarantees the third matches too. This is the most commonly used shortcut.
• SAS (Side-Angle-Side) Similarity: If the ratios of two pairs of corresponding sides are equal and the included angle (the angle between those sides) is congruent, the triangles are similar. The angle must be the one between the two sides you're comparing.
• SSS (Side-Side-Side) Similarity: If the ratios of all three pairs of corresponding sides are equal, the triangles are similar. No angle information is needed.

For polygons with more than three sides, there's no shortcut. You must verify that all corresponding angles are congruent and all corresponding sides are proportional. Having just one of those conditions isn't enough. A square and a non-square rectangle have all right angles (congruent corresponding angles), but their sides aren't proportional, so they aren't similar. Conversely, a square and a rhombus can have all sides proportional, but if the rhombus's angles aren't $90°$, the angles don't match, so they aren't similar either.

Calculations in Similar Shapes

Once you know two polygons are similar, the ratio of any pair of corresponding sides equals the scale factor. If $\triangle ABC \sim \triangle DEF$ with a scale factor of $k$, then:

$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = k$

Corresponding angles are congruent:

$\angle A \cong \angle D, \quad \angle B \cong \angle E, \quad \angle C \cong \angle F$

Two useful results follow from the scale factor:

• Perimeters: The ratio of the perimeters equals the scale factor. If $k = 3$, the larger polygon's perimeter is 3 times the smaller's.
• Areas: The ratio of the areas equals the square of the scale factor. If $k = 3$, the larger polygon's area is $3^2 = 9$ times the smaller's.

The area relationship trips people up on tests. Think of it this way: area is two-dimensional, so the scale factor gets applied in two directions, which is why you square it.

For example, if two similar triangles have a scale factor of $\frac{2}{5}$, their perimeters have a ratio of $\frac{2}{5}$ and their areas have a ratio of $\left(\frac{2}{5}\right)^2 = \frac{4}{25}$.

Problem-Solving with Similar Polygons

Finding missing side lengths: Set up a proportion using corresponding sides. If $\triangle ABC \sim \triangle DEF$ with $AB = 6$, $DE = 9$, and $BC = 8$, find $EF$:

1. Write the proportion: $\frac{AB}{DE} = \frac{BC}{EF}$
2. Substitute: $\frac{6}{9} = \frac{8}{EF}$
3. Cross-multiply: $6 \cdot EF = 9 \cdot 8 = 72$
4. Solve: $EF = 12$

Always double-check that you've matched corresponding sides correctly. $AB$ pairs with $DE$ (not $DF$) because $A \leftrightarrow D$ and $B \leftrightarrow E$.

Indirect measurement: Similar triangles let you find heights or distances you can't measure directly. For instance, if a 1.5 m stick casts a 2 m shadow at the same time a tree casts a 14 m shadow, the stick and tree form similar triangles (by AA, since the sun's rays hit at the same angle and both objects stand perpendicular to the ground). Set up $\frac{1.5}{2} = \frac{h}{14}$ and cross-multiply: $2h = 21$, so $h = 10.5$ m.

Geometric mean in right triangles: When you draw the altitude from the right angle to the hypotenuse of a right triangle, it creates two smaller triangles that are each similar to the original and to each other (by AA). This gives rise to the geometric mean relationship:

$\frac{a}{x} = \frac{x}{b}$

Here $a$ and $b$ are the lengths of the two segments the altitude creates on the hypotenuse, and $x$ is the length of the altitude. Cross-multiplying gives $x^2 = ab$, so $x = \sqrt{ab}$.

For example, if the altitude divides the hypotenuse into segments of length 4 and 9, the altitude's length is $\sqrt{4 \cdot 9} = \sqrt{36} = 6$.

There are actually two other geometric mean relationships in this same configuration. If the hypotenuse has total length $c = a + b$, and the legs of the original triangle are $p$ and $q$, then each leg is the geometric mean of the hypotenuse and the adjacent segment:

$p = \sqrt{a \cdot c} \quad \text{and} \quad q = \sqrt{b \cdot c}$

These all come from setting up proportions between the similar triangles, so if you forget the formulas, you can always re-derive them by identifying corresponding sides.