Same shape for any polygon — and you need BOTH equal angles and proportional sides.
Triangles were spoiled. Get their angles right and the sides had to follow — that was the gift of AA. For a polygon with more than three sides, that shortcut vanishes. A square and a long rectangle share all four right angles yet are obviously different shapes; a square and a leaning rhombus have all sides in the same ratio yet clearly different shapes. So "similar polygon" must demand both things at once: corresponding angles equal and corresponding sides proportional. Settle that, and the friendly scaling laws — perimeter ×k, area ×k² — carry straight over, ready to power scale drawings and maps.
In 17.4 we cashed in similarity for triangles: lengths grow by k, area by k². Here we lift the whole idea from triangles to any polygon — and discover that for four or more sides, equal angles alone is no longer enough.
Two polygons are similar when their vertices can be matched in order so that
(1) every pair of corresponding angles is equal, and (2) every pair of corresponding sides is proportional (the same scale factor k).
We write ABCDE ~ A′B′C′D′E′ with a tilde ~, and the order of the letters fixes the matching: A↔A′, B↔B′, and so on. The scale factor is
k = A′B′AB = B′C′BC = C′D′CD = … (one ratio, shared by every side pair).
For triangles, condition (1) forces condition (2): two equal angles (AA) make the sides proportional automatically. For four or more sides the two conditions are genuinely independent — you must check both. A polygon has "room" to keep the angles while sliding a side longer or shorter.
Here is the cleanest way to feel why. Below you can toggle through three pairs of quadrilaterals. One pair is truly similar; the other two each pass one test and flunk the other — and the figure shows you exactly which.
How do you actually prove two polygons are similar? Reduce the polygon to the one shape you already understand — the triangle. Draw the diagonals from one corresponding vertex in each polygon, slicing both into the same number of triangles. If every pair of corresponding triangles is similar with the same scale factor k, then the whole polygons are similar.
Why does this work? The diagonal AC is itself a length, so if both triangles scale by k, then AC and A′C′ agree on that k at the shared seam. The angles of the polygon are simply sums of the angles of its triangles, so equal triangle-angles rebuild equal polygon-angles, and every side belongs to one of the triangles. Triangle similarity (AA, SAS, SSS from 17.3) does all the heavy lifting.
Suppose △ABC ~ △A′B′C′ with k = 3∶2, and the second triangle △ACD ~ △A′C′D′ also at 3∶2. The shared diagonal forces a consistent k, so the four sides AB, BC, CD, DA all scale by 3∶2 and every angle matches. Conclusion: ABCD ~ A′B′C′D′ with k = 3∶2.
Cut from matching vertices. Slicing ABCD from A but A′B′C′D′ from B′ compares the wrong triangles and proves nothing. Keep the letter order honest, just as you do inside the tilde.
Once two polygons are similar, the two scaling laws you learned for triangles return unchanged, and for the same reason.
Perimeter scales by k. The perimeter is a sum of sides, and each side becomes k times longer:
P′ = ka + kb + kc + … = k(a + b + c + …) = k·P.
Area scales by k². Slice the polygon into triangles; each triangle's area is ½·base·height, and both the base and the height stretch by k, so each triangle's area grows by k·k = k². Add them up — total area grows by k² too.
Two regular hexagons are similar (all regular n-gons of the same n are). If k = 3∶2, then their perimeters are in ratio 3∶2 and their areas are in ratio 3²∶2² = 9∶4. A side of 4 in the small one matches 6 in the large; an area of 16 matches 36.
| quantity | how it scales | k = 3∶2 gives |
|---|---|---|
| any length (side, diagonal) | ×k | 3 ∶ 2 |
| perimeter | ×k | 3 ∶ 2 |
| area | ×k² | 9 ∶ 4 |
| every angle | unchanged | same |
Going the other way, if two similar polygons have areas in ratio m∶n, their sides (and perimeters) are in ratio √m∶√n. Areas 16∶25 ⇒ sides 4∶5. The most common slip in this whole stage is scaling area by k instead of k² — double a shape and its area quadruples.
This is where similarity earns its keep. The moment you know k — read off from one pair of corresponding sides — every other length is one multiplication away:
unknown side = (known corresponding side) × k.
A scale drawing is exactly a similar figure: a floor plan, a model car, an architect's blueprint. And a map is the grandest scale drawing of all. A 1∶25000 map means every length on paper is k = 1/25000 of the real length — the map and the land are similar figures.
Two similar quadrilaterals match a side of 6 in the small one to 9 in the large, so k = 9/6 = 3∶2. Then a side of 8 in the small one matches 8 × 3/2 = 12 in the large. No need to re-measure — k does it.
Slide the scale factor below and watch one "unknown" side solve itself.
Every scale drawing — a map, a plan, a model — is a pair of similar figures. Find k from the legend (or one known pair), multiply, and you have read the real world off the paper.
Two similar pentagons have scale factor k = 2. A side of length 7 in the small one corresponds to which length in the large one?
7 × 2 = 14. Every length scales by k.
Is a 2 × 4 rectangle similar to a 3 × 6 rectangle? Explain.
Yes. All angles are 90° (equal), and the sides are 2∶3 and 4∶6 = 2∶3 (proportional). Both conditions hold, so they are similar, k = 3∶2.
Is a square similar to a non-square rhombus (a "diamond" with the same four side lengths)?
No. The sides are in ratio 1∶1 (proportional), but the angles differ — the square has 90° corners, the rhombus does not. Condition (1) fails, so they are not similar.
Two similar polygons have areas in ratio 16∶25. Find the ratio of their sides and the ratio of their perimeters.
Area ratio is k², so k = √16 ∶ √25 = 4∶5. Sides and perimeters are both in ratio 4∶5 (perimeter scales by k, same as any length).
On a 1∶50000 map, two towns are 6 cm apart. How far apart are they on the ground?
6 cm × 50000 = 300000 cm = 3 km. The map is a similar figure with k = 1/50000, so real distance = map distance ÷ k.
A quadrilateral floor plan is enlarged by k = 1.5 for a poster. A wall drawn 10 cm long on the plan becomes how long on the poster?
10 × 1.5 = 15 cm. Lengths scale by k. (The poster's area, by the way, is 1.5² = 2.25× the plan's.)
Six questions to lock it in. Tap the answer you think is right.
The whole lesson turns on one correction to a tempting half-truth. After a stage on triangles, students internalize "same angles means same shape" — and for triangles that is exactly right (AA). The big idea here is that polygons with four or more sides break that rule: a square and a 1×3 rectangle share every angle but are plainly different shapes, and a square and a leaning rhombus share every side length but are plainly different shapes. Similar polygons require both equal corresponding angles and proportional corresponding sides, matched vertex-by-vertex in order. The Both conditions widget is built to make that failure visible — each non-example honestly shows the one condition that breaks.
The misconception to watch: assuming equal angles alone (or equal/proportional sides alone) makes polygons similar — that shortcut is true only for triangles. A close second is scaling area by k instead of k² when comparing similar figures; have students count the tiles in the area figure (k = 2 → 4 tiles) until "double the length, quadruple the area" feels obvious. When they reverse it, remind them to take a square root: areas 16∶25 give sides 4∶5, not 16∶25.
Common Core: this lesson is G-SRT.A.2 (define similarity via equal angles and proportional sides), 7.G.A.1 (solve problems involving scale drawings), and G-MG.A.1 / A.3 (model real situations — maps and scale drawings — with geometric figures).