Ⅲ Plane Geometry · Stage 17 — Similarity & Dilation · 17.4 Properties of Similar TrianglesAll lessons →
Stage 17 · Similarity & Dilation

17.4  Properties of Similar Triangles

Once they're similar: angles hold, every length scales by k, and area scales by k².

Ages 12–15 · Reasoning, one step at a time
A small triangle and its scaled copy at k = 2. Every length doubles — but the big triangle holds four copies of the small one, so its area is ×4 = k².

Testing for similarity was the hard part; now we collect the rewards. If two triangles are similar with scale factor k, then everything that is a length — each side, the altitude, the median, the angle bisector, the whole perimeter — grows by exactly k. But an area is built from two lengths multiplied together, so it grows by k × k = k². Double a triangle and its perimeter doubles while its area quadruples. That single fact — area scales by k², not by k — is the most useful, and the most often forgotten, idea in all of similarity.

Recall the notation from the last lesson. △ABC ~ △A′B′C′ is read "triangle ABC is similar to triangle A-prime-B-prime-C-prime," and the order of the letters fixes the correspondence: A↔A′, B↔B′, C↔C′. The scale factor is the ratio of a side in the image to its match in the original, A′B′AB= k. Everything in this lesson is a consequence of that one ratio being the same for all three pairs of sides.

17.4.1 Sides proportional, angles equal

Start by writing down exactly what similarity gives us. If △ABC ~ △A′B′C′ with scale factor k = A′B′/AB, then two things hold at once:

Every pair of corresponding sides is in the ratio k. Naming the small triangle's sides a, b, c, the big triangle's matching sides are ka, kb, kc:

A′B′AB= B′C′BC= C′A′CA= k.

Every pair of corresponding angles is equal. Magnifying a figure stretches its lengths but never touches its angles — an angle is a pure measure of turning, with no units to scale. So

∠A = ∠A′,   ∠B = ∠B′,   ∠C = ∠C′.

The same shape at two sizes. Sides a, b, c become ka, kb, kc; the matching angles (one, two, three ticks) stay equal. Here k = 1.5.
Key idea

Similar triangles trade in two kinds of fact: the angles are equal and the sides are proportional (one shared ratio k). The first is exact equality; the second is a constant multiple. Congruent triangles are simply the case k = 1 — equal angles and equal sides.

From here the lesson is short and sweet. Anything you can measure inside a triangle is some length or some angle — and we have just settled what happens to both. Lengths scale by k; angles don't move. The rest is following that rule into the altitude, the median, the perimeter, and finally the area.

17.4.2 The ratio of altitudes, medians, and bisectors

A triangle hides many special segments inside it. From a vertex you can drop the altitude (the perpendicular to the opposite side), draw the median (to the midpoint of the opposite side), or swing the angle bisector (cutting the vertex angle in half). Each is built by a recipe that uses only the triangle's own points. So when we build the same segment in two similar triangles, the two copies are corresponding lengths — and corresponding lengths scale by k.

Here is the reasoning out loud for the altitude. In △ABC drop the altitude CH to side AB; in △A′B′C′ drop C′H′ to A′B′. Look at the two little right triangles △ACH and △A′C′H′: each has a right angle at the foot, and ∠A = ∠A′ because the big triangles are similar. Two equal angles ⇒ △ACH ~ △A′C′H′ by AA. Therefore

C′H′CH= A′C′AC= k   ⇒   the altitude scales by k.

The exact same argument works for the median and the bisector (and even for the radius of the inscribed or circumscribed circle): each forms a smaller similar pair, so each is in ratio k.

Worked example

Two similar triangles have scale factor k = 3/2. The smaller one has an altitude of 4. The larger triangle's matching altitude is k × 4 = 32 × 4 = 6. Its median scales the same way; if the small median were 5, the large one would be 7.5.

Slide the scale factor below and watch one shared k stretch the side, the perimeter, and the chosen inner segment all at once.

Try it One slider, three laws
Drag k. A blue base triangle and its green copy. Pick which corresponding segment to highlight; watch the side scale by ×k, the perimeter by ×k, and the area by ×k².
scale factor k
highlight
Corresponding altitudes in two similar triangles. The little right triangles at each foot are themselves similar (AA), so C′H′ = k·CH.

17.4.3 The perimeter ratio equals k

The perimeter is nothing but the three sides added up, so the scaling carries through the sum with no surprises. If the small triangle has sides a, b, c, its perimeter is P = a + b + c. The big triangle's sides are ka, kb, kc, so its perimeter is

P′ = ka + kb + kc = k(a + b + c) = k·P.

Factor the k out and there it is: the perimeters are in the same ratio k as the sides. This works precisely because every piece of the sum scaled by the same factor — which is exactly what similarity guarantees.

Worked example

A triangle has perimeter 12. Scale it by k = 2. The new perimeter is 2 × 12 = 24 — doubled, just like every side. Run it backward: two similar triangles with perimeters 15 and 25 have scale factor k = 25/15 = 5/3.

Any length you can add up around the figure — perimeter, the total of two sides, the distance once around — follows the same one-line rule. It is only when we multiply two lengths together that the story changes.

17.4.4 The area ratio equals k²

Area is the first quantity in this stage that is not a single length. The area of a triangle is ½ · base · height — a product of two lengths. Under a similar copy, the base scales by k and the height scales by k as well (it's an altitude — section 17.4.2). So the area picks up two factors of k:

S′ = ½ · (k·base) · (k·height) = k² · (½ · base · height) = k² · S.

Similar triangles with sides in ratio k have areas in ratio . Double the sides and the area is multiplied by 2² = 4; triple the sides and it is multiplied by 3² = 9. Run the rule backward and you take a square root: if two similar triangles have areas in ratio m : n, their sides are in ratio √m : √n.

Why the square appears: the k = 3 copy is paved by 3² = 9 tiles, each congruent to the small triangle. Nine times the area for three times the side.
Watch out

The single most common mistake in all of similarity is scaling area by k instead of . If a model is built at half scale (k = ½), each length is halved — but each area (and each amount of paint, or fabric) is only a quarter, because (½)² = ¼. The same warning, squared again, will apply to volume later: volume scales by .

Worked example

A model triangle has area 6 cm² and is enlarged by k = 3. The new area is k² × 6 = 9 × 6 = 54 cm² — not 18. And two similar triangles with areas 16 and 25 have sides in ratio √16 : √25 = 4 : 5, so k = 5/4.

Build the area law with your own eyes below. Set the scale factor and the larger triangle fills with congruent small copies — count them, and you'll always count .

Try it Why area grows by k²
Step k up. The big triangle is paved by congruent copies of the small one. The perimeter is ×k, but the count of tiles — the area ratio — is .
scale factor k 2
quantitywhat it isscales by
sideone lengthk
altitude · median · bisectorone lengthk
perimetera sum of lengthsk
anglea measure of turningunchanged (×1)
areaa product of two lengths

Recap

The whole lesson in five lines

For similar triangles △ABC ~ △A′B′C′ with scale factor k:

Angles are equal — magnifying never bends an angle.

Every length — each side, altitude, median, angle bisector — scales by k.

Perimeter (a sum of sides) scales by k.

Area (a product of two lengths) scales by — and so areas in ratio m:n mean sides in ratio √m : √n.

One slogan: lengths grow by k, area grows by k² (volume, later, by k³).

Exercises

  1. △ABC ~ △DEF with scale factor k = 2 and BC = 5. Find EF.

    Answer

    EF corresponds to BC, so EF = k · BC = 2 × 5 = 10. (Every side scales by k.)

  2. Two similar triangles have perimeters 15 and 25. Find the scale factor k.

    Answer

    Perimeter ratio equals k, so k = 25 ⁄ 15 = 5/3 (about 1.67). Each side of the larger triangle is 53 of its match.

  3. Two similar triangles have scale factor k = 3. By what factor is the area multiplied?

    Answer

    Area scales by k², so by k² = 3² = 9. (Not 3 — area is a product of two lengths.)

  4. Two similar triangles have areas 9 and 16. Find the ratio of their corresponding sides.

    Answer

    Areas are in ratio k², so k² = 16 ⁄ 9 and k = √(16/9) = 4 ⁄ 3. The sides are in ratio 3 : 4 (= √9 : √16).

  5. A model triangle of area 6 cm² is enlarged by k = 2.5. Find the new area.

    Answer

    New area = k² × 6 = 2.5² × 6 = 6.25 × 6 = 37.5 cm². (Scaling by 2.5 multiplies area by 6.25, not 2.5.)

  6. Two similar triangles have scale factor k = 4. The smaller one's corresponding median is 3. Find the larger triangle's median, and explain why a median scales like a side.

    Answer

    Median = k × 3 = 4 × 3 = 12. A median is a length built the same way (vertex to opposite midpoint) in both triangles; it forms a smaller similar pair, so it scales by the same k as every other length.

🎯 Quick check

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§ For teachers and parents

This lesson is the "payoff" of similarity: having spent 17.3 proving two triangles similar, students now harvest the consequences. The big idea is a clean dependency on dimension. A length is one-dimensional and scales by k; an area is two lengths multiplied, so it scales by ; a volume (met later) is three, so . An angle is dimensionless and does not scale at all. If a student can sort each quantity by "how many lengths is it made of," every scaling fact in this lesson follows.

The misconception to watch is almost universal: scaling area (or, later, volume) by k instead of (or k³). A student who doubles a triangle confidently says the area doubles — it actually quadruples. The cure is the picture in 17.4.4: tile the big triangle and count the small copies (k = 2 gives 4, k = 3 gives 9). Counting tiles makes the square undeniable. A companion error is running the area rule backward without the square root: from an area ratio of 9 : 16 the sides are 3 : 4, not 9 : 16. Have students always ask, "is this a length or an area?" before they reach for k versus k².

Common Core: G-SRT.B.5 (use congruence and similarity criteria to solve problems and prove relationships), with connections to G-MG.A (modeling with geometry) and the area-scaling strand of 7.G and G-GMD. The k-versus-k² distinction is exactly the proportional-reasoning leap these standards target.

Next: 17.5 — Similar Polygons · Previous: 17.3 — Tests for Similar Triangles

eastmath.com · Stage 17 · 17.4 Properties of Similar Triangles · Reasoning, one step at a time