Similar Right Triangles: Rules, Ratios & Worked Examples

Two right triangles are similar when they share one pair of equal acute angles. That is the whole rule. The right angles already match by definition, so one matching acute angle gives you two pairs of equal angles, and because the three angles of any triangle sum to 180° the third pair is forced to match as well. This is AA similarity, and for right triangles it collapses to a single check: find one acute angle in each triangle that agrees, and the triangles are similar. Once they are similar, every pair of corresponding sides has the same ratio, and that ratio is the scale factor.

Figure 1

Figure 2

The 6-8-10 triangle is twice the size of the 3-4-5, yet the angles computed from its side lengths are identical: 36.87° and 53.13° in both. Doubling every length changed the size and nothing else.

AA similarity, stated precisely

AA similarity postulate: if two angles of one triangle are equal to two angles of another triangle, the triangles are similar.

That statement mentions no side lengths, and it does not need to. Angles fix shape. Once two angle pairs agree the third agrees automatically, because in both triangles you are subtracting the same two numbers from the same total of 180°.

Why one acute angle is enough for right triangles

In a general triangle you must verify two angle pairs. In a right triangle one of those pairs is free:

1. Both triangles have a 90° angle. That is the first matching pair, and it costs nothing to check.
2. Suppose one acute angle in each triangle measures 36.87°. That is the second matching pair.
3. The remaining angle in each must be 180° − 90° − 36.87° = 53.13°. Third pair matches.

So the working rule is: two right triangles are similar if and only if one acute angle of the first equals one acute angle of the second. The “only if” direction holds too. If the triangles are similar, corresponding angles are equal by definition, so such a pair exists.

Two other routes help when you are handed sides rather than angles:

• SSS similarity: all three side pairs share a common ratio. For right triangles you rarely need all three, since the third side follows from the Pythagorean theorem.
• SAS similarity: two side pairs share a ratio and the included angles are equal. The right angle sits between the two legs, so if the two legs are proportional, the triangles are similar. A 9-12 pair and a 3-4 pair both reduce to 3:4, so those triangles are similar without measuring a single angle.

What similarity preserves and what it does not

Similarity preserves exactly two things and destroys one.

Preserved: angles. A 36.87° angle stays 36.87° whether the triangle fits on a postage stamp or spans a football field.

Preserved: side ratios. The ratio of any two sides within one triangle equals the ratio of the corresponding two sides within the other. In the 3-4-5 triangle the short leg over the hypotenuse is 3/5 = 0.6; in the 6-8-10 triangle it is 6/10 = 0.6. This is why sine, cosine and tangent are well defined as functions of an angle alone, a point developed in the guide to SOHCAHTOA.

Not preserved: size. Lengths, perimeter and area all change. Similar is not congruent; congruent triangles are the special case where the scale factor equals 1.

Writing the similarity statement: vertex order carries the information

The symbol ~ means “is similar to”, but â–³ABC ~ â–³DEF says far more than that. The order of the letters declares the correspondence: A pairs with D, B with E, C with F. Every proportion you write afterwards reads straight off that order, so scrambling the letters makes every proportion wrong even when the triangles themselves are perfectly similar.

Take two concrete triangles:

• â–³ABC: right angle at C, angle A = 36.87°, angle B = 53.13°. Sides BC = 3, AC = 4, AB = 5.
• â–³DEF: right angle at F, angle D = 36.87°, angle E = 53.13°. Sides EF = 6, DF = 8, DE = 10.

Corresponding sides sit opposite corresponding angles, so BC (opposite A) pairs with EF, AC (opposite B) pairs with DF, and AB pairs with DE:

The wrong statement claims BC pairs with DF and AC pairs with EF:

The triangles did not change. Only the claimed pairing changed, and that was enough to make the statement false.

Reading corresponding parts off a diagram

This routine works no matter how the triangles are rotated or flipped on the page:

1. Mark the right angles. They always correspond to each other.
2. Find the equal acute angles. Either they are marked, or you compute them from the sides with a tangent ratio, or the problem supplies them (a shared angle, parallel lines, vertical angles).
3. Match sides by the angle they face, never by where they sit on the page. The side opposite the 36.87° angle in one triangle corresponds to the side opposite the 36.87° angle in the other, and the hypotenuse always corresponds to the hypotenuse.

Step 3 is the one people skip. “The bottom side goes with the bottom side” is not a rule; it holds only when both triangles happen to be drawn the same way up.

Setting up and solving proportions

Once the correspondence is fixed, a missing side is a one-line proportion. Put corresponding sides across from each other and keep the two triangles on consistent sides of the fraction bar.

Cross-multiplication is just multiplying both sides by both denominators at once. Starting from 3/6 = 4/x, multiply both sides by 6x: the left becomes 3x, the right becomes 24.

An equally valid set-up compares two sides within the same triangle:

Both layouts are correct. Mixing them halfway through is not, and it produces an inverted ratio. Pick a layout, write both fractions in it, then cross-multiply.

The scale factor k, and what it does to perimeter and area

The scale factor k is the single number that converts every length in the first triangle into the matching length in the second:

Perimeter is a sum of lengths, so multiplying each length by k multiplies the sum by k. Area is a length times a length, so it picks up k twice.

Worked demonstration: 3-4-5 against 6-8-10

The perimeter went from 12 to 24, a factor of exactly k = 2. The area went from 6 to 24, a factor of exactly 4 = 2². Doubling the sides quadrupled the area, which is the most commonly misremembered fact in this topic. If a scaled copy has nine times the area, its sides are three times as long, not nine times.

Check it once more with k = 5. Scaling 3-4-5 by 5 gives 15-20-25, perimeter 15 + 20 + 25 = 60 = 12 à - 5, and area ½ · 15 · 20 = 150 = 6 à - 5².

A table of similar right triangle pairs

Every row below is a genuinely similar pair. The acute angles are identical down each row and differ between rows, which is why triangles from different rows are not similar to each other.

A 3-4-5 and a 5-12-13 are both right triangles with whole-number sides, yet 36.87° ≠ 22.62°, so they are not similar and no proportion between them is valid. Confirm any row by entering the two leg lengths into the right triangle calculator and comparing the angles it returns.

Worked examples

Check: 12² + 9² = 225 = 15², so 9-12-15 is 3-4-5 with every side tripled.

Verification: 12.5² + 30² = 156.25 + 900 = 1056.25 = 32.5². The acute angles remain 22.62° and 67.38°, exactly as in the original 5-12-13 triangle.

The sun’s rays arrive at the same angle for both objects, so the angle of elevation is shared, and both objects stand at a right angle to the ground. One matching acute angle is all AA needs. Sanity check: the tree’s shadow is 18 / 2.4 = 7.5 times the person’s, and 1.7 Ã - 7.5 = 12.75.

Light reflects at an equal angle to the mirror, so the angle at the observer’s feet equals the angle at the pole’s base, and both triangles stand at a right angle to the same flat ground. Check: 4.5 / 1.2 = 3.75, and 1.6 Ã - 3.75 = 6.

These triangles are close in shape, which is why eyeballing a diagram fails. A gap of 1.34° is invisible at textbook scale but fatal to any proportion you write.

Figure 3

The rotation is a trap for the eye, not for the method. Angle S equals angle P and both triangles have a right angle, so AA gives △PQR ~ △STU, and the correspondence P↔S, Q↔T, R↔U comes from the letters, not the picture. Verification: 24² + 45² = 576 + 2025 = 2601 = 51², and k = 51/17 = 3 matches 24/8 and 45/15.

Direct recomputation agrees: 6 + 8 + 10 = 24 and ½ · 6 · 8 = 24. Both landing on 24 is a coincidence of these numbers; perimeter and area scale by different powers of k.

Check: ½ · 15 · 20 = 150 square metres, and 15² + 20² = 225 + 400 = 625 = 25². Take the square root of the area ratio to get the length ratio. Using the area ratio itself would give k = 25 and sides of 75-100-125, an area 25 times too large.

The altitude case: one right triangle, three similar triangles

Drop a perpendicular from the right-angle vertex onto the hypotenuse and you create two smaller right triangles, both similar to each other and to the original. That is three similar triangles in one figure.

Figure 4

Name the parts precisely. In △ABC the right angle is at C, vertex A sits at the bottom-right of Figure 4 and vertex B at the top-left, with leg AC = 12, leg BC = 9 and hypotenuse AB = 15. Let D be the foot of the altitude from C, so CD ⊥ AB, with AD = 9.6 and DB = 5.4.

The correspondence, written out carefully

Read it one triangle at a time:

• â–³ADC ~ â–³ACB. Angle A is shared. Angle ADC = 90° corresponds to angle ACB = 90°, so angle ACD corresponds to angle ABC. Correspondence: A↔A, D↔C, C↔B.
• â–³CDB ~ â–³ACB. Angle B is shared. Angle CDB = 90° corresponds to angle ACB = 90°, and angle DCB = 90° − B = angle A. Correspondence: C↔A, D↔C, B↔B.
• â–³ADC ~ â–³CDB follows, since both are similar to the same triangle.

Numerically, for △ADC ~ △ACB: AD/AC = 9.6/12 = 0.8, DC/CB = 7.2/9 = 0.8 and AC/AB = 12/15 = 0.8, so k = 0.8 throughout. For △CDB ~ △ACB: CD/AC = 7.2/12 = 0.6, DB/CB = 5.4/9 = 0.6 and CB/AB = 9/15 = 0.6, so k = 0.6. Two scale factors, one shared shape: all three triangles have acute angles of 36.87° and 53.13°.

Independent check: DB = 15 − 9.6 = 5.4, and 5.4² + 7.2² = 29.16 + 51.84 = 81 = 9², so △CDB closes correctly. The proportion AD/AC = AC/AB is a geometric mean relationship in disguise, proved in the article on the right triangle altitude theorem; for drills on those relationships, work through the geometric mean problems. This article stops at the similarity that makes them possible.

Common correspondence mistakes

Matching sides by position on the page

The most frequent error is pairing “the side along the bottom” with “the side along the bottom”. Position on the page is not a geometric property: rotate one triangle 90°, as in Problem 6, and every screen position changes while every angle stays put. Match each side to the side that faces the equal angle. Hypotenuse always pairs with hypotenuse, the one pairing that position never breaks.

Inverting the ratio

Having set up 3/9 = 4/x, it is easy to write the second fraction the other way round as 3/9 = x/4. Cross-multiplying gives 9x = 12 and x = 1.333, a larger triangle with a shorter side. Guard against it with a size check before the algebra: in Problem 1 the second triangle has 9 where the first had 3, so k = 3 and every answer must be three times its partner. An answer smaller than its partner means the fraction was flipped.

Assuming equal perimeters imply similarity

They do not. A 5-12-13 triangle has perimeter 30. A right triangle with legs 6 and 11.25 has hypotenuse √(36 + 126.5625) = √162.5625 = 12.75 and perimeter 6 + 11.25 + 12.75 = 30, the same total. But the acute angles are 22.62° and 67.38° for the first, 28.07° and 61.93° for the second, and the areas differ (30 against 33.75). Equal perimeters, equal areas and equal longest sides are all consistent with different shapes. Only equal angles, or a common ratio across corresponding sides, establishes similarity.

Assuming all right triangles are similar

A shared right angle is one matching pair of angles, and AA needs two. Compare the two most familiar right triangles:

36.87° ≠ 45° and 53.13° ≠ 45°, so these two triangles are not similar and never can be. No scale factor turns a 3-4-5 into an isosceles right triangle, because scaling cannot move an angle by 8.13°.

Figure 5

Figure 5 is similar to every other isosceles right triangle, whatever its size, and every 30-60-90 is similar to every other 30-60-90. Both families are set out in the guide to special right triangles. Across families, similarity fails.

Practice

Work these on paper, then check the side lengths and angles with the triangle solver on the homepage.

1. â–³ABC ~ â–³XYZ with right angles at C and Z. AC = 6 and BC = 8, and XZ = 15. Find YZ and XY.
2. A child 1.5 m tall casts a shadow 2 m long while a nearby building casts a shadow 46 m long. How tall is the building?
3. Is a right triangle with legs 7 and 24 similar to a right triangle with legs 14 and 48? Justify with a ratio and an angle.
4. A 5-12-13 triangle is scaled by k = 1.4. Find the new perimeter and the new area.
5. A right triangle has legs 5 and 12 and hypotenuse 13. Use similarity to find the two segments the altitude from the right angle cuts on the hypotenuse.

Answer key

1. YZ = 20, XY = 25. k = 15/6 = 2.5, so YZ = 8 à - 2.5 = 20. The first hypotenuse is √(6² + 8²) = 10, so XY = 25. Check: 15² + 20² = 625 = 25².
2. 34.5 m. h/46 = 1.5/2, so 2h = 69 and h = 34.5. The shadow is 23 times longer, and 1.5 Ã - 23 = 34.5.
3. Yes. 7/24 = 14/48 = 0.29167, and arctan(7/24) = arctan(14/48) = 16.26°. The hypotenuses are 25 and 50, so k = 2.
4. Perimeter 42, area 58.8. The original perimeter is 30 and the original area is ½ · 5 · 12 = 30, so P = 30 à - 1.4 = 42 and A = 30 à - 1.96 = 58.8. Check on the scaled sides 7, 16.8, 18.2: 7 + 16.8 + 18.2 = 42 and ½ · 7 · 16.8 = 58.8.
5. 25/13 ≈ 1.92 and 144/13 ≈ 11.08. Similarity gives (short segment)/5 = 5/13, so that segment is 25/13; the one next to the leg of 12 is 144/13. Check: 25/13 + 144/13 = 169/13 = 13. The altitude is 60/13 ≈ 4.62.

Frequently Asked Questions

Are all right triangles similar to each other?

No. Sharing a right angle gives only one matching pair, and AA similarity requires two. A 3-4-5 triangle has acute angles of 36.87° and 53.13°, while a 45-45-90 has two angles of 45°, so those two are not similar. Right triangles are similar only when an acute angle matches as well.

Is one pair of equal acute angles really enough?

Yes, for right triangles. The right angles supply the first matching pair for free, the acute angles supply the second, and the third follows because the angles sum to 180°. That satisfies AA completely, so no side measurement is needed.

Does the order of the letters in a similarity statement matter?

Yes. â–³ABC ~ â–³DEF declares that A corresponds to D, B to E and C to F, and every proportion is read from that order. For the triangles above, â–³ABC ~ â–³DEF is true while â–³ABC ~ â–³EDF is false, because the second pairs a 36.87° angle with a 53.13° angle.

If the sides double, does the area double?

No, the area quadruples. Lengths and perimeter scale by k, area scales by k². Doubling a 3-4-5 to 6-8-10 takes the perimeter from 12 to 24 but the area from 6 to 24. Working backwards, an area ratio of 9 means a length ratio of √9 = 3.

Are the two small triangles formed by the altitude similar to each other?

Yes, and to the original as well. With the right angle at C and the altitude foot at D on the hypotenuse, â–³ADC ~ â–³ACB ~ â–³CDB. All three share the same pair of acute angles; only the scale factors differ. In the 9-12-15 example the small triangles sit at k = 0.8 and k = 0.6.