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MathematicsGeometryNovember 7, 2026

Similar Triangles Without Guesswork: Missing Lengths and Area Scale Factors

Learn to match corresponding sides, solve missing-length problems, and use squared scale factors for area, with two worked examples, a labeled diagram, and practice.

A geometry diagram can make two triangles look alike without proving that they are similar. A reliable solution starts with a reason for similarity, a careful match between vertices, and a scale factor whose direction stays consistent. Once those three pieces are in place, finding a missing length becomes an algebra problem you can check.

Doubling every side does not double a figure's area. Two worked problems connect these ideas, followed by a practice challenge.

First establish which measurements correspond

When we write triangle ADE ∼ triangle ABC, the order matters: A corresponds to A, D to B, and E to C. Therefore AD matches AB, AE matches AC, and DE matches BC. A small-to-large ratio must use that order in every fraction:

AD/AB = AE/AC = DE/BC.

Similarity requires evidence, such as two pairs of equal angles, three proportional side pairs, or two proportional side pairs with equal included angles. Appearance alone is not evidence. For each problem, state why similarity applies before setting up a proportion.

Worked problem 1: a parallel segment inside a triangle

In triangle ABC, D lies on side AB and E lies on side AC. Segment DE is parallel to BC. The given lengths are AD = 6 cm, AB = 10 cm, AC = 15 cm, and BC = 20 cm. Find AE, EC, and DE.

Parallel segments create two similar triangles Triangle ABC has A at the top, B at the lower left, and C at the lower right. D is on AB and E is on AC. The horizontal segment DE is parallel to the horizontal base BC. AD is 6 centimeters, AB is 10 centimeters, AC is 15 centimeters, and BC is 20 centimeters. The diagram is not to scale. ABCDE AD = 6 cmAB = 10 cmAC = 15 cmBC = 20 cm
DE ∥ BC. Match the smaller triangle ADE to the whole triangle ABC. Diagram not to scale; use the stated lengths.

1. Explain the similarity

The triangles share the angle at A. Since DE ∥ BC, corresponding angles ADE and ABC are equal. Two equal angle pairs establish AA similarity, so triangle ADE ∼ triangle ABC.

2. Use one consistent scale factor

The smaller triangle's lengths are 6/10 = 3/5 of the corresponding whole-triangle lengths. Multiply the larger lengths by 3/5:

  • AE = (3/5)(15) = 9 cm.
  • DE = (3/5)(20) = 12 cm.

The question also asks for EC, which is only the remaining part of AC. Since AC = AE + EC, EC = 15 − 9 = 6 cm. Using 6/10 = AE/EC would incorrectly compare a smaller whole side with only a leftover segment.

3. Check the result

The three corresponding ratios agree: 6/10 = 9/15 = 12/20 = 3/5. The lengths also fit their segments: AE + EC = 9 + 6 = 15. These checks catch a reversed ratio or a whole-versus-part mistake.

Why area uses the square of the scale factor

If every length is multiplied by a positive factor k, a triangle's base and perpendicular height are both multiplied by k. Its new area is (1/2)(kb)(kh) = k2(1/2)bh. The two length factors produce an area factor of k2.

The same rule holds for similar polygons, which divide into corresponding triangles. Perimeter scales by k because it adds lengths; area scales by k2 because it measures two-dimensional size.

Worked problem 2: recover a length from an area ratio

Two similar quadrilaterals have areas of 50 cm2 and 98 cm2. A side of the smaller quadrilateral measures 10 cm. Find its corresponding side in the larger quadrilateral.

Define k as the larger-to-smaller length factor. Then k2 = 98/50 = 49/25. Lengths are positive, so k = 7/5. The required side is (7/5)(10) = 14 cm.

Check by squaring the length ratio: (14/10)2 = 1.96 = 98/50. Multiplying 10 by 98/50 would give 19.6 cm, incorrectly using an area factor as a length factor.

Try a problem before opening the answer

In triangle ABC, D lies on AB, E lies on AC, and DE ∥ BC. Given AD = 8 cm, DB = 4 cm, AC = 15 cm, BC = 9 cm, and area ABC = 54 cm2, find AE, EC, DE, and area ADE.

Show the worked answer

First find the whole side: AB = 8 + 4 = 12 cm. AA similarity gives a smaller-to-larger factor of 8/12 = 2/3. Therefore AE = 10 cm, EC = 15 − 10 = 5 cm, and DE = 6 cm. The area factor is (2/3)2 = 4/9, so area ADE = (4/9)(54) = 24 cm2. The area uses the squared factor; each length uses the unsquared factor.

A short routine for your next geometry assignment

  1. Identify the similarity theorem and write the vertices in matching order.
  2. Decide whether your ratio runs small-to-large or large-to-small.
  3. Separate whole sides from leftover segments.
  4. Use k for lengths and perimeters, and k2 for areas.
  5. Check corresponding ratios, units, and whether the answer is a plausible size.

Build this habit with Geometry Honors MCQ practice. For each answer, explain the matching sides before doing the arithmetic.