Similar Triangles Proportions Worksheet Answer Key – Geometry

Problem 1 Find a missing side

The scale factor is DE ÷ AB = 15 ÷ 6 = 2.5. Then EF = 2.5 × 8 = 20.

k = 2.5; EF = 20

Problem 2 AA similarity check

The first triangle’s third angle is 180° − 38° − 74° = 68°, so its angles are 38°, 74°, 68°. The second triangle’s third angle is 180° − 74° − 68° = 38°, so its angles are the same. Two angle pairs match, so the triangles are similar by AA.

Yes, similar by AA

Problem 3 SAS similarity

Compare the ratios: 6 ÷ 8 = 0.75 and 9 ÷ 12 = 0.75. The included angles are both 50°. By SAS similarity the triangles are similar.

Yes, similar by SAS ∼

Problem 4 SSS similarity

Divide each side of triangle 2 by the matching side of triangle 1: 10.5 ÷ 7 = 1.5, 13.5 ÷ 9 = 1.5 and 18 ÷ 12 = 1.5. The ratios are equal, so the triangles are similar by SSS ∼ with scale factor 1.5.

Similar; k = 1.5

Problem 5 Use the side-splitter theorem

By the Side-Splitter Theorem, AD/DB = AE/EC: 4/6 = 5/EC. Cross-multiplying, 4EC = 30, so EC = 7.5.

EC = 7.5

Problem 6 Side-splitter with algebra

2x/12 = (x + 3)/9. Cross-multiplying: 18x = 12x + 36, so 6x = 36 and x = 6.

AD = 12 and AE = 9. Check: 12/12 = 9/9 = 1.

x = 6; AD = 12; AE = 9

Problem 7 Is the line parallel

1. 6/9 = 2/3 and 8/12 = 2/3. The sides are divided proportionally, so DE ∥ BC.
2. 5/7 ≈ 0.714 and 6/8 = 0.75. The ratios differ, so DE is not parallel to BC.

(a) parallel, (b) not parallel

Problem 8 Length of the parallel segment

△ADE ∼ △ABC, and AB = 6 + 9 = 15. So DE/BC = AD/AB = 6/15, giving DE = 25 · 6/15 = 10.

DE = 10

Problem 9 Altitude to the hypotenuse

The altitude is the geometric mean of the two pieces: CD² = 4 · 9 = 36, so CD = 6.

CD = 6

Problem 10 Hypotenuse split into 9 and 16

The hypotenuse is 9 + 16 = 25. Altitude: h = √(9 · 16) = 12.

Legs: a = √(9 · 25) = 15 and b = √(16 · 25) = 20. Check: 15² + 20² = 225 + 400 = 625 = 25².

Altitude 12; legs 15 and 20

Problem 11 Geometric mean

The geometric mean of a and b is √(ab).

1. √(6 · 24) = √144 = 12.
2. √(5 · 20) = √100 = 10.
3. √(8 · 12) = √96 ≈ 9.80.

(a) 12, (b) 10, (c) ≈ 9.80

Problem 12 Decimal pieces of the hypotenuse

Altitude: √(3.6 · 6.4) = √23.04 = 4.8. Legs: √(3.6 · 10) = 6 and √(6.4 · 10) = 8.

Check: 6² + 8² = 100 = 10².

Altitude 4.8; legs 6 and 8

Problem 13 Pythagorean theorem by similarity

Add the equations: a² + b² = pc + qc = c(p + q). The pieces make up the hypotenuse, so p + q = c. Therefore a² + b² = c · c = c².

Check: 15² = 225 = 9 · 25 and 20² = 400 = 16 · 25, and 225 + 400 = 625 = 25².

a² + b² = c²

Problem 14 Mirror measurement

The light ray makes equal angles with the ground at the mirror, and both triangles have right angles at the ground, so the two right triangles are similar by AA.

Corresponding sides: height ÷ 18 = 1.6 ÷ 2.4. So height = 18 × 1.6 ÷ 2.4 = 12.

The building is 12 meters tall

Problem 15 River width

△TAB and △CDB are right triangles that share ∠B, so they are similar by AA. Corresponding sides: TA/CD = AB/DB.

TA = 9 · 40 ÷ 12 = 30.

The river is 30 meters wide

Problem 16 Scale model of a lot

The scale factor is 8 ÷ 120 = 1/15 inch per foot. The other sides: 150 ÷ 15 = 10 inches and 200 ÷ 15 ≈ 13.3 inches.

Perimeter = 8 + 10 + 13.3 = 31.3 inches (or 470 ÷ 15 ≈ 31.3).

Sides 10 in and ≈ 13.3 in; perimeter ≈ 31.3 in

Problem 17 Three parallel streets

Parallel lines cut two transversals proportionally: 120/180 = 95/x. So x = 95 · 180 ÷ 120 = 142.5.

142.5 meters

Problem 18 Altitude in a 12-16-20 triangle

The area is ½ · 12 · 16 = 96, and also ½ · 20 · h, so h = 192 ÷ 20 = 9.6.

Pieces: 12² ÷ 20 = 7.2 and 16² ÷ 20 = 12.8, which add to 20.

Check: 7.2 · 12.8 = 92.16 = 9.6².

h = 9.6; pieces 7.2 and 12.8

Problem 19 Prove similarity from a parallel line

△ADE ∼ △ABC

Problem 20 Prove AC² = AD · AB

AC² = AD · AB