← Back to the Dilations Scale Factor Similarity worksheet
Last updated: October 22, 2026 · Written by the Math Made Easy math teachers
Standards: G-SRT.A.1, G-SRT.A.2, G-SRT.A.3 · Level: Grade 10 Geometry
Problem 2 Shrink a point
Multiply each coordinate by ½: (4, −3).
B′(4, −3)
Problem 3 Find the scale factor
Divide each image coordinate by the original: 10 ÷ 4 = 2.5 and 15 ÷ 6 = 2.5.
k = 2.5
Problem 4 Length under a dilation
Lengths are multiplied by the scale factor: (a) 7 × 4 = 28 and (b) 7 × ½ = 3.5.
(a) 28 cm, (b) 3.5 cm
Problem 5 Distances from the center
CP′ = 2.5 × 6 = 15. For the second dilation, k = CP″ ÷ CP = 9 ÷ 6 = 1.5.
CP′ = 15; k = 1.5
Problem 6 Dilate a triangle
Multiply each coordinate by 2: A′(2, 2), B′(8, 2), C′(8, 6).
Side lengths: AB = 3, BC = 2, AC = √(9 + 4) = √13. The image sides are 6, 4 and 2√13, each twice as long, so the perimeter doubles from 5 + √13 to 10 + 2√13.
A′(2, 2), B′(8, 2), C′(8, 6); perimeter × 2
Problem 7 Dilation about another center
The vector from C to P is (5 − 2, 4 − 1) = (3, 3). Doubling gives (6, 6). Add to the center: (2 + 6, 1 + 6) = (8, 7).
Check: C, P and P′ lie on the line y = x − 1.
P′(8, 7)
Problem 8 A line not through the center
(0, 3) → (0, 9) and (1, 5) → (3, 15). The slope of the image line is (15 − 9) ÷ (3 − 0) = 2, and its y-intercept is 9, so it is y = 2x + 9.
The slope is unchanged, so the image line is parallel to the original.
y = 2x + 9, parallel to y = 2x + 3
Problem 9 A line through the center
(1, 3) → (4, 12). Check the equation: 3(4) = 12, so (4, 12) lies on y = 3x. A line through the center maps onto itself.
(4, 12) is on y = 3x
Problem 10 Perimeter and area
The perimeter is multiplied by 4: 20 × 4 = 80. The area is multiplied by 42 = 16: 15 × 16 = 240.
Perimeter 80; area 240
Problem 11 Write the similarity statement
The third angles are m∠C = 180° − 52° − 61° = 67° and m∠F = 180° − 52° − 67° = 61°. Both triangles have angles 52°, 61° and 67°, so they are similar by AA.
Matching: A ↔ D (52°), B ↔ F (61°), C ↔ E (67°).
Yes: △ABC ∼ △DFE
Problem 12 Proportional sides
The scale factor is DE ÷ AB = 10 ÷ 6 = 5/3. Then EF = (5/3)(9) = 15.
k = 5/3; EF = 15
Problem 13 Is it a dilation
(0, 0) → (1, 0), but a dilation about the origin keeps the origin fixed. So it is not a dilation about the origin.
The rule doubles every coordinate and then shifts right by 1, so it is a dilation about the origin with k = 2 followed by a translation by ⟨1, 0⟩. This is a similarity transformation.
Not a dilation about the origin; a dilation (k = 2) followed by a translation
Problem 14 A similarity sequence
The length AB = 2 and A′B′ = 4, so k = 2. Dilate about the origin: (0, 0), (4, 0), (0, 6).
Then translate by ⟨4, 1⟩: (4, 1), (8, 1), (4, 7). These match A′, B′, C′. Check BC: √13 and B′C′ = √(16 + 36) = 2√13.
Dilation with k = 2 about the origin, then translation by ⟨4, 1⟩
Problem 15 Scale model of a statue
The statue is 450 centimeters tall, so k = 18 ÷ 450 = 0.04 (or 1/25).
Model base = 120 × 0.04 = 4.8 centimeters. Areas scale by k2 = (1/25)2 = 1/625, so the model’s surface area is 625 times smaller.
k = 0.04; base 4.8 cm; area 1/625 of the statue’s
Problem 16 Projector enlargement
The screen image is 3.5 × 18 = 63 centimeters by 2.4 × 18 = 43.2 centimeters. Its area is 63 × 43.2 = 2,721.6 square centimeters.
The slide’s area is 3.5 × 2.4 = 8.4, so the area grew by 2,721.6 ÷ 8.4 = 324 = 182.
63 cm by 43.2 cm; area 2,721.6 cm²; area × 324
Problem 17 Resizing a photo
The scale factor is 5 ÷ 8 = 5/8. The new height is 10 × 5/8 = 6.25 inches.
The original perimeter is 2(8 + 10) = 36 inches, so the new perimeter is 36 × 5/8 = 22.5 inches. Check: 2(5 + 6.25) = 22.5.
k = 5/8; height 6.25 in; perimeter 22.5 in
Problem 18 Dilation with a center off the origin
A: vector from the center is (1, 2); tripled it is (3, 6); A′ = (1 + 3, −2 + 6) = (4, 4).
B: vector (3, 3); tripled (9, 9); B′ = (1 + 9, −2 + 9) = (10, 7).
AB = √(2² + 1²) = √5 and A′B′ = √(6² + 3²) = √45 = 3√5, which is 3 times AB.
A′(4, 4), B′(10, 7); A′B′ = 3 · AB
Problem 19 Prove the AA similarity
| Statement | Reason |
|---|---|
| 1. ∠A ≅ ∠D and ∠B ≅ ∠E | Given |
| 2. Let k = DE ÷ AB. Dilate △ABC about A by k to get △AB′C′ | Definition of dilation |
| 3. AB′ = k · AB = DE | Dilation multiplies lengths by k |
| 4. ∠A ≅ ∠A, ∠B′ ≅ ∠B ≅ ∠E | Dilations preserve angle measure; Given |
| 5. △AB′C′ ≅ △DEF | ASA (∠A ≅ ∠D, AB′ = DE, ∠B′ ≅ ∠E) |
| 6. A rigid motion maps △AB′C′ onto △DEF | Congruent triangles are related by rigid motions |
| 7. △ABC ∼ △DEF | A dilation followed by a rigid motion maps one onto the other |
△ABC ∼ △DEF
Problem 20 Shadow measurement
The sun’s rays are parallel, so each person or tree and its shadow form a right triangle with the same angle at the ground. Each also has a right angle at the base. Two pairs of congruent angles make the triangles similar by AA.
Scale factor from the person’s triangle to the tree’s triangle: 12 ÷ 2.55. The tree is taller by the same factor, or by ratio: height ÷ 12 = 1.7 ÷ 2.55 = 2/3, so height = (2/3)(12) = 8.
The tree is 8 meters tall