Unit 6: Similar Triangles Homework 1 – Ratio & Proportion Answers
Introduction
In geometry, similar triangles are a cornerstone concept that links shape, size, and proportion. Unit 6 focuses on the relationship between corresponding sides and angles, and how these relationships translate into practical ratio problems. This article presents a comprehensive set of answers for Homework 1, along with step‑by‑step explanations, key formulas, and common pitfalls to avoid. By the end, you’ll not only have the correct answers but also a deeper understanding of why those answers work Small thing, real impact..
1. Recap of Core Concepts
1.1 Definition of Similar Triangles
Two triangles are similar if their corresponding angles are equal and their corresponding sides are in proportion. Symbolically:
[ \triangle ABC \sim \triangle DEF \quad \Longleftrightarrow \quad \begin{cases} \angle A = \angle D \ \angle B = \angle E \ \angle C = \angle F \ \frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{FD} \end{cases} ]
1.2 Ratio & Proportion Basics
A ratio compares two quantities, while a proportion states that two ratios are equal. In the context of similar triangles, the ratio of any two sides in one triangle equals the ratio of the corresponding sides in the other triangle That alone is useful..
2. Homework 1 – Problem Set and Detailed Solutions
Problem 1
Given:
- Triangle ( \triangle ABC ) has sides ( AB = 8 ) cm, ( BC = 12 ) cm, ( AC = 16 ) cm.
- Triangle ( \triangle DEF ) is similar to ( \triangle ABC ).
- Side ( DE = 4 ) cm.
Find: Lengths of ( EF ) and ( FD ).
Solution
- Identify the scale factor ( k ) between the triangles.
[ k = \frac{DE}{AB} = \frac{4}{8} = 0.5 ] - Apply the scale factor to the other sides of ( \triangle ABC ): [ EF = k \times BC = 0.5 \times 12 = 6 \text{ cm} ] [ FD = k \times AC = 0.5 \times 16 = 8 \text{ cm} ] Answer: ( EF = 6 ) cm, ( FD = 8 ) cm.
Problem 2
Given:
- ( \triangle XYZ ) and ( \triangle PQR ) are similar.
- ( XY = 9 ) cm, ( YZ = 15 ) cm, ( XZ = 18 ) cm.
- ( PQ = 12 ) cm.
Find: The missing side ( QR ).
Solution
- Determine the scale factor using the known sides: [ k = \frac{PQ}{XY} = \frac{12}{9} = \frac{4}{3} ]
- Apply the factor to the side ( YZ ): [ QR = k \times YZ = \frac{4}{3} \times 15 = 20 \text{ cm} ] Answer: ( QR = 20 ) cm.
Problem 3
Given:
- Two similar triangles with a ratio of corresponding sides ( \frac{1}{3} ).
- One triangle has a perimeter of 30 cm.
Find: The perimeter of the other triangle.
Solution
Perimeter scales with the same factor as side lengths.
[
\text{Perimeter}_{\text{larger}} = \frac{1}{\frac{1}{3}} \times 30 = 3 \times 30 = 90 \text{ cm}
]
Answer: 90 cm.
Problem 4
Given:
- ( \triangle ABC ) is similar to ( \triangle DEF ).
- ( AB = 7 ) cm, ( BC = 14 ) cm, ( AC = 21 ) cm.
- ( DE = 5 ) cm.
Find: The ratio of similarity and the lengths of ( EF ) and ( FD ).
Solution
- Scale factor: [ k = \frac{DE}{AB} = \frac{5}{7} ]
- Compute other sides: [ EF = k \times BC = \frac{5}{7} \times 14 = 10 \text{ cm} ] [ FD = k \times AC = \frac{5}{7} \times 21 = 15 \text{ cm} ] Answer: Ratio ( \frac{5}{7} ); ( EF = 10 ) cm, ( FD = 15 ) cm.
Problem 5
Given:
- In ( \triangle PQR ), ( PQ = 6 ) cm, ( PR = 10 ) cm.
- ( \triangle XYZ ) is similar to ( \triangle PQR ) with ( XY = 9 ) cm.
Find: Length of ( YZ ) and ( XZ ).
Solution
- Scale factor: [ k = \frac{XY}{PQ} = \frac{9}{6} = 1.5 ]
- Apply to other sides: [ YZ = k \times QR = 1.5 \times QR ] Even so, we need ( QR ). Since we only have ( PR ), we can use the proportionality of sides: [ \frac{PR}{PQ} = \frac{QR}{QR} \quad \text{(not helpful directly)} ] Instead, assume the triangles are right triangles and use the Pythagorean theorem: [ QR = \sqrt{PR^2 - PQ^2} = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = \sqrt{64} = 8 \text{ cm} ] Now: [ YZ = 1.5 \times 8 = 12 \text{ cm} ] [ XZ = k \times PR = 1.5 \times 10 = 15 \text{ cm} ] Answer: ( YZ = 12 ) cm, ( XZ = 15 ) cm.
3. Common Mistakes & How to Avoid Them
| Mistake | Why It Happens | Fix |
|---|---|---|
| Using the wrong side for the scale factor | Confusing corresponding sides | Double‑check angle labels; match exact sides. Now, |
| Assuming similarity without angle verification | Relying solely on side ratios | Verify at least two angles are equal before applying ratios. |
| Forgetting to simplify fractions | Leaving unsimplified ratios | Reduce fractions to lowest terms to avoid calculation errors. |
| Mixing units | Mixing cm with in or mm | Keep all measurements in the same unit. |
| Misapplying the Pythagorean theorem | Using it in non‑right triangles | Only apply when a right angle is confirmed. |
4. FAQ
Q1: How do I confirm that two triangles are similar if only side lengths are given?
A1: If all three pairs of corresponding sides are in proportion, the triangles are similar by the Side‑Side‑Side (SSS) similarity criterion.
Q2: Can the ratio of similarity be greater than 1?
A2: Yes. A ratio greater than 1 indicates the second triangle is larger than the first.
Q3: What if one side length is missing?
A3: Use the ratio of the known sides to find the missing side:
[
\frac{a}{b} = \frac{c}{d} \quad \Rightarrow \quad d = \frac{b \times c}{a}
]
Q4: Are similar triangles always congruent?
A4: No. Congruent triangles are a special case of similarity where the ratio is exactly 1.
5. Real‑World Applications
- Engineering: Scaling models of bridges or aircraft.
- Architecture: Designing proportional facades.
- Computer Graphics: Resizing images without distortion.
- Navigation: Calculating distances using similar triangles in triangulation.
6. Conclusion
Mastering the ratio and proportion relationships in similar triangles equips you with a versatile tool for solving a wide array of geometry problems. By systematically identifying the scale factor, verifying angle congruence, and applying proportional reasoning, you can confidently tackle any homework question that involves similar triangles. Keep practicing with varied examples, and soon the process will become intuitive—making geometry both reliable and enjoyable Surprisingly effective..
7. Practice Problem Set
| # | Problem | Key Insight |
|---|---|---|
| 1 | Two triangles share an angle. Think about it: if (AB = 6) cm, (BC = 8) cm, and the corresponding side in the other triangle is (DE = 9) cm, find the missing side (EF). On top of that, | Use the ratio ( \frac{AB}{DE} = \frac{BC}{EF}). |
| 2 | Triangle (PQR) is similar to triangle (STU). Day to day, given (PQ = 5) cm, (QR = 7) cm, (ST = 10) cm, find (TU). | Scale factor (k = \frac{ST}{PQ} = 2). Then (TU = k \cdot QR). |
| 3 | A right triangle has legs 9 cm and 12 cm. On the flip side, its hypotenuse is in a triangle that is twice as large. What is the hypotenuse of the larger triangle? Even so, | Scale factor 2 → ( \sqrt{9^2+12^2} \times 2 = 15 \times 2 = 30) cm. Worth adding: |
| 4 | Two triangles are similar with a ratio of 5 : 3. If the perimeter of the smaller triangle is 42 cm, what is the perimeter of the larger? | Perimeter scales by the same ratio: (42 \times \frac{5}{3} = 70) cm. That said, |
| 5 | A triangle has sides 4 cm, 7 cm, and 9 cm. Now, a similar triangle has a side of 12 cm that corresponds to the 4 cm side. Find the other two sides. Practically speaking, | Scale factor (k = \frac{12}{4} = 3). Multiply 7 cm and 9 cm by 3 → 21 cm and 27 cm. |
Tip: When solving such problems, write down the ratio first. It becomes the backbone of the calculation, ensuring you don’t mix up corresponding sides Not complicated — just consistent..
8. Quick Reference Cheat Sheet
| Concept | Formula | Example |
|---|---|---|
| Scale factor | (k = \frac{\text{corresponding side of larger}}{\text{corresponding side of smaller}}) | (k = \frac{15}{10} = 1.Worth adding: 5 \times 10 = 15) cm |
| SSS similarity | (\frac{a}{a'} = \frac{b}{b'} = \frac{c}{c'}) | If all three ratios equal, triangles are similar |
| Perimeter scaling | (P_{\text{large}} = k \times P_{\text{small}}) | (P_{\text{large}} = 1. 5) |
| Side calculation | ( \text{Missing side} = k \times \text{known side}) | (XZ = 1.5 \times 30 = 45) cm |
| Area scaling | (A_{\text{large}} = k^2 \times A_{\text{small}}) | (A_{\text{large}} = 1. |
9. Final Thoughts
You’ve now seen how the humble ratio is the secret sauce behind all similarity problems. Whether you’re measuring a miniature model, designing a building, or simply solving a textbook question, the steps are the same:
- Confirm similarity (angles or side ratios).
- Determine the scale factor from any pair of corresponding sides.
- Apply the factor to compute missing sides, perimeters, or areas.
With practice, you’ll find that each new problem feels like a puzzle you’ve already solved before—just with different numbers. Keep experimenting with different configurations, and soon the patterns will become second nature. Happy geometry!
