Triangles · Answer guide

Lesson 1 · Match corresponding parts

Use AA, SAS and SSS similarity with correct correspondence.

1. Two triangles each have angles 40° and 60°. What can we conclude?

   Answer: They are similar by AA. Both third angles are 80°, so the corresponding angles match. Their sizes can still differ.

2. A triangle has angles 50° and 70°. Find the third angle.

   Answer: 60°. The third angle is 180° − 50° − 70° = 60°.

3. Triangles have side lengths 4, 6, 8 and 6, 9, 12. Which criterion establishes their similarity?

   Answer: SSS. The corresponding larger/smaller ratios are 6/4 = 9/6 = 12/8 = 3/2. Three proportional side pairs establish SSS similarity.

Discuss: Draw a triangle and a copy enlarged by factor 3. Label matching vertices and write all three corresponding side ratios.

Teaching note: Two sides in proportion alone are not enough for SAS; the angle between those sides must match. Open responses need discussion; they are not automatically marked.

Lesson 2 · Use a parallel line inside a triangle

Apply the basic proportionality theorem and its converse.

1. In △ABC, D is on AB, E is on AC and DE ∥ BC. If AD = 3, DB = 2, AE = 6, find EC.

   Answer: 4. The basic proportionality theorem gives AD/DB = AE/EC. Thus EC = 6 × 2/3 = 4.

2. In △ABC, D is on AB, E is on AC and DE ∥ BC. If AD = 4, DB = 3, AE = 8, find EC.

   Answer: 6. The basic proportionality theorem gives AD/DB = AE/EC. Thus EC = 8 × 3/4 = 6.

3. In △ABC, D is on AB, E is on AC and DE ∥ BC. If AD = 2, DB = 5, AE = 4, find EC.

   Answer: 10. The basic proportionality theorem gives AD/DB = AE/EC. Thus EC = 4 × 5/2 = 10.

Discuss: Draw the diagram and change DB while keeping AD fixed. Predict whether EC must grow or shrink if AE stays fixed.

Teaching note: AD/AB = AE/AC is also valid in this parallel configuration, but do not mix AD/DB with AE/AC. Open responses need discussion; they are not automatically marked.

Lesson 3 · Use similarity to measure indirectly

Justify the similar triangles before setting up a scale ratio.

1. At the same time on level ground, a 2 m pole has a 3 m shadow. A tree has a 12 m shadow. Assuming parallel sunlight, find the tree height.

   Answer: 8 m. The right triangles share the sun angle, so height/shadow is constant: h/12 = 2/3. Thus h = 8 m.

2. At the same time on level ground, a 3 m pole has a 4 m shadow. A tree has a 20 m shadow. Assuming parallel sunlight, find the tree height.

   Answer: 15 m. The right triangles share the sun angle, so height/shadow is constant: h/20 = 3/4. Thus h = 15 m.

3. At the same time on level ground, a 1 m pole has a 2 m shadow. A tree has a 14 m shadow. Assuming parallel sunlight, find the tree height.

   Answer: 7 m. The right triangles share the sun angle, so height/shadow is constant: h/14 = 1/2. Thus h = 7 m.

Discuss: Use safe tabletop objects and a single lamp to test which conditions are needed before comparing shadows. Record measurements and assumptions.

Teaching note: The shadow ratio works only when corresponding angles match. Similarity should be justified, not assumed from a drawing. Open responses need discussion; they are not automatically marked.

Final check · Set 1

Try one set after the lessons. Use the other set after reviewing any errors.

1. For SAS similarity, two corresponding side pairs are proportional. What angle condition is also required?

   The included angles must be equal. SAS requires equality of the angle between the two compared sides.

2. Similar triangles have AB = 4, PQ = 6 and BC = 8, with A↔P, B↔Q, C↔R. Find QR.

   12. The scale factor is PQ/AB = 6/4 = 3/2. Hence QR = 8 × 3/2 = 12.

3. In △ABC, D is on AB, E is on AC and DE ∥ BC. If AD = 6, DB = 3, AE = 4, find EC.

   2. The basic proportionality theorem gives AD/DB = AE/EC. Thus EC = 4 × 3/6 = 2.

4. In △ABC, D is on AB, E is on AC and DE ∥ BC. If AD = 3, DB = 4, AE = 9, find EC.

   12. The basic proportionality theorem gives AD/DB = AE/EC. Thus EC = 9 × 4/3 = 12.

5. At the same time on level ground, a 4 m pole has a 3 m shadow. A tree has a 9 m shadow. Assuming parallel sunlight, find the tree height.

   12 m. The right triangles share the sun angle, so height/shadow is constant: h/9 = 4/3. Thus h = 12 m.

6. At the same time on level ground, a 3 m pole has a 2 m shadow. A tree has a 10 m shadow. Assuming parallel sunlight, find the tree height.

   15 m. The right triangles share the sun angle, so height/shadow is constant: h/10 = 3/2. Thus h = 15 m.

Final check · Set 2

Try one set after the lessons. Use the other set after reviewing any errors.

1. Two triangles have one equal angle but no further measurements. Is that alone enough for similarity?

   No. AA needs two corresponding equal angles; a single equality does not determine the triangle shape.

2. Corresponding side lengths in two similar triangles are 5 and 15. What is the larger/smaller length scale factor?

   3. The length scale factor is 15/5 = 3. A factor of 9 would describe the area ratio, not the length ratio.

3. In △ABC, D is on AB, E is on AC and DE ∥ BC. If AD = 2, DB = 3, AE = 8, find EC.

   12. The basic proportionality theorem gives AD/DB = AE/EC. Thus EC = 8 × 3/2 = 12.

4. In △ABC, D is on AB, E is on AC and DE ∥ BC. If AD = 4, DB = 5, AE = 12, find EC.

   15. The basic proportionality theorem gives AD/DB = AE/EC. Thus EC = 12 × 5/4 = 15.

5. At the same time on level ground, a 2 m pole has a 4 m shadow. A tree has a 18 m shadow. Assuming parallel sunlight, find the tree height.

   9 m. The right triangles share the sun angle, so height/shadow is constant: h/18 = 2/4. Thus h = 9 m.

6. At the same time on level ground, a 5 m pole has a 6 m shadow. A tree has a 24 m shadow. Assuming parallel sunlight, find the tree height.

   20 m. The right triangles share the sun angle, so height/shadow is constant: h/24 = 5/6. Thus h = 20 m.