Mathematics • Year 10 • Unit 3 • Lesson 12
Similar Triangles — Mixed Challenge
Pull every Lesson 12 idea together: prove similarity using the right test (AAA, SSS or SAS), find unknown sides, solve shadow / sighting problems, spot the classic SSA-style mistake, and design your own similar-triangle setup that lets you measure something unreachable.
1. Mixed problems — choose the right test
Each problem pulls on a different idea from Lesson 12. Before you start, decide which test (AAA, SSS, SAS) applies. 2-3 marks each
1.1 △ABC has angles 50°, 60°, 70°. △XYZ has angles 50°, 60°, 70°, and AB = 6 cm, XY = 18 cm, BC = 9 cm. State the similarity test and find YZ. 3 marks
1.2 Two triangles have sides (4, 6, 8) and (10, 15, 20). Are they similar? State the test and the scale factor. 2 marks
1.3 △ABC has AB = 3 cm, AC = 4 cm and ∠A = 70°. △PQR has PQ = 9 cm, PR = 12 cm and ∠P = 70°. Are they similar? State the test. 2 marks
1.4 A 1.5 m stick casts a 2.5 m shadow. At the same time a building casts a 75 m shadow. Find the building's height. 2 marks
1.5 In △ABC, D lies on AB and E lies on AC with DE ∥ BC. AD = 5 cm, DB = 10 cm and BC = 24 cm. Find DE. 3 marks
1.6 Triangles ABC and DEF have AB = 4, BC = 6, AC = 8 and DE = 6, EF = 9, DF = 12. Show they are similar and state the scale factor (large / small). 3 marks
2. Find the mistake
Another Year 10 student has tried to prove two triangles are similar by SAS. Their working is shown. Exactly one line contains a mistake. Spot it, explain why it is wrong, and re-do the working correctly. 3 marks
Triangles: △ABC has AB = 4 cm, AC = 6 cm and ∠B = 50°. △DEF has DE = 8 cm, DF = 12 cm and ∠E = 50°. Student's working:
Line 1: AB / DE = 4 / 8 = 1/2.
Line 2: AC / DF = 6 / 12 = 1/2.
Line 3: Two sides in same ratio, plus ∠B = ∠E = 50°.
Line 4: Therefore △ABC ~ △DEF (SAS).
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) What extra information would be needed to legitimately apply SAS here? (Or, if no SAS proof is possible, state why.)
Stuck? Revisit lesson § "SAS Similarity" — the equal angle must be the included angle between the two proportional sides.3. Open-ended challenge — design your own field measurement
This question has many valid answers. Be creative but show every number. 4 marks
3.1 Design a real-world setup in which you can measure the height of an inaccessible object (your choice: the school flagpole, a tree, a goal post, a building) using only a metre ruler and a tape measure.
In your answer:
(i) Describe the setup in 2-3 sentences (include the role of the sun or a sighting line).
(ii) State which similarity test makes the two triangles similar and give the reason.
(iii) Make up realistic numbers (stick height, stick shadow, object shadow) and use them to calculate the object's height.
(iv) State one practical limitation (cloudy day? wind? slope of ground?) and how you would handle it.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Same angles, two side lengths
AAA (all three angles equal). k = XY / AB = 18 / 6 = 3.
YZ = BC × k = 9 × 3 = 27 cm.
1.2 — (4,6,8) and (10,15,20)
Match shortest to shortest: 10/4 = 2.5, 15/6 = 2.5, 20/8 = 2.5. All three ratios equal.
SSS similarity, k = 2.5 ✓.
1.3 — Two sides + included angle
PQ / AB = 9/3 = 3; PR / AC = 12/4 = 3. Two sides in ratio 1:3, and the included angle ∠A = ∠P = 70°.
Yes, similar by SAS, k = 3.
1.4 — Stick and building shadow
h / 1.5 = 75 / 2.5 = 30. h = 1.5 × 30 = 45 m. (AAA similarity, parallel sun rays.)
1.5 — DE in △ADE
AB = AD + DB = 5 + 10 = 15.
△ADE ~ △ABC by AAA (DE ∥ BC). k = AD / AB = 5/15 = 1/3.
DE = BC × k = 24 × (1/3) = 8 cm.
1.6 — Sides 4,6,8 and 6,9,12
Match shortest to shortest: 6/4 = 1.5, 9/6 = 1.5, 12/8 = 1.5. All ratios equal.
△ABC ~ △DEF (SSS), k = 3/2 (= 1.5).
2 — Find the mistake
(a) The mistake is on Line 3 (which then breaks Line 4).
(b) The 50° angle given is ∠B in the first triangle and ∠E in the second. ∠B is not the included angle between AB and AC — the included angle would be ∠A. Similarly ∠E is between DE and EF, not between DE and DF. So the SAS test does not actually apply, even though two sides are in the same ratio.
(c) To legitimately use SAS you would need ∠A = ∠D (the angle included between AB and AC, and between DE and DF). Otherwise, you would need to check the third pair of sides (BC and EF) and use SSS, or find another angle to use AAA.
3 — Open-ended (sample solution)
(i) Setup: Push a 1 m stick vertically into level ground near the school flagpole on a sunny day. Measure the stick's shadow with a tape measure, then measure the flagpole's shadow.
(ii) Similarity: Both stick and flagpole are vertical, so both small triangles have a right angle at the base. The sun is so far away that its rays are essentially parallel, so the angle of elevation is the same for both. Two pairs of equal angles → AAA, so the triangles are similar.
(iii) Numbers: Stick = 1 m. Stick shadow = 0.8 m. Flagpole shadow = 8 m. Proportion: h / 1 = 8 / 0.8 = 10. Flagpole height = 10 m.
(iv) Limitation: If the ground slopes, the two "right angles at the base" are not actually at the same horizontal, and the shadow lengths are off. I would re-do the measurement on the flattest patch I could find, and ideally average two readings.
Marking: 1 for a sensible setup, 1 for naming AAA with the parallel-rays reason, 1 for showing realistic numbers and a correct proportion, 1 for a sensible practical limitation. Any reasonable object and any consistent set of numbers is acceptable.