Mathematics • Year 10 • Unit 3 • Lesson 12
Measuring the Unreachable — Similar Triangles in the Field
Apply AAA similarity to real Australian outdoor problems — the height of a gum tree at Centennial Park, the width of the Hawkesbury River from one bank, the height of the Westfield clock tower from its shadow. Every problem is a "set up two similar right triangles, then solve a proportion" task.
1. Word problems
For each problem: (i) draw the two similar right triangles, (ii) state the similarity test (almost always AAA for shadow / sighting problems), (iii) write a proportion, (iv) solve for the unknown.
1.1 — Gum tree at Centennial Park. On a sunny morning, a 2 m vertical stake casts a 1.5 m shadow on flat ground. At the same moment, a gum tree casts a 12 m shadow.
(a) Sketch the two similar right triangles.
(b) State the similarity test used.
(c) Find the height of the gum tree. 4 marks
1.2 — Westfield clock tower shadow. A 1.6 m student stands beside a clock tower at Bondi Junction Westfield. Her shadow is 1 m long, and the tower's shadow is 25 m long.
(a) Set up a proportion.
(b) Find the height of the clock tower. 3 marks
1.3 — Width of the Hawkesbury River. Standing on the south bank at point A, a hiker sights a marker M directly across the river. She walks 8 m along the bank to point B (so AB = 8 m, perpendicular to AM). From B she walks 3 m further along the bank to point C and sights M again — her sighting line cuts AB extended at the foot of a small post at point P where BP = 3 m. The post is 4 m from M' (the foot of the perpendicular to her line of sight). Using similar triangles, find the width AM of the river. (Assume AM ⊥ AC and PM' ⊥ AC, with AP = 11 m, BP = 3 m and PM' / BP = AM / AP gives the river width.) 4 marks
1.4 — Ramp at Parramatta Aquatic Centre. A 1 m walking stick is held vertically at the bottom of an accessibility ramp. Its tip touches the underside of the ramp at a point 1.5 m up the slope from the base. The total slope length of the ramp is 9 m. The ramp is a straight slope.
(a) Use similar triangles to find the vertical rise of the entire ramp.
(b) Briefly state which similarity test you used. 3 marks
1.5 — Triangle in a triangle. In △ABC, D is on AB and E is on AC such that DE ∥ BC. Given AD = 4 cm, DB = 6 cm and BC = 15 cm, find DE. 3 marks
2. Explain your thinking
This question is about communication, not just numbers. Use full sentences. 4 marks
2.1 A friend says: "Shadow problems use Pythagoras." Using the language of Lesson 12 (AAA, parallel sun rays, scale factor, proportion), explain in 4-6 sentences (i) why the friend's claim is wrong, (ii) which similarity test actually applies and why, and (iii) what setting up the proportion looks like in symbols. Include a specific example (stick height, stick shadow, tree shadow) and the height of the tree to make your point concrete.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Gum tree shadow
(a) Both are right triangles with horizontal shadow as base and vertical height as the other leg.
(b) AAA: both triangles have a right angle at the base, and the sun's angle of elevation is the same for both (parallel rays).
(c) h / 2 = 12 / 1.5 = 8. So h = 2 × 8 = 16 m.
1.2 — Westfield clock tower
(a) h / 1.6 = 25 / 1 = 25.
(b) h = 1.6 × 25 = 40 m.
This is AAA similarity again — same sun angle.
1.3 — Hawkesbury River width
By AAA similarity, AM / AP = PM' / BP.
AM / 11 = 4 / 3, so AM = 11 × 4 / 3 = 44/3 ≈ 14.7 m.
The two right triangles share the angle at the sighting point C, plus both have a right angle, so they are similar.
1.4 — Parramatta ramp
(a) The small triangle has slope length 1.5 m and the stick's vertical height 1 m. The full ramp has slope length 9 m. Scale factor k = 9 / 1.5 = 6.
Vertical rise of the ramp = 1 × 6 = 6 m.
(b) AAA: both right triangles share the slope angle and both have a right angle where the vertical meets the ground.
1.5 — Triangle in a triangle
AB = AD + DB = 4 + 6 = 10 cm.
△ADE ~ △ABC by AAA (DE ∥ BC gives equal corresponding angles, ∠A is common).
Scale factor k = AD / AB = 4 / 10 = 2/5.
DE = BC × k = 15 × (2/5) = 6 cm.
2.1 — Explain your thinking (sample response)
My friend is wrong: Pythagoras tells us the third side of a right triangle when we know the other two, but it does not link two different right triangles. Shadow problems link two different triangles — the small one made by the stick and its shadow, and the large one made by the tree and its shadow. These triangles are similar by AAA because both have a right angle at the ground and the sun's parallel rays make the same angle of elevation in both. So the correct setup is a proportion: height of tree / height of stick = shadow of tree / shadow of stick. For example, a 2 m stick casting a 1.5 m shadow at the same time as a tree casting a 12 m shadow gives h / 2 = 12 / 1.5, so the tree is 16 m tall.
Marking: 1 for naming AAA as the test, 1 for the parallel-rays reason, 1 for writing the proportion in symbols, 1 for a worked numerical example with a height result.