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.