Mathematics • Year 10 • Unit 3 • Lesson 17
Triangle Areas — Real Australian Land
Apply A = ½ ab sin C and Heron's formula to authentic land, design and pricing problems — a Riverina paddock, a school sail-shade, a glass coffee-table top, a turf lawn, and a council park. Decide SAS vs SSS, calculate the area, then turn it into a real number (hectares, square metres, dollars).
1. Word problems
For each problem: (i) sketch the triangle with all known information, (ii) decide whether SAS or SSS is given, (iii) write the right formula, (iv) calculate, (v) convert to the requested final unit (m², hectares, dollars). Use 1 hectare = 10 000 m².
1.1 — Riverina paddock. A triangular paddock near Griffith has two sides 100 m and 150 m with an included angle of 70°.
(a) Find the area of the paddock in m² and in hectares.
(b) If wheat yields 4.5 tonnes per hectare, how many tonnes of wheat could this paddock produce? 4 marks
1.2 — Primary-school sail shade. A triangular shade-sail at Bondi Primary has sides 6 m, 8 m, and 10 m.
(a) Use Heron's formula to find the sail area.
(b) Shade-sail fabric costs $42 per m². Find the fabric cost (ignore offcuts). 3 marks
1.3 — Glass coffee-table top. A custom glass top is cut as a triangle with two sides 80 cm and 60 cm and an included angle of 110°.
(a) Find the area of the glass in cm² and in m².
(b) The glazier charges $0.85 per cm². Find the cost. 3 marks
1.4 — Turf lawn at a Perth home. A backyard lawn has three sides 9 m, 11 m, and 14 m.
(a) Use Heron's formula to find the lawn area to 1 dp.
(b) Buffalo turf is sold in rolls covering 1.6 m². How many rolls must the homeowner buy (round up)? 3 marks
1.5 — Council pocket park. A small triangular Sydney pocket park has two boundary fences of 28 m and 35 m meeting at an angle of 95°.
(a) Find the area to the nearest m².
(b) If new mulch costs $14 per m², find the cost of mulching the whole park. 3 marks
2. Explain your thinking
This question is about reasoning, not just numbers. Use full sentences. 4 marks
2.1 A real-estate agent claims: "Two paddocks with the same perimeter must have the same area." A neighbour points out the agent is wrong. In 4-6 sentences, explain (i) why two triangles with the same perimeter can have very different areas, (ii) which combination of perimeter and angles tends to give the largest area, and (iii) give a concrete numerical example of two triangles with perimeter 24 cm but different areas. Reference the formula A = ½ ab sin C and/or Heron's formula in your explanation.
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Riverina paddock (100 m, 150 m, 70°)
(a) A = ½ × 100 × 150 × sin 70° = 7 500 × 0.9397 ≈ 7 048 m² ≈ 0.705 ha.
(b) Wheat: 0.705 × 4.5 ≈ 3.17 tonnes.
1.2 — Bondi sail (6, 8, 10)
(a) s = (6 + 8 + 10) / 2 = 12. A = √(12 × 6 × 4 × 2) = √576 = 24 m². (Matches ½ × 6 × 8 because the 6-8-10 triangle is right-angled.)
(b) Cost = 24 × $42 = $1 008.
1.3 — Glass top (80 cm, 60 cm, 110°)
(a) A = ½ × 80 × 60 × sin 110° = 2 400 × 0.9397 ≈ 2 255.3 cm² ≈ 0.226 m².
(b) Cost = 2 255.3 × $0.85 ≈ $1 917.
1.4 — Perth lawn (9, 11, 14)
(a) s = (9 + 11 + 14) / 2 = 17. A = √(17 × 8 × 6 × 3) = √2 448 ≈ 49.5 m².
(b) Rolls needed = 49.5 / 1.6 = 30.9375 → round up to 31 rolls.
1.5 — Pocket park (28 m, 35 m, 95°)
(a) A = ½ × 28 × 35 × sin 95° = 490 × 0.9962 ≈ 488 m².
(b) Mulch cost = 488 × $14 = $6 832.
2.1 — Explain your thinking (sample response)
The agent is wrong. Heron's formula A = √[s(s−a)(s−b)(s−c)] shows that the area depends on the individual side lengths a, b, c — not just their sum s. Two triangles can share a perimeter (and hence the same s) but have wildly different products (s−a)(s−b)(s−c). The area is largest when the triangle is equilateral (or, equivalently, when the sides are as close to equal as possible), and shrinks toward zero as the triangle becomes long and thin (one side approaches the sum of the other two).
Concrete example with perimeter 24 cm:
· 6-8-10 right triangle: s = 12; A = √(12 × 6 × 4 × 2) = √576 = 24 cm².
· 11-11-2 long thin triangle: s = 12; A = √(12 × 1 × 1 × 10) = √120 ≈ 11.0 cm².
Same perimeter, very different areas — proving the agent's claim is false.
Marking: 1 for explaining area depends on individual sides not just s, 1 for naming equilateral/equal-sided as the maximum, 1 for the worked numerical contrast, 1 for explicit reference to Heron's formula or A = ½ ab sin C.