Mathematics • Year 10 • Unit 1 • Lesson 15
Area of Composite Shapes — Mixed Challenge
Pull together every idea from Lesson 15: the six core formulas (rectangle, triangle, parallelogram, trapezium, circle, sector), addition and subtraction strategies, and unit consistency. Spot a plausible Year 10 mistake from the Misconceptions card, then design your own composite-shape problem.
1. Mixed problems — choose the right tool
Each question uses a different formula and/or strategy from Lesson 15. Decide which approach applies before writing. Show your working. 3 marks each
1.1 Calculate the area of a triangle with base 12 cm and perpendicular height 5 cm, and the area of a parallelogram with base 9 m and perpendicular height 4 m.
1.2 A trapezium has parallel sides 8 m and 5 m, with perpendicular height 3 m. Calculate its area.
1.3 A sector has radius 6 cm and angle 60°. Calculate the area in terms of π, then to 1 d.p.
1.4 A composite shape is made from a rectangle 9 m by 4 m with a semicircle of diameter 4 m on one end. Find the total area in terms of π. State which strategy you used.
1.5 A square of side 10 cm has a smaller square of side 4 cm cut out from one corner. Find the remaining area and state your strategy.
1.6 A driveway is shaped as a rectangle 8 m by 2 m, with a quarter-circle of radius 2 m on each end (a "stadium" shape). Find the total area in terms of π. (Hint: two quarter-circles combine into one semicircle, or use the addition strategy with one rectangle + two quarter-circles.)
2. Find the mistake
Another Year 10 student tried to calculate the area of a parallelogram. Their working is shown below. Exactly one line contains a mistake the lesson's Misconceptions card warns about. Spot it, explain why, then re-do correctly. 3 marks
Student's working — find the area of a parallelogram with base 10 cm, slanted side 6 cm, and perpendicular height 4 cm:
Line 1: Area of parallelogram = base × side length
Line 2: = 10 × 6
Line 3: = 60 cm²
(a) Which line contains the mistake?
(b) Explain in one or two sentences why that line is wrong.
(c) Write out the corrected working in full, including the corrected final answer.
Stuck? The lesson's Misconceptions card explicitly states: parallelogram area = base × perpendicular height, NOT base × slanted side length.3. Open-ended challenge — design your own composite-shape problem
This question has many valid answers. Be creative but show every number. 4 marks
3.1 Design one realistic composite-shape problem drawn from a real Australian context (a backyard, a community pool, a school courtyard, a council park). Your problem must:
(i) involve at least three different basic shapes from the lesson's formula list (rectangle, triangle, parallelogram, trapezium, circle, sector);
(ii) require using both the addition strategy AND the subtraction strategy somewhere in the solution (e.g. "add A and B, then subtract C");
(iii) result in a final area between 100 m² and 500 m²;
(iv) include a follow-up real-world cost calculation at a stated $/m² rate.
Write out: (a) the problem statement with a small labelled sketch (or a clear text description of dimensions), (b) your full worked solution showing both strategies, and (c) the final cost.
Bonus: the dimensions must all be realistic for the context (a backyard pool is not 90 m wide).
How did this worksheet feel?
What I'll revisit before next class:
1.1 — Triangle and parallelogram
Triangle: A = ½ × 12 × 5 = 30 cm².
Parallelogram: A = 9 × 4 = 36 m².
1.2 — Trapezium
A = ½(a + b) × h = ½(8 + 5) × 3 = ½ × 13 × 3 = 19.5 m².
1.3 — Sector (60°, radius 6 cm)
A = (60/360) × π × 6² = (1/6) × 36π = 6π cm² ≈ 18.8 cm² (1 d.p.).
1.4 — Rectangle + semicircle
Strategy: addition. Rectangle = 9 × 4 = 36 m². Semicircle radius = 2 m → area = ½ × π × 4 = 2π m².
Total = (36 + 2π) m².
1.5 — Square with cut-out
Strategy: subtraction. Big square = 100 cm². Small square = 16 cm². Remaining = 84 cm².
1.6 — Stadium driveway
Rectangle (middle) = 8 × 2 × 2 = wait — width is 2 m × 2 (diameter spans width), so rectangle = 8 × 4 = 32 m²? Re-read carefully: width of rectangle = 2 × radius = 2 × 2 = 4 m. So rectangle = 8 × 4 = 32 m².
Two quarter-circles of radius 2 m combine to one semicircle of radius 2 m: area = ½ × π × 2² = 2π m². But the stadium shape needs two full ends, each a semicircle: 2 × (½π r²) = π × 4 = 4π m².
Total = (32 + 4π) m² ≈ 44.6 m².
Alternative reading: if "rectangle 8 × 2" means the full footprint length is 8 and width 2, then ends are quarter-circles of radius 1 — also acceptable. Either reading scores full marks if the calculation is internally consistent.
2 — Find the mistake
(a) The mistake is on Line 1 (Lines 2 and 3 inherit it).
(b) The parallelogram formula uses the perpendicular height, NOT the slanted side length. The student multiplied by 6 (the slant) instead of 4 (the perpendicular height). The Misconceptions card warns against using diagonal or slant lengths.
(c) Corrected working:
Area = base × perpendicular height = 10 × 4 = 40 cm².
3 — Open-ended challenge (sample solution)
Problem: A community park is a 20 m × 15 m rectangular grass area, with a semicircular flowerbed of radius 4 m built onto one of the 15 m sides, and a triangular sand pit of base 6 m and height 4 m cut out from one corner. Calculate the area of mowable grass, and the cost to maintain it at $8/m² per year.
Solution: Three basic shapes used: rectangle, semicircle, triangle. Strategies used: addition (rectangle + semicircle for total park) AND subtraction (subtract sand pit).
Rectangle = 20 × 15 = 300 m².
Semicircle = ½ × π × 4² = 8π m² ≈ 25.13 m².
Triangle = ½ × 6 × 4 = 12 m².
Total park area = 300 + 8π ≈ 325.13 m². Grass area = 325.13 − 12 = 313.13 m² (within 100–500 m² range ✓).
Cost = 313.13 × $8 = $2,505.04 per year.
Marking: 1 for three different basic shapes; 1 for using both addition AND subtraction in the same problem; 1 for a realistic Australian context with sensible dimensions; 1 for a final area in [100, 500] m² and a correctly computed cost. Any valid problem meeting all four criteria scores full marks.