Advanced Area and Perimeter Reasoning

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Part 1: Composite Shapes and Algebraic Reasoning

Formula reference: Rectangle area = length × width. Rectangle perimeter = 2 × (length + width). Triangle area = 1/2 × base × perpendicular height. Circle area = πr². Circle circumference = 2πr. Use π = 3.14 unless otherwise stated. Show all working and include appropriate units.

Success criteria: I can decompose complex shapes, find missing dimensions, form and solve equations, and explain why my method works.

1. An L-shaped courtyard is formed from a 14 m × 11 m rectangle with a 6 m × 4 m rectangle removed from one corner. Find the area and perimeter of the courtyard.

2. A rectilinear garden has consecutive outer side lengths of 18 m, 7 m, 5 m, 9 m, 13 m and 16 m as you travel clockwise around its boundary. The 7 m and 9 m sides are parallel, and the 5 m and 13 m sides are parallel. Determine the missing horizontal and vertical lengths needed to sketch the shape, then find its perimeter. Explain how you know your missing lengths are correct.

3. A rectangle has length (3x + 2) cm and width (x − 1) cm. Its perimeter is 50 cm. Find x, the dimensions of the rectangle and its area.

4. A 20 m × 12 m rectangular lawn has a uniform path of width x metres around the outside. The total area of the lawn and path is 384 m². Find x and the area of the path.

5. A 10 cm × 8 cm rectangle has a semicircle attached to each of its 8 cm sides. Find the total area and the outside perimeter of the resulting stadium shape. Use π = 3.14.

6. A circular pond has radius 4.5 m. A square deck just surrounds the pond, with each side of the square tangent to the circle. Find the area of the deck. Give your answer to the nearest square metre.

7. A rectangular sheet measures 30 cm by 18 cm. A circle of radius 6 cm and a right-angled triangle with base 10 cm and perpendicular height 8 cm are cut out. Find the remaining area. Explain why the two cut-out areas can be subtracted directly.

Part 2: Multi-Step Applications and Justification

Success criteria: I can choose efficient strategies, interpret scale and design information, check the reasonableness of answers, and justify my conclusions.

8. A floor plan is drawn at a scale of 1:50. A rectangular room measures 9.6 cm by 7.2 cm on the plan. A 1.2 m wide doorway is not covered by skirting board. Find the actual area of the room and the length of skirting board required.

9. A rectangular swimming pool is 18 m by 8 m. A semicircular shallow end is attached along one 8 m side. Find the total surface area of the pool and the length of the pool’s outside boundary. Use π = 3.14.

10. A 24 m by 16 m rectangular paddock contains a circular vegetable garden of radius 5 m and a triangular compost area with base 8 m and perpendicular height 6 m. What percentage of the paddock remains for grass? Give your answer to 1 decimal place.

11. A rectangular sign has area 1800 cm². Its length is 15 cm more than twice its width. Find its dimensions and perimeter. Show the equation you use.

12. A 12 m by 9 m rectangle has a quarter-circle of radius 9 m removed from one corner. Find the remaining area and perimeter. Explain which parts of the original rectangle’s perimeter must be replaced by the curved edge.

13. Two designs have the same perimeter of 48 cm. Design A is a square. Design B is a rectangle whose length is 4 cm greater than its width. Find the area of each design and determine which has the greater area. Justify your conclusion.

14. A triangular roof has sides 13 m, 14 m and 15 m. Its perpendicular height to the 14 m base is 12 m. Find its area and perimeter. Optional extension: use Pythagoras to verify one of the internal lengths if the triangle is split into two right-angled triangles.

Answer Key, Extension and Support

Answer key

1. Original area = 154 m²; removed area = 24 m². Area = 130 m². Perimeter = 14 + 11 + 8 + 4 + 6 + 7 = 50 m.

2. Missing lengths are found by balancing opposite horizontal and vertical movements: missing horizontal length = 10 m; missing vertical length = 5 m. Perimeter = 18 + 7 + 5 + 9 + 13 + 16 + 10 + 5 = 83 m. A correct sketch and explanation should show that total movement right equals total movement left, and total movement up equals total movement down.

3. 2[(3x + 2) + (x − 1)] = 50, so 8x + 2 = 50 and x = 6. Dimensions are 20 cm by 5 cm; area = 100 cm².

4. (20 + 2x)(12 + 2x) = 384. Solving gives x = 2 m. Outer area = 384 m²; lawn area = 240 m²; path area = 144 m².

5. Rectangle area = 80 cm². Two semicircles form one circle of radius 4 cm: area = 3.14 × 4² = 50.24 cm². Total area = 130.24 cm². Outside perimeter = 2 × 10 + circumference of circle = 20 + 25.12 = 45.12 cm.

6. The square side is the circle’s diameter: 9 m. Square area = 81 m²; circle area = 3.14 × 4.5² = 63.585 m². Deck area = 17.415 m², approximately 17 m².

7. Rectangle area = 540 cm². Circle area = 3.14 × 6² = 113.04 cm². Triangle area = 1/2 × 10 × 8 = 40 cm². Remaining area = 386.96 cm². The areas are subtracted because the cut-outs do not overlap and both lie completely inside the sheet.

8. Actual dimensions: 9.6 × 50 = 480 cm = 4.8 m; 7.2 × 50 = 360 cm = 3.6 m. Area = 17.28 m². Skirting = 2(4.8 + 3.6) − 1.2 = 15.6 m.

9. Rectangle area = 144 m². Semicircle radius = 4 m, area = 1/2 × 3.14 × 4² = 25.12 m². Total area = 169.12 m². Boundary = 18 + 18 + 8 + semicircle arc (3.14 × 4) = 56.56 m.

10. Paddock area = 384 m². Circle area = 78.5 m²; triangle area = 24 m². Grass area = 281.5 m². Percentage remaining = 281.5 ÷ 384 × 100 = 73.3%.

11. Let width = w and length = 2w + 15. Then w(2w + 15) = 1800. Solving gives w = 30 cm; length = 75 cm. Perimeter = 210 cm.

12. Rectangle area = 108 m². Quarter-circle area = 1/4 × 3.14 × 9² = 63.585 m². Remaining area = 44.415 m². Perimeter = 12 + 9 + 3.14 × 9 ÷ 2 = 35.13 m. The 9 m and 9 m corner edges are replaced by the quarter-circle arc.

13. Square side = 48 ÷ 4 = 12 cm; area = 144 cm². For Design B, 2L + 2W = 48 and L = W + 4, giving W = 10 cm and L = 14 cm. Area = 140 cm². Design A is greater by 4 cm².

14. Area = 1/2 × 14 × 12 = 84 m². Perimeter = 13 + 14 + 15 = 42 m. Extension: splitting the triangle gives right-triangle lengths 5 m and 9 m; 5² + 12² = 13² and 9² + 12² = 15².

Extension challenge

A rectangle has perimeter 60 cm. A semicircle is attached to one of its shorter sides. Find the rectangle’s dimensions that produce the greatest total area if the shorter side is 12 cm. Compare this with a square of perimeter 60 cm and explain which design is more efficient.

Differentiation supports

Use squared paper to redraw composite shapes. Mark every known length and use matching parallel sides to find missing dimensions. Write the formula before substituting numbers. Convert all measurements to the same unit before calculating. For algebraic questions, create a table of possible values or use a calculator to check the final solution. For circles, label the radius or diameter clearly and separate straight edges from curved edges.