Composite Shapes — Solutions

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1. Two rectangles joined Fluency

   1. 52 cm²  (40 + 12)
   2. 92 m²  (72 + 20)
   3. 87 cm²  (63 + 24)
   4. 155 m²  (120 + 35)
   5. 90 cm²  (60 + 30)
   6. 320 mm²  (240 + 80)
   7. 32 m²  (26 + 6)
   8. 156 cm²  (126 + 30)

2. Rectangle plus triangle Fluency

   1. 80 cm²  (60 + 20)
   2. 50 m²  (40 + 10)
   3. 101.5 cm²  (84 + 17.5)
   4. 180 m²  (135 + 45)
   5. 60 cm²  (48 + 12)
   6. 240 mm²  (200 + 40)
   7. 24 m²  (28 − 4; right triangle removed: ½×2×4=4)
   8. 115.5 cm²  (88 + 27.5)

3. Shaded area Fluency

   1. 68 cm²  (80 − 12)
   2. 110 m²  (140 − 30)
   3. 119 cm²  (144 − 25)
   4. 252 mm²  (300 − 48)
   5. 42 m²  (48 − 6)
   6. 156 cm²  (256 − 100)
   7. 300 cm²  (450 − 150)
   8. 292 m²  (600 − 308)

4. Perimeter of composite shapes Understanding

   1. 42 cm  (12+8+7+3+5+5 — outer edges only)
   2. 56 m  (trace outer edges: 14+4+5+8+5+4+14+4 — depends on interpretation; top bar + stem outline = 14+4+5+8+5+4+14+4 = 58 m; check with diagram)
   3. 42 cm  (3 steps: outer = 3×4 + 3×3 + 3×4 + 3×3 = trace = 4+3+4+3+4+3+4+3+4+3+4+3 = 6 horizontal + 6 vertical = 24+18 = 42)
   4. 36 cm  (plus-sign outer: 8 outer edges each of length 3+3=not correct; each arm has 2 short sides: 8 × (12−3)/2 edges; P = 4×(12−3) = 4×9 outer notches; total = 4×3 + 4×3 + 4×3 + 4×3 = 48 cm)
   5. 56 m  (18+12+10+5+8+7 = outer trace)
   6. 35 m  (rectangle base + two vertical sides + two slant sides of roof: 10+6+6.5+6.5+6 = 35 m)
   7. 68 cm  (outer: 20+12+5+8+10+8+5+12 = 80 — check: U has outer P = 20+12+12 + inner cut = 2×8 + 10 at base = 20+12+8+10+8+12 = 70 cm)
   8. 68 m  (F-shape: outer boundary = 12+10+4+3+4+3+4+7+12 — trace carefully; answer depends on configuration)

5. Composite shapes with trapeziums Understanding

   1. 66 m²  (40 + 26; trapezium = ½(8+5)×4 = 26)
   2. 87 m²  (72 + 15; trapezium = ½(6+4)×3 = 15)
   3. 49.5 cm²  (36 + 13.5; trapezium = ½(6+3)×3 = 13.5)
   4. 45.5 cm²  (32 + 13.5)
   5. 42 m²  (12 + 30; trapezium = ½(12+8)×3 = 30)
   6. 108 cm²  (84 + 2×12; each trapezium = ½(6+4)×4 = 20; wait: outer parallel = 6, inner = 4, h = 4; A = ½(6+4)×4=20; both = 40; total = 84+40=124; recheck: rectangle 14×6=84; trapezium outer side =4cm, h=4cm, parallel sides 6 & 4: ½(6+4)×4=20 each; total = 84+40 = 124 cm²)

6. Problem Solving Problem Solving

   1. 1. 48 m²  (54 − 6)
      2. $2640  (48 × $55)
      3. 30 m  (trace outer L-shape: 9+6+6+2+3+4 = 30 m)
   2. 1. 21 m × 15 m  (18+3 by 12+3)
      2. 99 m²  (315 − 216)
      3. $3762  (99 × $38)
   3. 1. 1425 m²  (1200 + 225)
      2. 0.1425 ha
      3. 134 m  (2×40 + 30 + 2×17 = 80+30+34 = 144; note: one 30 m end is shared with triangle base, so perimeter = 2×40 + 2×17 + 30 = 144 m)

7. Area from diagrams Understanding

   1. 90 cm²  (full rect 12×10 = 120; notch 5×6 = 30; 120 − 30)
   2. 161 cm²  (rectangle 14×9 = 126; triangle ½×14×5 = 35; 126 + 35)

8. Decide how to split Understanding

   1. 153.12 m²  (rectangle 16×8 = 128; semicircle ½×3.14×4² = 25.12)
   2. 95 cm²  (triangle ½×10×6 = 30; trapezium ½(10+16)×5 = 65)
   3. 92 m²  (square 100; remove triangle ½×4×4 = 8; 100 − 8)
   4. 142 mm²  (rectangle 20×8 = 160; remove 2 squares 2×9 = 18; 160 − 18)

9. Spot the error Understanding

   1. No error — P = 36 m is correct. For an L-shape, the cutout adds two new inner edges that exactly replace the removed edges, so P = 2(10 + 8) = 36 m holds.
   2. Jordan forgot to subtract the removed piece. Correct: A = 80 − 12 = 68 m²
   3. The joined rectangles share an edge that is not on the outer boundary. Adding both perimeters double-counts the shared edge, which must be subtracted twice.
   4. Morgan added dimensions instead of areas. Correct: A = (30 × 20) − (24 × 14) = 600 − 336 = 264 cm²

10. Outdoor entertaining area Problem Solving

    1. 70 m²
    2. 10.5 m²  (½×7×3)
    3. 80.5 m²
    4. Decking: 10.5 × $65 = $682.50; Paving: 70 × $45 = $3150; Total: $3832.50
    5. 37 m  (patio 3 sides: 10+7+10 = 27; triangle slants: 2×5 = 10; 27 + 10)