Every solution shows the formula, the substitution, and the final area with units. Use the answers to check your work, and compare your method on the composite shapes, since more than one way of cutting a polygon gives the same total area.
Problem 1: Right triangle
The legs are perpendicular, so use them as base and height.
\(A = \frac{1}{2} \times 9 \times 4 = \frac{1}{2} \times 36 = 18\)
The area is \(18\text{ ft}^2\).
Problem 2: Triangle area
\(A = \frac{1}{2} \times 14 \times 6 = 7 \times 6 = 42\)
The area is \(42\text{ m}^2\).
Problem 3: Parallelogram area
\(A = bh = 12 \times 7 = 84\)
The area is \(84\text{ in}^2\).
Problem 4: Missing height
\(45 = \frac{1}{2} \times 10 \times h\), so \(45 = 5h\).
\(h = 45 : 5 = 9\)
The height is \(9\text{ yd}\).
Problem 5: Trapezoid area
\(A = \frac{1}{2}(b_1 + b_2)h = \frac{1}{2}(6 + 10)(4)\)
\(= \frac{1}{2} \times 16 \times 4 = 8 \times 4 = 32\)
The area is \(32\text{ ft}^2\).
Problem 6: Missing base
\(72 = b \times 9\)
\(b = 72 : 9 = 8\)
The base is \(8\text{ cm}\).
Problem 7: Rectangle plus triangle
Rectangle: \(10 \times 6 = 60\text{ ft}^2\).
Triangle: \(\frac{1}{2} \times 10 \times 4 = 20\text{ ft}^2\).
Total: \(60 + 20 = 80\)
The area is \(80\text{ ft}^2\).
Problem 8: Garden beds
Area: \(\frac{1}{2}(9 + 15)(8) = \frac{1}{2} \times 24 \times 8 = 96\text{ m}^2\).
Bags: \(96 : 12 = 8\)
Sofia needs 8 bags.
Problem 9: Painting a gable wall
Rectangle: \(16 \times 8 = 128\text{ ft}^2\).
Triangle: \(\frac{1}{2} \times 16 \times 5 = 40\text{ ft}^2\).
Total area: \(128 + 40 = 168\text{ ft}^2\).
Gallons: \(168 : 84 = 2\). Cost: \(2 \times \$32 = \$64\).
The wall is \(168\text{ ft}^2\) and the paint costs \(\$64\).
Problem 10: Patio pavers
Full rectangle: \(14 \times 10 = 140\text{ ft}^2\).
Corner: \(6 \times 4 = 24\text{ ft}^2\).
Patio: \(140 – 24 = 116\text{ ft}^2\).
Cost: \(116 \times \$3.50 = \$406\).
The patio is \(116\text{ ft}^2\) and the pavers cost \(\$406\).
See every chapter on the Grade 6 math page.