Every solution lists each face, shows the area of every face, and adds them to find the surface area in square units. Use the answers to check your work, and compare your count of faces with the nets drawn in the solution figures for the triangular prism and the pyramid.
Problem 1: Cube net
Each face: \(5 \times 5 = 25\text{ in}^2\).
\(SA = 6 \times 25 = 150\)
The surface area is \(150\text{ in}^2\).
Problem 2: Rectangular prism
Faces come in pairs: \(6 \times 4 = 24\), \(6 \times 3 = 18\), and \(4 \times 3 = 12\).
\(SA = 2(24 + 18 + 12) = 2(54) = 108\)
The surface area is \(108\text{ ft}^2\).
Problem 3: Name the solid
Two triangles and three rectangles fold into a triangular prism.
\(2 + 3 = 5\)
It is a triangular prism with 5 faces.
Problem 4: Triangular prism
Each triangle: \(\frac{1}{2} \times 3 \times 4 = 6\text{ in}^2\).
Rectangles: \(3 \times 10 = 30\), \(4 \times 10 = 40\), \(5 \times 10 = 50\).
\(SA = 6 + 6 + 30 + 40 + 50 = 132\)
The surface area is \(132\text{ in}^2\).
Problem 5: Square pyramid
Base: \(6 \times 6 = 36\text{ m}^2\).
One triangle: \(\frac{1}{2} \times 6 \times 5 = 15\text{ m}^2\). Four triangles: \(4 \times 15 = 60\text{ m}^2\).
\(SA = 36 + 60 = 96\)
The surface area is \(96\text{ m}^2\).
Problem 6: Open-top box
Base: \(10 \times 8 = 80\). Two long sides: \(2 \times (10 \times 4) = 80\). Two short sides: \(2 \times (8 \times 4) = 64\).
\(80 + 80 + 64 = 224\)
The surface area is \(224\text{ in}^2\).
Problem 7: Missing edge
\(6s^2 = 216\), so \(s^2 = 36\).
\(6 \times 6 = 36\), so \(s = 6\).
The edge is \(6\text{ in}\).
Problem 8: Gift wrap
\(12 \times 9 = 108\), \(12 \times 3 = 36\), \(9 \times 3 = 27\).
\(SA = 2(108 + 36 + 27) = 2(171) = 342\)
She needs \(342\text{ in}^2\) of paper.
Problem 9: Toy chest paint
\(4 \times 2 = 8\), \(4 \times 2 = 8\), \(2 \times 2 = 4\).
\(SA = 2(8 + 8 + 4) = 40\text{ ft}^2\).
Two coats: \(2 \times 40 = 80\text{ ft}^2\). Quarts: \(80 : 20 = 4\). Cost: \(4 \times \$9 = \$36\).
The surface area is \(40\text{ ft}^2\) and the paint costs \(\$36\).
Problem 10: Camping tent
Two ends: \(2 \times (\frac{1}{2} \times 6 \times 4) = 24\text{ ft}^2\).
Two slanted sides: \(2 \times (5 \times 8) = 80\text{ ft}^2\). Floor: \(6 \times 8 = 48\text{ ft}^2\).
\(24 + 80 + 48 = 152\). Cost: \(152 \times \$2.50 = \$380\).
The tent needs \(152\text{ ft}^2\) of fabric costing \(\$380\).
See every chapter on the Grade 6 math page.