This answer key gives a worked solution to each of the 20 area, volume, and surface area problems. Each solution shows how the figure is split, every formula, and the arithmetic, and ends with the answer in bold with correct units. Use it to check your work and fix mistakes.
Problem 1: Parallelogram and triangle
- \(A = bh = 9 \times 6 = 54\).
- \(A = \frac{1}{2}bh = \frac{1}{2} \times 14 \times 5 = \frac{1}{2} \times 70 = 35\).
(a) \(54\text{ in}^2\). (b) \(35\text{ in}^2\).
Problem 2: Volume of a rectangular prism
\(V=\ell wh=8\times 5\times 3\).
\(8\times 5=40\) and \(40\times 3=120\).
\(120\text{ in}^3\)
Problem 3: Cube surface and volume
A cube has 6 identical square faces. One face: \(7 \times 7 = 49\). Surface area: \(6 \times 49 = 294\).
Volume: \(7 \times 7 \times 7 = 49 \times 7 = 343\).
Surface area \(294\text{ ft}^2\); volume \(343\text{ ft}^3\).
Problem 4: Area of a trapezoid
Use \(A=\frac{1}{2}(b_1+b_2)h\) with \(b_1=8\), \(b_2=14\), \(h=5\).
\(A=\frac{1}{2}(8+14)(5)=\frac{1}{2}(22)(5)=55\).
Check by splitting: two triangles with base \(3\) and height \(5\) have area \(7.5\) each, and the middle rectangle is \(8\times 5=40\). \(7.5+7.5+40=55\).
\(55\text{ in}^2\)
Problem 5: Trapezoid ramp
\(A=\frac{1}{2}(b_1+b_2)h=\frac{1}{2}(9+15)(6)\)
\(=\frac{1}{2}(24)(6)=12 \times 6=72\).
Check: the average of the bases is \(12\), and \(12 \times 6 = 72\).
\(72\text{ ft}^2\)
Problem 6: Area of an L-shape
Split the patio into a bottom rectangle \(12\text{ ft}\) by \(5\text{ ft}\) and a top rectangle \(7\text{ ft}\) by \(3\text{ ft}\).
\(12\times 5=60\) and \(7\times 3=21\), so \(60+21=81\).
Check: the full \(12\times 8=96\) minus the notch \(5\times 3=15\) is \(81\).
\(81\text{ ft}^2\)
Problem 7: Lawn with a pool
Whole yard: \(20 \times 12 = 240\).
Pool: \(5 \times 4 = 20\).
Grass: \(240 – 20 = 220\).
\(220\text{ ft}^2\)
Problem 8: Rectangle and triangle
Rectangle: \(10\times 6=60\).
Triangle: \(\frac{1}{2}\times 10\times 4=20\).
Total: \(60+20=80\).
\(80\text{ cm}^2\)
Problem 9: Volume of a triangular prism
Base area: \(B=\frac{1}{2}\times 6\times 5=15\).
\(V=Bh=15\times 12=180\).
\(180\text{ cm}^3\)
Problem 10: Right triangle prism
Base area: \(B=\frac{1}{2}\times 7\times 6=21\).
\(V=Bh=21\times 8=168\).
\(168\text{ in}^3\)
Problem 11: Surface area of a box
The faces come in three pairs.
Top and bottom: \(2\times (12\times 7)=2\times 84=168\).
Front and back: \(2\times (12\times 5)=2\times 60=120\).
Two sides: \(2\times (7\times 5)=2\times 35=70\).
\(168+120+70=358\).
\(358\text{ in}^2\)
Problem 12: Surface area from a net
Find the area of each face. Top and bottom: \(6 \times 4 = 24\) each. Front and back: \(6 \times 3 = 18\) each. Two sides: \(4 \times 3 = 12\) each.
Add the six faces: \(24+24+18+18+12+12=108\).
Check with pairs: \(2(24+18+12)=2(54)=108\).
\(108\text{ in}^2\)
Problem 13: Surface area of a triangular prism
Each base: \(B=\frac{1}{2}\times 12\times 5=30\), so the two bases total \(60\).
Rectangles: \(12\times 8=96\), \(5\times 8=40\), \(13\times 8=104\). Their sum is \(240\).
Surface area: \(60+240=300\).
\(300\text{ cm}^2\)
Problem 14: Missing prism length
Base area: \(B=\frac{1}{2}\times 10\times 6=30\).
\(V=Bh\), so \(270=30h\) and \(h=270: 30=9\).
Check: \(30\times 9=270\).
\(9\text{ cm}\)
Problem 15: Missing dimension
\(V=Bh\) and the base area is \(B=24\times 10=240\).
\(240h=3{,}600\), so \(h=3{,}600: 240=15\).
Check: \(24\times 10\times 15=3{,}600\).
\(15\text{ in}\)
Problem 16: Painting a toy chest
Lid: \(4 \times 2 = 8\). Front and back: \(2 \times (4 \times 2) = 16\). Two ends: \(2 \times (2 \times 2) = 8\).
Area to paint: \(8+16+8=32\text{ ft}^2\) (the bottom is left out).
Cans: \(32 : 16 = 2\). Cost: \(2 \times 9 = 18\).
\(32\text{ ft}^2\) to paint; the paint costs \(\$18\).
Problem 17: Volume of a composite solid
First find the area of the L-shaped face: \(8\times 6-3\times 2=48-6=42\).
Then \(V=Bh=42\times 5=210\).
\(210\text{ cm}^3\)
Problem 18: House-shaped solid
Rectangular part: \(V=6\times 4\times 10=240\text{ ft}^3\).
Roof: the triangle has area \(\frac{1}{2}\times 6\times 3=9\), so \(V=9\times 10=90\text{ ft}^3\).
Total: \(240+90=330\).
\(330\text{ ft}^3\)
Problem 19: Tiling a pool
(a) \(V=20\times 10\times 5=1{,}000\text{ ft}^3\).
(b) Floor: \(20\times 10=200\). Two long walls: \(2\times (20\times 5)=200\). Two short walls: \(2\times (10\times 5)=100\). The pool has no top, so the area is \(200+200+100=500\text{ ft}^2\).
(c) \(500\times 3=1{,}500\).
(a) \(1{,}000\text{ ft}^3\). (b) \(500\text{ ft}^2\). (c) \(\$1{,}500\).
Problem 20: Tent problem
- Base area: \(\frac{1}{2}\times 6\times 4=12\). Volume: \(12\times 8=96\). \(96\text{ ft}^3\)
- Two ends: \(2\times 12=24\). Two slanted sides: \(2\times (5\times 8)=80\). Total: \(24+80=104\). \(104\text{ ft}^2\)
The floor, \(6\times 8=48\text{ ft}^2\), is not counted.
See every chapter on the Grade 7 math page.