Appendix A — Answer Key, Chapter 12: Area and Perimeter of Composite Figures

SOL 8.MG.5 · Covers textbook Chapter 12 and the companion workbook. Item numbers match the textbook; workbook items are the same problems, so this key serves both. Item numbers run continuously from 1 to 104 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Conventions used throughout: every answer containing $\pi$ is given exactly first and then approximately, using $\pi \approx 3.14$. Areas carry square units, perimeters carry linear units.

Two checks are used repeatedly and are worth teaching as habits:

• Check an area by subdividing a second, different way. Two different cuts of the same figure must give the same number.
• Check a perimeter by tracing the boundary. Every interior subdivision line — including the diameter where a semicircle joins — is left out.

Lesson 12.1 — Subdividing a Figure and Adding the Pieces

Guided practice

1. $A = 30 + 18 = 48$ cm$^2$
2. $A = 36 + 12 = 48$ cm$^2$
3. $A = 60 - 12 = 48$ cm$^2$
4. Pieces: a rectangle $12$ cm by $9$ cm and a triangle with base $12$ cm and height $8$ cm. $A = 108 + \tfrac{1}{2}(12)(8) = 108 + 48 = 156$ cm$^2$. The $10$ cm sides are slant lengths, not the height.
5. Rectangle $= (14)(5) = 70$; trapezoid $= \tfrac{1}{2}(14 + 6)(6) = 60$; total $130$ cm$^2$.
6. Base $10$ cm, height $3$ cm, so the parallelogram is $(10)(3) = 30$ cm$^2$. With the rectangle, $A = 40 + 30 = 70$ cm$^2$. The $5$ cm slanted sides are not used for area.

Independent practice

7. $34$ cm$^2$. Way 1 (cut across): $(8)(3) + (5)(2) = 24 + 10 = 34$. Way 2 (subtraction): $(8)(5) - (3)(2) = 40 - 6 = 34$. A third correct cut: $(5)(5) + (3)(3) = 25 + 9 = 34$.
8. A rectangle $10$ cm by $4$ cm and a right triangle with legs $3$ cm and $4$ cm: $A = 40 + \tfrac{1}{2}(3)(4) = 40 + 6 = 46$ cm$^2$. Check as a single trapezoid with parallel sides $13$ and $10$ and height $4$: $\tfrac{1}{2}(13 + 10)(4) = 46$ ✓
9. Trapezoid formula: $\tfrac{1}{2}(16 + 10)(6) = \tfrac{1}{2}(26)(6) = 78$ cm$^2$. As pieces: a $10$ by $6$ rectangle plus two triangles of base $3$ and height $6$, so $60 + 9 + 9 = 78$ cm$^2$ ✓
10. Vertical strips: $(3)(9) + (3)(6) + (3)(3) = 27 + 18 + 9 = 54$ cm$^2$. Horizontal strips: $(9)(3) + (6)(3) + (3)(3) = 27 + 18 + 9 = 54$ cm$^2$ ✓
11. Area measures how much surface the figure covers. A subdivision line is imaginary — it takes nothing away and adds nothing — so it cannot change how much surface is there. Different cuts only group the same surface differently, so the totals must match.
12. Subtraction is faster when the figure is a full rectangle with one piece missing, as in Figure 2, because it takes only two products instead of finding unlabeled side lengths first. Adding is better when the figure is not a rectangle with a bite out of it — Figure 3, a rectangle with a triangle on top, has nothing natural to subtract from.
13. The student used $9$, the rectangle's height, as the triangle's height. The triangle's own height is $8$ cm, marked inside the triangle. Correct: $A = 108 + \tfrac{1}{2}(12)(8) = 108 + 48 = 156$ cm$^2$.
14. The student used $11$, the height of the whole figure, as the trapezoid's height. The trapezoid's height is $6$ cm, the distance between its two parallel sides. Correct: $\tfrac{1}{2}(14 + 6)(6) = 60$, and the total is $70 + 60 = 130$ cm$^2$.
15. Area $= 34$ ft$^2$; cost $= 34 \times 6 = \$204$.
16. Two pieces that meet along a line share only that line, and a line has no area — it has length but no width. Nothing is created at the seam, so the two pieces' areas add to exactly the area of the whole. (This is precisely the opposite of what happens with perimeter, where the shared edge does have to be accounted for.)

Exit ticket 12.1

17. $A = (12)(9) + \tfrac{1}{2}(12)(8) = 108 + 48 = 156$ cm$^2$
18. $A = (14)(5) + \tfrac{1}{2}(14 + 6)(6) = 70 + 60 = 130$ cm$^2$
19. Cut across: $30 + 18 = 48$ cm$^2$. Cut down: $36 + 12 = 48$ cm$^2$.
20. To subdivide a figure is to cut it, on paper, into simpler shapes whose area formulas you already know; we do it because the whole figure has no formula of its own but every piece does.

Lesson 12.2 — Circles, Semicircles, and Pieces Removed

Guided practice

21. $A = \pi(4)^2 = 16\pi$ cm$^2$, or about $50.24$ cm$^2$
22. $A = \tfrac{1}{2}\pi(6)^2 = \tfrac{1}{2}\pi(36) = 18\pi$ cm$^2$, or about $56.52$ cm$^2$
23. Rectangle $(10)(6) = 60$ cm$^2$; semicircle $\tfrac{1}{2}\pi(5)^2 = 12.5\pi \approx 39.25$ cm$^2$; total $60 + 12.5\pi$ cm$^2$, or about $99.25$ cm$^2$.
24. $A = (20)(12) - \pi(4)^2 = 240 - 16\pi$ m$^2$, or about $189.76$ m$^2$
25. $A = (12)(8) - \tfrac{1}{2}\pi(3)^2 = 96 - 4.5\pi$ cm$^2$, or about $81.87$ cm$^2$
26. The flat side of the semicircle lies along the top of the rectangle, so the $10$ cm is the diameter; the radius is half of it, $5$ cm.

Independent practice

27. $A = (14)(6) + \tfrac{1}{2}\pi(3)^2 = 84 + 4.5\pi$ cm$^2$, or about $98.13$ cm$^2$
28. $A = (12)(12) - \tfrac{1}{2}\pi(6)^2 = 144 - 18\pi$ cm$^2$, or about $87.48$ cm$^2$
29. $A = (16)(10) + 2\left(\tfrac{1}{2}\pi(5)^2\right) = 160 + 25\pi$ cm$^2$, or about $238.5$ cm$^2$
30. $A = (18)(12) - \pi(3)^2 = 216 - 9\pi$ cm$^2$, or about $187.74$ cm$^2$
31. The two semicircles are congruent, each with radius $5$, so together they are one whole circle of radius $5$: $\pi(5)^2 = 25\pi$. Adding halves gives $12.5\pi + 12.5\pi = 25\pi$ as well, so both methods give $160 + 25\pi \approx 238.5$ cm$^2$ ✓
32. The student used the $6$ cm width of the rectangle as the radius. That $6$ cm is the semicircle's diameter, so $r = 3$. Correct: $84 + \tfrac{1}{2}\pi(3)^2 = 84 + 4.5\pi \approx 98.13$ cm$^2$.
33. The circle is a hole, so its area must be subtracted, not added. Correct: $216 - 9\pi \approx 187.74$ cm$^2$.
34. $A = (4)(5) + \tfrac{1}{2}\pi(2)^2 = 20 + 2\pi$ ft$^2$, or about $26.28$ ft$^2$
35. $A = 144 - 18\pi \approx 87.48$ in$^2$; cost $= 87.48 \times 0.40 = \$34.99$ (rounded from $\$34.992$).
36. $60 + 12.5\pi$ is the exact value. Writing $99.25$ requires replacing $\pi$ with $3.14$, which is itself only an approximation of $\pi = 3.14159\ldots$, so the decimal is already slightly off before any further rounding. Keeping $\pi$ also lets later steps cancel or combine exactly.

Exit ticket 12.2

37. $96 - 4.5\pi$ cm$^2$, or about $81.87$ cm$^2$
38. $60 + 12.5\pi$ cm$^2$, or about $99.25$ cm$^2$
39. $216 - 9\pi$ cm$^2$, or about $187.74$ cm$^2$
40. Attaching a semicircle adds $\tfrac{1}{2}\pi r^2$ to the area; removing one subtracts $\tfrac{1}{2}\pi r^2$. The formula for the piece is identical either way — only the sign in front of it changes, and the figure tells you which it is: material added on the outside, or material taken out.

Lesson 12.3 — The Perimeter of a Composite Figure

Guided practice

41. $2(12 + 4) = 32$ units and $2(4 + 6) = 20$ units; together $52$ units of edge.
42. Tracing: $4 + 6 + 4 + 4 + 12 + 4 + 4 + 6 = 44$ units. It is not $52$ because the $4$-unit edge where the stem meets the bar is now inside the figure. Each rectangle's own perimeter counted that edge once, so the sum counted it twice when the boundary counts it zero times: $52 - 2(4) = 44$.
43. $P = 10 + 3 + 4 + 3 + 6 + 6 = 32$ cm
44. $P = 10 + 6 + 6 + \pi(5) = 22 + 5\pi$ cm, or about $37.7$ cm. The $10$ cm segment where the semicircle meets the rectangle — the semicircle's diameter — was left out, because it is a seam inside the figure, not part of the outside boundary.
45. $P = 12 + 8 + 3 + \pi(3) + 3 + 8 = 34 + 3\pi$ cm, or about $43.42$ cm. Cutting the notch out increased the perimeter: $6$ cm of straight top edge was replaced by an arc about $9.42$ cm long.
46. $P = 16 + 6 + 6 + 4 + 10 + 10 = 52$ ft

Independent practice

47. $P = 12 + 4 + 5 + 3 + 7 + 7 = 38$ cm
48. Tracing: $P = 4 + 5 + 3 + 3 + 10 + 3 + 3 + 5 = 36$ cm. The two rectangles apart: $2(10 + 3) = 26$ and $2(4 + 5) = 18$, totaling $44$ cm. The difference is $44 - 36 = 8$ cm, and the shared edge is $4$ cm, so the difference is exactly $2(4) = 8$ ✓
49. $P = 8 + 5 + 5 + \pi(4) = 18 + 4\pi$ cm, or about $30.56$ cm
50. $P = 12 + 9 + 10 + 10 + 9 = 50$ cm. The $8$ cm triangle height is interior and is not walked.
51. $P = 14 + 11 + 6 + 10 + 5 = 46$ cm
52. $P = 10 + 4 + 5 + 10 + 5 + 4 = 38$ cm. The $3$ cm parallelogram height is interior; the $5$ cm slants are what you walk.
53. The student added the two rectangles' perimeters, which counts the $4$ cm shared edge twice even though it is not on the boundary at all. Correct: trace the outside, $P = 36$ cm — or correct the sum, $44 - 2(4) = 36$ cm.
54. The student included the $8$ cm top of the rectangle. That segment is the semicircle's diameter, a seam inside the figure. The arc replaces it. Correct: $P = 8 + 5 + 5 + 4\pi = 18 + 4\pi \approx 30.56$ cm.
55. Each piece's perimeter includes the shared edge once, so the sum includes it twice. The joined figure's boundary includes it zero times, because the edge now has material on both sides of it. The sum therefore exceeds the true perimeter by $2 \times$ (length of the shared edge), every time.
56. The rectangle's perimeter is $2(6 + 8) = 28$ cm, which is less than the L-shape's $32$ cm even though both have area $48$ cm$^2$. Area and perimeter are independent measurements: knowing one never determines the other, and among figures of a given area the more "spread out" or notched shapes have the larger perimeter.

Exit ticket 12.3

57. $P = 16 + 7 + 10 + 10 + 7 = 50$ cm. The $6$ cm triangle height is not part of the boundary.
58. $P = 4 + 5 + 5 + \pi(2) = 14 + 2\pi$ ft, or about $20.28$ ft
59. $P = 80 + 80 + 2\pi(25) = 160 + 50\pi$ m, or about $317$ m. The two $50$ m ends of the rectangle are seams and are not part of the boundary.
60. A subdivision line lies inside the figure with material on both sides of it, so it is not on the outside boundary — and the perimeter measures only the outside boundary.

Lesson 12.4 — Composite Figures in Context

Guided practice

61. $A = (10)(10) + (6)(6) = 136$ ft$^2$ (check: $(16)(10) - (6)(4) = 160 - 24 = 136$ ✓); cost $= 136 \times 9 = \$1224$.
62. $P = 14 + 2\pi \approx 20.28$ ft, so buy $21$ ft — $20$ ft would leave a gap. Cost $= 21 \times 3 = \$63$.
63. $P = 160 + 50\pi \approx 317$ m; two laps $\approx 634$ m.
64. $A = (80)(50) + \pi(25)^2 = 4000 + 625\pi$ m$^2$, or about $5962.5$ m$^2$
65. $A = 240 - 16\pi$ m$^2$, or about $189.76$ m$^2$

Independent practice

66. Walk only $= (26)(18) - (20)(12) = 468 - 240 = 228$ ft$^2$; cost $= 228 \times 8 = \$1824$.
67. $P = 2(26 + 18) = 88$ ft; cost $= 88 \times 6 = \$528$.
68. Soil (area) $= (15)(8) + \tfrac{1}{2}\pi(4)^2 = 120 + 8\pi$ ft$^2$, or about $145.12$ ft$^2$. Edging (perimeter) $= 15 + 15 + 8 + \pi(4) = 38 + 4\pi$ ft, or about $50.56$ ft.
69. Glass $= (6)(3) + 2\left(\tfrac{1}{2}\pi(1.5)^2\right) = 18 + 2.25\pi$ ft$^2$, or about $25.07$ ft$^2$. Trim $= 6 + 6 + 2\left(\pi(1.5)\right) = 12 + 3\pi$ ft, or about $21.42$ ft.
70. $(26)(18) = 468$ ft$^2$ is the area of the walk and the pool together. The pool is not part of the walk, so its $240$ ft$^2$ must come out. Correct: $228$ ft$^2$.
71. The $8$ ft side is where the semicircular end joins the rectangle — it is the semicircle's diameter, an interior seam, so no edging runs along it. The arc runs there instead. Correct: $38 + 4\pi \approx 50.56$ ft.
72. Glass covers a surface, so it is an area, in square feet. Trim runs around the outside edge, so it is a perimeter, in feet.
73. Soil: $145.12 \times 4 = \$580.48$. Edging: $50.56 \times 3 = \$151.68$. Total $= \$732.16$.
74. A rate's units must cancel the measurement's units and leave dollars. Dollars per foot cancels feet, so it needs a length — a perimeter. Dollars per square foot cancels square feet, so it needs an area. If your measurement's units do not match the rate's, you have computed the wrong measure.

Exit ticket 12.4

75. $136 \times 0.75 = \$102.00$
76. $2(160 + 50\pi) = 320 + 100\pi \approx 634$ m
77. $12 + 3\pi$ ft, or about $21.42$ ft
78. A fence runs along the edge of a yard, and edges are measured by length, in feet. Sod covers the surface inside the edge, and surfaces are measured by area, in square feet — one measures a path, the other measures a covering.

Chapter 12 Review

Part A — Subdividing and finding area (8.MG.5a)

79. $96$ ft$^2$. Way 1 (cut across): $(14)(4) + (8)(5) = 56 + 40 = 96$. Way 2 (subtraction): $(14)(9) - (6)(5) = 126 - 30 = 96$ ✓
80. Triangle base $12$ ft, height $8$ ft. $A = (12)(5) + \tfrac{1}{2}(12)(8) = 60 + 48 = 108$ ft$^2$.
81. $A = (20)(10) - \tfrac{1}{2}\pi(5)^2 = 200 - 12.5\pi$ ft$^2$, or about $160.75$ ft$^2$
82. Trapezoid formula: $(10)(4) + \tfrac{1}{2}(10 + 4)(8) = 40 + 56 = 96$ ft$^2$. Second way (a $10$ by $12$ rectangle with a triangle of legs $6$ and $8$ removed): $120 - \tfrac{1}{2}(6)(8) = 120 - 24 = 96$ ft$^2$ ✓
83. $A = (10)(6) - (4)(3) = 48$ cm$^2$
84. $60 + 12.5\pi$ cm$^2$, or about $99.25$ cm$^2$
85. $240 - 16\pi$ m$^2$, or about $189.76$ m$^2$
86. $96 - 4.5\pi$ cm$^2$, or about $81.87$ cm$^2$
87. The cuts are imaginary lines with no area, so they neither add nor remove surface. Both subdivisions describe the same surface, just grouped differently, and grouping does not change a total.
88. $13$ ft is the height of the whole figure, rectangle plus triangle. The triangle's own height is $8$ ft — the distance from its base up to the apex. Correct: $A = 60 + \tfrac{1}{2}(12)(8) = 108$ ft$^2$.

Part B — Using the subdivisions to find perimeter (8.MG.5b)

89. $P = 14 + 4 + 6 + 5 + 8 + 9 = 46$ ft
90. $P = 12 + 5 + 10 + 10 + 5 = 42$ ft. The $8$ ft height is interior.
91. $P = 20 + 20 + 10 + \pi(5) = 50 + 5\pi$ ft, or about $65.7$ ft. The removed semicircle's $10$ ft diameter is gone from the boundary; the arc took its place.
92. $P = 10 + 12 + 4 + 10 + 4 = 40$ ft
93. Tracing: $4 + 6 + 4 + 4 + 12 + 4 + 4 + 6 = 44$ units. By pieces, corrected: $32 + 20 - 2(4) = 44$ units ✓
94. $P = 14 + 2\pi$ ft, or about $20.28$ ft
95. The student added the perimeters of the two rectangles the figure was cut into, which counts the $8$ ft shared edge twice; the boundary contains it zero times. Correct: $62 - 2(8) = 46$ ft, which matches tracing the outside.
96. The arc is on the outside of the figure — it is the part you would walk along or run trim around. The diameter is the seam where the semicircle was attached, so it has material on both sides of it and lies inside the figure. Perimeter measures only the outside boundary, so the arc counts and the diameter does not.

Part C — Contextual problems (8.MG.5c)

97. $A = 96$ ft$^2$; cost $= 96 \times 7 = \$672$.
98. $P = 46$ ft; cost $= 46 \times 2.50 = \$115$.
99. $A = 108$ ft$^2$; $108 \div 25 = 4.32$, so 5 whole pints are needed — $4$ pints would cover only $100$ ft$^2$.
100. $A = 200 - 12.5\pi \approx 160.75$ ft$^2$; cost $= 160.75 \times 1.20 = \$192.90$.
101. Area $= 96$ ft$^2$; trim $= 40$ ft.
102. Resurfacing (area): $4000 + 625\pi \approx 5962.5$ m$^2$, at $8 per m$^2$ gives $\$47{,}700$. Rope (perimeter): $160 + 50\pi \approx 317$ m, at $1.50 per m gives $\$475.50$.
103. For fencing, measure the perimeter — the distance around the edge of the yard, in feet — because fence runs along a line. For sod, measure the area — the surface inside that edge, in square feet — because sod covers a region. The two answers are different kinds of numbers: one is a length, counted in feet, and the other is a covering, counted in squares one foot on a side. That is why they can never be compared or substituted for one another.
104. Two errors. First, trim goes around the boundary, so the problem needs the perimeter, not the area. Second, the rate $3 per foot cannot be applied to a quantity in square feet — the units do not match, which is itself the signal that the wrong measure was chosen. Correct: $P = 14 + 2\pi \approx 20.28$ ft, so buy $21$ ft, and $21 \times 3 = \$63$.

Workbook-only items

Page 2, area formulas. rectangle $A = lw$ · square $A = s^2$ · triangle $A = \tfrac{1}{2}bh$ · parallelogram $A = bh$ · trapezoid $A = \tfrac{1}{2}(b_1 + b_2)h$ · circle $A = \pi r^2$ · semicircle $A = \tfrac{1}{2}\pi r^2$

Page 2, why did both give the same number? The subdivision line is imaginary and has no area, so both cuts describe the same surface; only the grouping changed.

Page 3, item 4 frame. Pieces: a rectangle and a triangle. Rectangle $12 \times 9 = 108$. Triangle $\tfrac{1}{2}(12)(8) = 48$. Total $156$ cm$^2$.

Page 3, item 5 frame. Rectangle $70$. Trapezoid $\tfrac{1}{2}(14 + 6)(6) = 60$. Total $130$ cm$^2$.

Page 3, item 6 frame. Base $10$, height $3$, area $30$. Total $70$ cm$^2$.

Page 7, complete. $A_{\text{circle}} = \pi r^2$ and $A_{\text{semicircle}} = \tfrac{1}{2}\pi r^2$. The straight edge is the diameter; the curved edge is the arc. The diameter is $10$ cm and the radius is $5$ cm.

Page 8, item 24 frame. $(20)(12) - \pi(\mathbf{4})^2 = 240 - 16\pi \approx 189.76$ m$^2$

Page 8, item 25 frame. $(12)(8) - \tfrac{1}{2}\pi(\mathbf{3})^2 = 96 - 4.5\pi \approx 81.87$ cm$^2$

Page 12, item 41 frame. $2(12 + 4) = \mathbf{32}$ and $2(4 + 6) = \mathbf{20}$, total $\mathbf{52}$ units of edge.

Page 12, item 42 frame. $4 + 6 + 4 + 4 + 12 + 4 + 4 + 6 = \mathbf{44}$ units. The shared edge is $\mathbf{4}$ units long, and it was counted $\mathbf{2}$ times instead of $\mathbf{0}$.

Page 13, item 43 frame. $P = \mathbf{32}$ cm.

Page 13, item 44 frame. Arc length of a semicircle $= \pi r$. $P = 10 + 6 + 6 + \pi(\mathbf{5}) = 22 + 5\pi \approx 37.7$ cm. Left out: the $10$ cm diameter, because it is the seam inside the figure.

Page 13, item 45 frame. $P = 34 + 3\pi \approx 43.42$ cm.

Page 13, item 46 frame. $P = 52$ ft.

Page 14, item 48 frame. Apart: $26 + 18 = 44$. Difference $44 - 36 = 8$. Shared edge $4$. Yes — the difference is twice the shared edge.

Page 17, which measure table. Tile, sod, paint, glass, mulch, sealer → area, in square units. Fence, trim, edging, ribbon, running → perimeter, in linear units. Dollars per square foot multiplies an area; dollars per foot multiplies a perimeter.

Page 18, item 63 frame. $80 + 80 + 2\pi(\mathbf{25}) = 160 + 50\pi \approx 317$ m. Two laps: about $634$ m.

Page 18, item 64 frame. $(80)(50) + \pi(\mathbf{25})^2 = 4000 + 625\pi \approx 5962.5$ m$^2$

Page 20, item 72 frame. Glass: area, units ft$^2$. Trim: perimeter, units ft.

Page 20, item 73 frame. Soil $580.48, edging $151.68, total $732.16.

Page 22, item 79 frame. Way 1: $56 + 40 = 96$. Way 2: $126 - 30 = 96$. They agree.

Page 22, item 80 frame. Base $12$, height $8$, total $108$ ft$^2$.

Page 23, item 93 frame. By tracing: $44$. By pieces, corrected: $32 + 20 - 2(4) = 44$.