Appendix A — Answer Key, Chapter 14: Circles: Circumference and Area

SOL 6.MG.1 · Covers textbook Chapter 14 and the companion workbook. Item numbers match the textbook; where the workbook repeats a textbook problem, this key serves both, and workbook-only items appear in the final section. All values use $\pi \approx 3.14$, so every circumference and area is an approximation. Reasoning answers show an acceptable response, not the only wording.

Lesson 14.1 — Parts of a Circle

Guided practice

1. A radius.
2. A diameter.
3. $d = 2 \times 9 = 18$ m
4. $r = 26 \div 2 = 13$ ft
5. Yes, every diameter is a chord, because both of its endpoints are on the circle. No, not every chord is a diameter, because a chord only counts as a diameter if it passes through the center.

Independent practice

6. Radius: $\overline{OA}$. Diameter: $\overline{BC}$. Chord that is not a diameter: $\overline{DE}$.
7. a) $12$ cm b) $9$ in c) $25$ m
8. a) $15$ ft b) $3.5$ cm c) $7.5$ in
9. Circumference is a length, measured in units such as centimeters or feet. Area is a surface, measured in square units such as $\text{cm}^2$ or $\text{ft}^2$.
10. The radius always runs from the center to the circle, so an $8$ cm radius means every point of the circle is $8$ cm from the center. An $8$ cm chord just connects two points on the circle and does not have to touch the center. In this circle the diameter is $16$ cm, so an $8$ cm chord is a short one that misses the center.
11. The $40$ ft measurement is the diameter. The distance from the center to the edge is $40 \div 2 = 20$ ft, which is the radius.
12. Any chord that misses the center takes a shortcut across the circle, so it is shorter than the full width. Only the chord through the center spans the full width, and that chord is the diameter. So the diameter is the longest chord, and nothing can beat it.

Exit ticket 14.1

1. $22$ cm
2. $4.5$ in
3. A chord is a segment whose two endpoints both sit on the circle.
4. Circumference measures a distance — how far it is once around — so it uses length units. Area measures how much flat surface is covered, which means multiplying a length by a length, so it uses square units.

Lesson 14.2 — Discovering Pi

Guided practice

1. $15.7 \div 5.0 = 3.14$
2. $25.3 \div 8.0 = 3.1625 \approx 3.16$
3. $75.2 \div 24.0 = 3.1333\ldots \approx 3.13$
4. $(3.14 + 3.16 + 3.13) \div 3 = 9.43 \div 3 = 3.1433\ldots \approx 3.14$
5. Measuring is never perfect. The string stretches or slips, the ruler mark falls between lines, and objects are not perfectly round. Those small errors move the ratio a few hundredths, so measured values land near $3.14$ rather than exactly on it.

Independent practice

6. a) $12.6 \div 4.0 = 3.15$ b) $47.1 \div 15.0 = 3.14$ c) $29.8 \div 9.5 = 3.1368\ldots \approx 3.14$ d) $69.1 \div 22.0 = 3.1409\ldots \approx 3.14$
7. $(3.15 + 3.14 + 3.14 + 3.14) \div 4 = 12.57 \div 4 = 3.1425 \approx 3.14$. It is close to pi.
8. $C \approx 3.14 \times 20 = 62.8$ cm
9. About $3.14$. The ratio of circumference to diameter is the same for every circle, no matter its size, so a circle the size of Earth has the same ratio as a bottle cap.
10. The wheel travels exactly its own circumference in one full turn, because the part of the rim that touches the ground unrolls along the ground. For $d = 0.5$ m, $C \approx 3.14 \times 0.5 = 1.57$ m.
11. The student's recorded diameter was $50 \div 3.60 \approx 13.9$ cm. The tray's diameter should be about $50 \div 3.14 \approx 15.9$ cm. The recorded diameter is about $2$ cm too small, which usually means the measurement was taken along a chord that missed the center rather than straight through it.
12. One circle gives you one ratio, and you cannot tell whether it is off because of your measuring or because of the circle. Measuring many circles of very different sizes shows that the ratio stays the same no matter the size, and averaging the results cancels out much of the random measurement error.

Exit ticket 14.2

1. $18.9 \div 6.0 = 3.15$
2. About $3.14$; it is called pi, written $\pi$.
3. A little more than $3$ diameters.
4. Because measuring introduces small errors — slipping string, ruler estimates, objects that are not perfectly round — so the computed ratios scatter slightly around $3.14$.

Lesson 14.3 — Circumference

Guided practice

1. $C \approx 3.14 \times 10 = 31.4$ cm
2. $C \approx 3.14 \times 20 = 62.8$ in
3. $C \approx 2 \times 3.14 \times 4 = 25.12$ ft
4. $C \approx 2 \times 3.14 \times 9 = 56.52$ m
5. $d = 2 \times 6 = 12$ cm; $C \approx 3.14 \times 12 = 37.68$ cm

Independent practice

6. a) $C \approx 3.14 \times 5 = 15.7$ cm b) $C \approx 3.14 \times 30 = 94.2$ mm c) $C \approx 2 \times 3.14 \times 11 = 69.08$ in d) $C \approx 2 \times 3.14 \times 2.5 = 15.7$ m
7. $C \approx 3.14 \times 7.5 = 23.55$ cm
8. $d \approx 94.2 \div 3.14 = 30$ ft; $r = 30 \div 2 = 15$ ft
9. Circle A: $C \approx 2 \times 3.14 \times 3 = 18.84$ cm. Circle B: $C \approx 2 \times 3.14 \times 6 = 37.68$ cm. Doubling the radius doubled the circumference, since $2 \times 18.84 = 37.68$.
10. One turn: $C \approx 3.14 \times 26 = 81.64$ in. Ten turns: $10 \times 81.64 = 816.4$ in.
11. $C \approx 2 \times 3.14 \times 3.5 = 21.98$ ft. Edging is sold by the whole foot, so the gardener should buy $22$ ft; $21$ ft would leave a gap.
12. She used the radius in the formula meant for the diameter. Either double the radius first ($d = 16$, so $C \approx 3.14 \times 16 = 50.24$ cm) or use $C = 2\pi r \approx 2 \times 3.14 \times 8 = 50.24$ cm. The correct circumference is about $50.24$ cm. Rule: $\pi$ multiplies the diameter, so with a radius you must also multiply by $2$.

Exit ticket 14.3

1. $C \approx 3.14 \times 15 = 47.1$ cm
2. $C \approx 2 \times 3.14 \times 5 = 31.4$ in
3. $d \approx 31.4 \div 3.14 = 10$ m
4. Because $d = 2r$ for every circle. Substituting $2r$ for $d$ in $C = \pi d$ gives $C = \pi(2r) = 2\pi r$, so the two formulas are the same statement written with different given information.

Mid-chapter check (Lessons 14.1–14.3)

1. $d = 2 \times 14 = 28$ cm
2. $r = 11 \div 2 = 5.5$ in
3. The diameter.
4. $34.5 \div 11.0 = 3.1363\ldots \approx 3.14$
5. $C \approx 3.14 \times 25 = 78.5$ ft
6. $C \approx 2 \times 3.14 \times 12 = 75.36$ m
7. $d \approx 18.84 \div 3.14 = 6$ cm, so $r = 3$ cm.
8. $C \approx 3.14 \times 4 = 12.56$ ft of ribbon.

Lesson 14.4 — Area of a Circle

Guided practice

1. $A \approx 3.14 \times 5^2 = 3.14 \times 25 = 78.5\ \text{cm}^2$
2. $A \approx 3.14 \times 10^2 = 3.14 \times 100 = 314\ \text{in}^2$
3. $r = 8 \div 2 = 4$ ft; $A \approx 3.14 \times 16 = 50.24\ \text{ft}^2$
4. $r = 20 \div 2 = 10$ m; $A \approx 3.14 \times 100 = 314\ \text{m}^2$
5. $A \approx 3.14 \times 1.5^2 = 3.14 \times 2.25 = 7.065\ \text{cm}^2$

Independent practice

6. a) $A \approx 3.14 \times 9 = 28.26\ \text{cm}^2$ b) $A \approx 3.14 \times 49 = 153.86\ \text{in}^2$ c) $r = 6$, $A \approx 3.14 \times 36 = 113.04\ \text{ft}^2$ d) $r = 15$, $A \approx 3.14 \times 225 = 706.5\ \text{m}^2$
7. $A \approx 3.14 \times 2.5^2 = 3.14 \times 6.25 = 19.625\ \text{in}^2$
8. $C \approx 2 \times 3.14 \times 6 = 37.68$ cm and $A \approx 3.14 \times 36 = 113.04\ \text{cm}^2$. The circumference is a single distance around the edge, so it stays in centimeters. The area comes from multiplying two lengths together, so its unit is centimeters times centimeters, or square centimeters.
9. Circle A: $A \approx 3.14 \times 16 = 50.24\ \text{m}^2$. Circle B: $A \approx 3.14 \times 64 = 200.96\ \text{m}^2$. Doubling the radius multiplied the area by $4$, since $4 \times 50.24 = 200.96$. Doubling inside a square doubles twice.
10. $r = 16 \div 2 = 8$ in; $A \approx 3.14 \times 64 = 200.96\ \text{in}^2$
11. $r = 5 \div 2 = 2.5$ ft; $A \approx 3.14 \times 6.25 = 19.625\ \text{ft}^2$
12. Devon squared the diameter instead of the radius. The radius is $10 \div 2 = 5$ cm, so $A \approx 3.14 \times 25 = 78.5\ \text{cm}^2$. Devon's answer was $4$ times too large, because $314 \div 78.5 = 4$; squaring a doubled length multiplies the result by $2^2 = 4$.

Exit ticket 14.4

1. $A \approx 3.14 \times 81 = 254.34\ \text{cm}^2$
2. $r = 2$ in; $A \approx 3.14 \times 4 = 12.56\ \text{in}^2$
3. $r^2 \approx 28.26 \div 3.14 = 9$, so $r = 3$ m.
4. Area is found by multiplying a length by a length, which makes the unit a unit times itself — a square unit. Circumference is a single distance around the edge, so it keeps a plain length unit.

Lesson 14.5 — Circle Problems in Context

Guided practice

1. $A \approx 3.14 \times 12^2 = 3.14 \times 144 = 452.16\ \text{ft}^2$
2. $C \approx 3.14 \times 40 = 125.6$ ft of fence.
3. $A \approx 3.14 \times 16 = 50.24\ \text{ft}^2$
4. $C \approx 3.14 \times 14 = 43.96$ ft of padding.
5. $r = 24 \div 2 = 12$ in; $A \approx 3.14 \times 144 = 452.16\ \text{in}^2$

Independent practice

6. $C \approx 3.14 \times 18 = 56.52$ ft around. $r = 9$ ft, so $A \approx 3.14 \times 81 = 254.34\ \text{ft}^2$.
7. $C \approx 3.14 \times 2 = 6.28$ ft per turn; $100 \times 6.28 = 628$ ft.
8. $C \approx 2 \times 3.14 \times 10 = 62.8$ ft; cost $\approx 62.8 \times 4 = \$251.20$.
9. $A \approx 3.14 \times 49 = 153.86\ \text{in}^2$
10. Each $10$-inch pizza has $r = 5$ in and area $\approx 3.14 \times 25 = 78.5\ \text{in}^2$, so two of them give $157\ \text{in}^2$. The $16$-inch pizza has $r = 8$ in and area $\approx 3.14 \times 64 = 200.96\ \text{in}^2$. The one $16$-inch pizza gives more, by $200.96 - 157 = 43.96\ \text{in}^2$.
11. A $6$ in overhang is $0.5$ ft on each side, adding $1$ ft to the width, so the cloth diameter is $5 + 1 = 6$ ft. Then $r = 3$ ft and $A \approx 3.14 \times 9 = 28.26\ \text{ft}^2$ of fabric.
12. For $d = 6$ cm: $C \approx 3.14 \times 6 = 18.84$ cm, and with $r = 3$, $A \approx 3.14 \times 9 = 28.26\ \text{cm}^2$. For $d = 12$ cm: $C \approx 3.14 \times 12 = 37.68$ cm, and with $r = 6$, $A \approx 3.14 \times 36 = 113.04\ \text{cm}^2$. The circumference doubled ($2 \times 18.84 = 37.68$) and the area was multiplied by $4$ ($4 \times 28.26 = 113.04$). Circumference depends on the diameter to the first power, so doubling the diameter doubles it once. Area depends on the radius squared, so the doubling happens twice: $2 \times 2 = 4$.

Exit ticket 14.5

1. $C \approx 3.14 \times 6 = 18.84$ ft of trim.
2. $A \approx 3.14 \times 81 = 254.34\ \text{ft}^2$
3. $C \approx 3.14 \times 3 = 9.42$ ft per turn; $50 \times 9.42 = 471$ ft.
4. Area, $A = \pi r^2$, because sod covers the surface inside the lawn rather than going around its edge. The answer will be in square feet.

Chapter 14 Review

Part A — Identifying and describing the parts of a circle (6.MG.1a)

1. a) radius b) chord c) circumference d) area
2. Both are segments with endpoints on the circle. A diameter must also pass through the center, which makes it the longest chord. A chord that misses the center is shorter.
3. Area, because it is found by multiplying a length by a length, so the unit becomes a square unit.
4. The longest chord is the diameter: $d = 2 \times 15 = 30$ cm.

Part B — Relationships among radius, diameter, and circumference (6.MG.1b)

5. a) $16$ in b) $7$ ft
6. a) $25$ m b) $4.5$ cm
7. $r = 2$: $C \approx 2 \times 3.14 \times 2 = 12.56$ cm. $r = 6$: $C \approx 2 \times 3.14 \times 6 = 37.68$ cm. Tripling the radius tripled the circumference, since $3 \times 12.56 = 37.68$.
8. About $3.14$ times. Wrapping a string around a circle and comparing it to the diameter always fits three diameters plus a little more, and that constant multiplier is pi.

Part C — Approximating pi from data (6.MG.1c)

9. a) $25.1 \div 8.0 = 3.1375 \approx 3.14$ b) $44.0 \div 14.0 = 3.1428\ldots \approx 3.14$
10. $(3.14 + 3.14) \div 2 = 3.14$. It approximates pi.
11. Every careful measurement lands near $3.14$, and $3.55$ is far outside that range, so the cause is an error rather than a different ratio. A likely cause is measuring the diameter along a chord that missed the center, making it too short, or letting the string sag so the circumference came out too long.

Part D — Developing and using the circumference formula (6.MG.1d)

12. Start with $\pi = \dfrac{C}{d}$ and multiply both sides by $d$. On the right, $\dfrac{C}{d} \times d = C$, leaving $\pi d = C$, or $C = \pi d$.
13. Every diameter is two radii long, so $d = 2r$. Substituting into $C = \pi d$ gives $C = \pi(2r) = 2\pi r$.
14. a) $C \approx 3.14 \times 22 = 69.08$ cm b) $C \approx 2 \times 3.14 \times 15 = 94.2$ in
15. $d \approx 43.96 \div 3.14 = 14$ ft; $r = 14 \div 2 = 7$ ft

Part E — Solving problems involving circumference and area (6.MG.1e)

16. a) $A \approx 3.14 \times 144 = 452.16\ \text{cm}^2$ b) $r = 13$ in, $A \approx 3.14 \times 169 = 530.66\ \text{in}^2$
17. $C \approx 2 \times 3.14 \times 6 = 37.68$ ft around; $A \approx 3.14 \times 36 = 113.04\ \text{ft}^2$
18. One turn: $C \approx 3.14 \times 30 = 94.2$ in. Twenty turns: $20 \times 94.2 = 1{,}884$ in.
19. $r = 10$ ft, so $A \approx 3.14 \times 100 = 314\ \text{ft}^2$ and the concrete costs $314 \times 5 = \$1{,}570$. The border is the circumference: $C \approx 3.14 \times 20 = 62.8$ ft, costing $62.8 \times 8 = \$502.40$. Total: $1{,}570 + 502.40 = \$2{,}072.40$.
20. Circumferences: $C \approx 2 \times 3.14 \times 5 = 31.4$ cm and $C \approx 2 \times 3.14 \times 10 = 62.8$ cm, which is exactly double. Areas: $A \approx 3.14 \times 25 = 78.5\ \text{cm}^2$ and $A \approx 3.14 \times 100 = 314\ \text{cm}^2$, which is four times as much. In $C = 2\pi r$ the radius appears once, so doubling it doubles the answer. In $A = \pi r^2$ the radius is used twice, so doubling it multiplies the answer by $2 \times 2 = 4$.

Workbook-only items

Page 2, definitions. A radius goes from the center to a point on the circle. A diameter passes through the center and has both endpoints on the circle. A chord has both endpoints on the circle. Circumference is the distance around the circle; units: units of length (cm, in, ft, m). Area is the surface inside the circle; units: square units ($\text{cm}^2$, $\text{in}^2$, $\text{ft}^2$, $\text{m}^2$).

Page 2, true or false. Every diameter is a chord — T. Every chord is a diameter — F. All radii of one circle are the same length — T. The diameter is the longest chord — T.

Page 3, radius and diameter table. $6$ cm → $12$ cm; $4.5$ in → $9$ in; $12.5$ m → $25$ m; $15$ ft ← $30$ ft; $3.5$ cm ← $7$ cm; $7.5$ in ← $15$ in; $14$ cm → $28$ cm; $5.5$ in ← $11$ in.

Page 3, apply it. The $40$ ft measurement is the diameter. Center to edge: $20$ ft, called the radius.

Page 3, explain. A chord that misses the center cuts across less than the full width of the circle, so it is shorter than the diameter, which spans the full width.

Page 5, string. About 3 diameters and a little more. Student data tables vary; any three ratios computed correctly and averaging near $3.14$ are acceptable. The number is pi, and $\pi \approx 3.14$.

Page 6, ratio table. Jar lid $3.14$; Cup rim $3.16$; Platter $3.13$; Bottle cap $3.15$; Salad bowl $3.14$; Mug $3.14$; Wastebasket $3.14$.

Page 6, what do they have in common? Every ratio is close to $3.14$, even though the objects are very different sizes.

Page 6, explain. Measurement error. String can slip or stretch, rulers are read to the nearest mark, and objects are not perfectly round.

Page 8, derivation. $\pi = \dfrac{C}{\mathbf{d}}$, so $C = \pi \times \mathbf{d}$. Because $d = 2r$, $C = \pi(\mathbf{2r}) = \mathbf{2}\pi r$.

Page 8, which formula. Radius $= 7$ m → $C = 2\pi r$. Diameter $= 7$ m → $C = \pi d$.

Page 8, items 1–4. 1. $3.14 \times 10 = 31.4$ cm 2. $62.8$ in 3. $2 \times 3.14 \times 4 = 25.12$ ft 4. $56.52$ m

Page 9, circumference grid. a) $15.7$ cm b) $94.2$ mm c) $69.08$ in d) $15.7$ m e) $23.55$ cm f) $37.68$ cm g) $78.5$ ft h) $75.36$ m

Page 9, work backward. 5. $d = 30$ ft, $r = 15$ ft 6. $d = 10$ m 7. $r = 3$ cm

Page 9, doubling. Circle A $18.84$ cm; Circle B $37.68$ cm. The circumference doubled.

Page 9, error hunt. Priya used the radius with the diameter formula. Correct: $C \approx 2 \times 3.14 \times 8 = 50.24$ cm.

Page 11, reasoning blanks. The height is the radius, $r$. The base is $\dfrac{2\pi r}{2} = \boldsymbol{\pi r}$. So $A = (\pi r)(r) = \boldsymbol{\pi r^2}$. Square the radius first, then multiply by $\pi$. For $r = 5$: $5^2 = 25$, then $3.14 \times 25 = 78.5$. Counting squares gives about $78.5$ square units.

Page 12, area grid. a) $78.5\ \text{cm}^2$ b) $314\ \text{in}^2$ c) $28.26\ \text{cm}^2$ d) $153.86\ \text{in}^2$ e) $19.625\ \text{in}^2$ f) $7.065\ \text{cm}^2$ g) $50.24\ \text{ft}^2$ h) $314\ \text{m}^2$ i) $113.04\ \text{ft}^2$ j) $706.5\ \text{m}^2$

Page 12, work backward. 1. $r^2 = 50.24 \div 3.14 = 16$, so $r = 4$ ft. 2. $r^2 = 28.26 \div 3.14 = 9$, so $r = 3$ m.

Page 12, same circle. $C \approx 37.68$ cm; $A \approx 113.04\ \text{cm}^2$. Circumference is one length, so it uses centimeters; area multiplies two lengths, so it uses square centimeters.

Page 12, error hunt. Devon squared the diameter instead of the radius. Correct area: $78.5\ \text{cm}^2$. Devon's answer was $4$ times too large.

Page 14, circle the measurement. Fencing a round pen — C. Mulching a round flower bed — A. Trim around a tabletop — C. Grass a sprinkler waters — A. One full turn of a wheel — C. Glass to cover a round table — A.

Page 14, radius or diameter. "Reaches $12$ ft in every direction" gives the radius. "A pizza is $16$ inches" gives the diameter.

Page 14, items 1–4. 1. $452.16\ \text{ft}^2$ 2. $125.6$ ft 3. $50.24\ \text{ft}^2$ 4. $43.96$ ft

Page 15, items 5–10. 5. Around: $56.52$ ft; surface: $254.34\ \text{ft}^2$. 6. $628$ ft. 7. $\$251.20$. 8. $153.86\ \text{in}^2$. 9. Two $10$-in: $157\ \text{in}^2$; one $16$-in: $200.96\ \text{in}^2$; the $16$-inch gives more, by $43.96\ \text{in}^2$. 10. Cloth diameter $6$ ft; fabric area $28.26\ \text{ft}^2$.

Page 15, item 11 semicircle window. Rectangle: $4 \times 3 = 12\ \text{ft}^2$. Full circle of radius $2$: $3.14 \times 4 = 12.56\ \text{ft}^2$, so the semicircle is $12.56 \div 2 = 6.28\ \text{ft}^2$. Total glass: $12 + 6.28 = 18.28\ \text{ft}^2$.

Page 15, item 12 doubling table.

Doubling the diameter multiplies $C$ by $2$ and $A$ by $4$. The circumference formula uses the radius once, so it doubles once. The area formula squares the radius, so the doubling happens twice.

Pages 17–18, Chapter 14 review. Same items as the textbook review; see the Chapter 14 Review key above.