Circles: area and circumference

1. The figure shows the radius: r = 5 in.
2. Area of a circle: A = \pi r^2 (on the formula sheet).
3. Square the radius first: 5^2 = 5 \times 5 = 25.
4. Multiply by \pi: A = 3.14 \times 25 = 78.5.

The area is 78.5 in².

Check: estimate with \pi \approx 3: 3 \times 25 = 75, a little less than 78.5 ✓.

Tip: In \pi r^2, only the radius is squared. Square it before multiplying by 3.14, and remember that area is in square units.

1. The distance around a circle is the circumference: C = 2\pi r (on the formula sheet).
2. The radius is r = 6 in.
3. Compute: C = 2 \times 3.14 \times 6 = 6.28 \times 6 = 37.68.

The ribbon is 37.68 inches long.

Check: the diameter is 12 in, and C = \pi d = 3.14 \times 12 = 37.68 ✓. A circumference is always a bit more than 3 times the diameter: 3 \times 12 = 36 ✓.

Tip: Circumference = edge (ribbon, trim, fence). Area = surface (paint, fabric, grass). If you get a number with "square" units, you used the wrong formula.

1. The diameter goes all the way across through the center, so it is twice the radius: d = 2r = 2 \times 7 = 14 cm.
2. Test the other statements:
   • "Diameter 3.5 cm" halves the radius instead of doubling it ✗.
   • Circumference is C = 2\pi r = 14\pi cm, not 7\pi ✗.
   • Area is A = \pi r^2 = 49\pi\text{ cm}^2, not 14\pi ✗.

The true statement: the diameter is 14 cm.

Check: a radius is the center-to-edge distance; laying two radii end to end across the circle makes one diameter: 7 + 7 = 14 ✓.

Tip: Before any circle problem, write r and d next to each other. Most wrong answers come from using one in place of the other.

1. "16 inches in diameter" → d = 16, so the radius is r = 16 \div 2 = 8 in.
2. Area: A = \pi r^2 = 3.14 \times 8^2.
3. 8^2 = 64, and 3.14 \times 64 = 200.96.
4. Round to the nearest square inch: 200.96 \approx 201.

The area is about 201 in².

Check: estimate 3 \times 64 = 192; the answer is a bit more ✓. (With the calculator's \pi key: \pi \times 64 \approx 201.06, which also rounds to 201 ✓.)

Tip: Always halve a diameter before using A = \pi r^2. Because the radius is squared, using the diameter makes the area 4 times too large (804 instead of 201).

1. A = \pi r^2, so 113.04 = 3.14 r^2.
2. Divide by 3.14: r^2 = 113.04 \div 3.14 = 36.
3. Undo the square with a square root: r = \sqrt{36} = 6 ft.
4. The question asks for the diameter: d = 2r = 2 \times 6 = 12 ft.

The diameter is 12 ft.

Check: with d = 12, r = 6 and A = 3.14 \times 36 = 113.04 ✓.

Tip: Working backwards from area takes three moves: divide by \pi, square-root, then double if the diameter is asked. Stopping early gives the trap answers 36 and 6.

1. Edging goes around the garden → circumference.
2. With the radius, use C = 2\pi r = 2 \times 3.14 \times 7 = 43.96 ft.
3. Cost: 43.96 \times 2.50 = 109.90.

The edging costs $109.90.

Check: estimate C \approx 6 \times 7 = 42 ft and 42 \times 2.5 = 105; our answer is close ✓.

Tip: Decide area or circumference from the job: edging, fencing and trim go around (circumference); mulch, sod and seed cover the inside (area). Then make sure you used 2\pi r (not \pi r) when you have the radius.

1. Use C = \pi d (on the formula sheet) and substitute: 94.2 = 3.14 \times d.
2. Divide both sides by 3.14: d = 94.2 \div 3.14 = 30.

The diameter is 30 inches.

Check: 3.14 \times 30 = 94.2 ✓. Estimate: the distance around is a bit more than 3 times the diameter, and 94.2 \div 3 \approx 31 — close to 30 ✓.

Tip: Work backwards from circumference by dividing by \pi (to get d) or by 2\pi (to get r). Read the question twice: here it asks for the diameter, so do not halve it.

1. Radius: r = 18 \div 2 = 9 in.
2. Area: A = \pi r^2 = \pi \times 9^2 = 81\pi.

The area is 81\pi\text{ in}^2.

Check: 81\pi \approx 81 \times 3.14 = 254.34\text{ in}^2. A square 18 in by 18 in has area 324, and the circle fits inside it, so the circle's area should be a bit less — 254 < 324 ✓.

Tip: "In terms of \pi" means leave \pi as a letter: do not multiply by 3.14. Just compute the number in front of \pi.

1. "Per square foot" means you need area: A = \pi r^2.
2. A = 3.14 \times 6^2 = 3.14 \times 36 = 113.04\text{ ft}^2.
3. Cost: 113.04 \times \$4 = \$452.16.

The pavers cost $452.16.

Check: estimate 3 \times 36 = 108\text{ ft}^2, and 108 \times 4 = 432. The exact answer is a little more because \pi > 3 ✓.

Tip: Match the price unit to the measurement: "per foot" → circumference, "per square foot" → area.

1. The region is a quarter of a circle with radius r = 20 ft.
2. Whole circle: A = \pi r^2 = 3.14 \times 20^2 = 3.14 \times 400 = 1{,}256\text{ ft}^2.
3. One quarter: 1{,}256 \div 4 = 314\text{ ft}^2.

The watered area is 314 ft².

Check: \tfrac{1}{4} \times 3.14 \times 400 = 3.14 \times 100 = 314 ✓.

Tip: For part of a circle, find the whole-circle area, then take the fraction: half → \div 2, quarter → \div 4. A right-angle corner at the center means a quarter.

1. Old area: A = 3.14 \times 5^2 = 3.14 \times 25 = 78.5\text{ ft}^2.
2. New area: A = 3.14 \times 10^2 = 3.14 \times 100 = 314\text{ ft}^2.
3. Increase: 314 - 78.5 = 235.5\text{ ft}^2.

The area increases by 235.5 ft².

Check: doubling the radius multiplies the area by 2^2 = 4: 4 \times 78.5 = 314 ✓, so the increase is 3 \times 78.5 = 235.5 ✓.

Tip: "Increase" or "how much more" means subtract new − old. And remember: double the radius → 4 times the area.

1. In one turn, a wheel rolls a distance equal to its circumference.
2. C = \pi d = 3.14 \times 24 = 75.36 in.
3. In 50 turns: 50 \times 75.36 = 3{,}768 in.
4. Convert to feet: 3{,}768 \div 12 = 314 ft.

The chair travels 314 ft.

Check: each turn is about 75.36 \div 12 = 6.28 ft, and 50 \times 6.28 = 314 ✓.

Tip: Watch for unit changes in the last line of the question. The math can be perfect and the answer still wrong if you report inches when feet were asked.

1. One turn moves the scooter one circumference: C = \pi d = 3.14 \times 21 = 65.94 in.
2. 400 turns: 400 \times 65.94 = 26{,}376 in.
3. Convert to feet: 26{,}376 \div 12 = 2{,}198 ft.

The scooter travels 2,198 feet.

Check: estimate 3 \times 21 \approx 63 in per turn ≈ just over 5 ft; 400 \times 5.5 = 2{,}200 ft ✓.

Tip: Wheel problems: distance = (number of turns) × circumference. Then check whether the question wants inches, feet or miles.

1. Use C = \pi d with d = 42 cm.
2. C = \tfrac{22}{7} \times 42.
3. Divide first: 42 \div 7 = 6. Then 22 \times 6 = 132.

The circumference is 132 cm.

Check: with \pi \approx 3.14: 3.14 \times 42 = 131.88 \approx 132 ✓. The two values of \pi are both approximations, so tiny differences like this are expected.

Tip: Use \tfrac{22}{7} when the radius or diameter is a multiple of 7 — the 7 cancels and the arithmetic is easy without a calculator.

1. From the circumference: 25.12 = 2 \times 3.14 \times r = 6.28r, so r = 25.12 \div 6.28 = 4 ft.
2. Area: A = \pi r^2 = 3.14 \times 4^2 = 3.14 \times 16 = 50.24\text{ ft}^2.

The area of the top is 50.24 ft².

Check: with r = 4: C = 6.28 \times 4 = 25.12 ✓, and A = 50.24 ✓.

Tip: Circumference → radius → area. The two most common slips: using the diameter (8) in \pi r^2, and forgetting to square the radius.

1. 12-inch pizza: r = 6; A = 3.14 \times 6^2 = 3.14 \times 36 = 113.04 in².
2. Price per in²: 10 \div 113.04 \approx 0.0885, about $0.088 (8.8¢).
3. 16-inch pizza: r = 8; A = 3.14 \times 8^2 = 3.14 \times 64 = 200.96 in².
4. Price per in²: 16 \div 200.96 \approx 0.0796, about $0.080 (8.0¢).
5. 8.0¢ is less than 8.8¢.

The 16-inch pizza is the better buy, at about $0.080 per square inch.

Check: the 16-inch costs 1.6 times as much as the 12-inch, but its area is 200.96 \div 113.04 \approx 1.78 times as large — you get more extra pizza than extra price ✓.

Tip: "How much pizza" is an area question, and area grows with the square of the size. Comparing diameters or circumferences (straight lengths) underestimates how much bigger a large pizza really is.

1. Use A = \pi r^2: 31{,}400 = 3.14r^2.
2. Divide by 3.14: r^2 = 31{,}400 \div 3.14 = 10{,}000.
3. Square root: r = \sqrt{10{,}000} = 100 m.
4. Diameter: d = 2r = 200 m.

The diameter is 200 m.

Check: with r = 100: 3.14 \times 100^2 = 3.14 \times 10{,}000 = 31{,}400 ✓.

Tip: Three steps from area to diameter: \div \pi, \sqrt{\phantom{x}}, \times 2. Write down which quantity you have after each step so you do not stop early.

1. C = \pi d = 10\pi, so d = 10 in.
2. r = d \div 2 = 5 in.
3. Area: A = \pi r^2 = \pi(5)^2 = 25\pi\text{ in}^2.
4. Test the choices: radius is 5 (not 10), diameter is 10 (not 5), area is 25\pi (not 100\pi).

True statement: the area is 25\pi square inches.

Check: C = 2\pi(5) = 10\pi ✓ and A = 25\pi ✓.

Tip: From a circumference in terms of \pi, the number in front of \pi is the diameter. Halve it for the radius, then square the radius for the area.

1. Distance in inches: 1{,}000 \times 12 = 12{,}000 in.
2. One turn: C = \pi d = 3.14 \times 25 = 78.5 in.
3. Number of turns: 12{,}000 \div 78.5 \approx 152.9, which is about 153 turns.

The tire makes about 153 turns.

Check: 153 \times 78.5 = 12{,}010.5 in \approx 1{,}001 ft ✓.

Tip: Work-backwards wheel problems: turns = distance ÷ circumference — but only after both are in the same unit. Dividing feet by inches gives a number 12 times too small.

1. One 18-inch pizza: r = 9, so A = 3.14 \times 81 = 254.34\text{ in}^2.
2. One 12-inch pizza: r = 6, so A = 3.14 \times 36 = 113.04\text{ in}^2. Two of them: 2 \times 113.04 = 226.08\text{ in}^2.
3. Compare: 254.34 - 226.08 = 28.26\text{ in}^2.

The one 18-inch pizza has more pizza, by 28.26 in².

Check: compare r^2 values without \pi: 81 vs 2 \times 36 = 72. Since 81 > 72, the single large pizza wins ✓; 3.14 \times 9 = 28.26 ✓.

Tip: Area grows with the square of the size, so "two small" is often less than "one large" even though 12 + 12 = 24 > 18. Never compare circles by adding diameters.

1. Full circumference: C = 2\pi r = 2 \times 3.14 \times 6 = 37.68 ft.
2. Curved edge (one quarter): 37.68 \div 4 = 9.42 ft.
3. Straight edges: two radii, 6 + 6 = 12 ft.
4. Total: 9.42 + 12 = 21.42 ft.

21.42 ft of fence are needed.

Check: the curve should be a bit longer than the straight line joining its ends (\approx 8.5 ft for a right triangle with legs 6 and 6), and 9.42 > 8.5 ✓.

Tip: Perimeter of a sector = curved part + two radii. The fraction (½, ¼) applies only to the curved part.