Apply Circle Area and Circumference Formulas - 7th Grade Math

A circular pool has a diameter of $20\text{ ft}$ . About how much area does a pool cover need? Give an exact answer in terms of $\pi$ and an approximate answer using $\pi\approx 3.14$ .
Answer: $A=100\pi\text{ ft}^2\approx 314\text{ ft}^2$ This question tests applying the area formula A=πr² after converting diameter to radius, r=d/2. For d=20 ft, r=10 ft, A=π(10)²=100π ft², and approximating with π≈3.14 gives 100×3.14=314 ft². For example, if d=10 ft, r=5, A=25π≈78.5 ft². First convert d to r, then square r and multiply by π. A common mistake is using diameter in the area formula, like π(20)²=400π, which is four times too large since (2r)²=4r². Steps include: (1) r=20/2=10, (2) r²=100, (3) A=100π exactly, (4) approximate 314, (5) units ft². Area quadruples when radius doubles, reflecting the quadratic relationship.
A circle has radius $r$ . If the radius doubles (becomes $2r$ ), how do the circumference and area change?
Answer: Circumference doubles; area quadruples. This question tests understanding how circumference and area change when radius doubles from r to 2r, using C=2πr and A=πr². New C=2π(2r)=4πr=2×original C, so doubles; new A=π(2r)²=4πr²=4×original A, so quadruples. For example, if original r=1, C=2π, A=π; doubled r=2, C=4π (doubles), A=4π (quadruples). Correct application: recognize the linear scaling for C and quadratic for A. Common errors include thinking both double or both quadruple, confusing the exponents. Steps: (1) recall formulas, (2) substitute 2r, (3) compare to originals. This highlights the key relationship: C ∝ r, A ∝ r².
A circular sign has a diameter of $14\text{ cm}$ . What is the area of the sign? Give the exact answer in terms of $\pi$ and the approximate answer using $\pi\approx 3.14$ .
Answer: $49\pi\text{ cm}^2\approx 153.86\text{ cm}^2$ This question tests calculating area A=πr² from diameter, converting r=d/2, with exact π and approximate using π≈3.14. For d=14 cm, r=7 cm, A=π×49=49π≈153.86 cm²; ensure to halve diameter, and remember area is quadratic in r. For example, with d=14 cm, r=7, A=π(7)²=49π≈153.86 cm² for the sign. Correctly find r=7, then A=49π, approximate 49×3.14=153.86. Common errors include using d in place of r like π(14)²=196π, or halving incorrectly. Steps: (1) identify diameter, (2) convert r=14/2=7, (3) use A=πr², (4) calculate 49π and 153.86, (5) include units cm². Mistakes: not dividing diameter by 2, squaring diameter instead, or using π≈3.14 as 3 giving ≈147.
A circular track has a radius of $7\text{ m}$ . About how far is it around the track one time (the circumference), using $\pi\approx 3.14$ ?
Answer: $43.96\text{ m}$ This question tests applying circumference C=2πr for r=7 m, with approximation using π≈3.14, no exact form required. C=2π×7=14π ≈14×3.14=43.96 m. For example, a track with r=7 m has circumference about 43.96 m. Correct application: multiply radius by 2π, then approximate. Common errors include using C=πr=7π≈22, or using area formula πr²=49π≈154. Common mistakes also involve poor π approximation like using 3, giving 42. Steps: (1) given r=7, (2) C=2πr, (3) approximate 14×3.14=43.96 m. Circumference doubles if radius doubles, unlike area.
A circular pizza has a radius of $8\text{ in}$ . If the radius doubles to $16\text{ in}$ , how do the pizza's circumference (the distance around its edge) and area change?
Answer: Circumference doubles; area quadruples. Circumference is C=2*pi*r, which depends linearly on r, so doubling the radius doubles the circumference. Area is A=pi* $r^2$ , which depends on r squared, so doubling the radius quadruples the area (since 2 squared is 4). This matches choice C. Choice A incorrectly assumes circumference also quadruples. Choice B has both relationships backwards. Choice D incorrectly assumes area only doubles, missing the squaring effect.
A circle's radius is doubled from $r$ to $2r$ . Which statement best describes what happens to the circumference and the area?
Answer: The circumference doubles and the area quadruples. This question tests understanding the relationship between radius changes and how circumference C=2πr and area A=πr² scale, specifically when radius doubles to 2r. New C=2π(2r)=4πr=2×original C, new A=π(2r)²=4πr²=4×original A; thus, circumference doubles (linear), area quadruples (quadratic). For example, if original r gives C=2πr and A=πr², doubling to 2r gives C=4πr (double) and A=4πr² (quadruple). The correct statement is that circumference doubles and area quadruples. Common errors include thinking both double or both quadruple, or mixing linear and quadratic scaling. Steps: (1) recall formulas, (2) substitute 2r into each, (3) compare to originals: C new=2×old, A new=4×old, (4) identify the description. Mistakes: assuming area is linear, confusing circumference with area scaling, or incorrect multiplication like 2r gives 2×A instead of 4×A.
Refer to the diagram. A rectangular piece of metal with dimensions 20 cm by 30 cm has four identical circular holes cut out from its corners. If each circular hole has a diameter of 8 cm, what is the remaining area of the metal piece?
Answer: $600 - 64\pi$ square cm Rectangle area = $20 \times 30 = 600$ square cm. Each circular hole has diameter 8 cm, so radius = 4 cm. Area of each hole = $\pi(4)^2 = 16\pi$ square cm. Total area of 4 holes = $4 \times 16\pi = 64\pi$ square cm. Remaining area = $600 - 64\pi$ square cm. Choice A uses only one hole. Choice B uses radius as 4 but calculates for 2 holes. Choice D uses diameter as radius.
A circular pool has a diameter of $20\text{ ft}$ . About how much area does a pool cover need? (Give the exact answer in terms of $\pi$ and the approximate answer using $\pi\approx 3.14$ .)
Answer: $100\pi\text{ ft}^2\approx 314\text{ ft}^2$ This question tests applying the area formula A=πr² after converting diameter to radius r=d/2, with exact π and approximate using π≈3.14. For diameter d=20 ft, r=10 ft, A=π×100=100π≈314 ft²; remember to halve the diameter for radius, and area quadruples if radius doubles. For example, with d=20 ft, r=10, A=π(10)²=100π≈314 ft² for the pool cover area. Correctly convert d to r=10, then A=π×100=100π, approximate 100×3.14=314. Common errors include using diameter in area formula like π(20)²=400π, or forgetting to square the radius. Steps: (1) identify diameter, (2) convert r=d/2=10, (3) use A=πr², (4) calculate 100π and 314, (5) include units ft². Mistakes: not converting diameter to radius, using C=πd for area, or poor π approximation like 3.0 giving 300.
A circular sticker has a radius of $5\text{ cm}$ . What is the area of the sticker? Give the exact answer in terms of $\pi$ and the approximate answer using $\pi\approx 3.14$ .
Answer: $25\pi\text{ cm}^2\approx 78.5\text{ cm}^2$ This question tests applying the circle area formula A=πr², where you calculate the exact value with π and approximate using π≈3.14, focusing on using the given radius directly. The area A=πr² requires squaring the radius (for r=5 cm, A=π×25=25π≈78.5 cm²), and since the radius is given, no conversion from diameter is needed; remember that area scales quadratically with radius, so doubling the radius would quadruple the area. For example, if a circle has r=5 cm, the area is π(25)=25π≈78.54 cm², which matches the calculation for this sticker. To find the area, substitute r=5 into A=πr² to get π×25=25π, and approximate as 25×3.14=78.5 cm². Common errors include using the circumference formula instead, like 2πr=10π for area, or forgetting to square the radius and doing π×5=5π. Steps: (1) identify the given radius, (2) select the area formula A=πr², (3) substitute r=5, (4) calculate exact 25π and approximate 78.5, (5) include units cm². Mistakes: confusing area with circumference, not squaring the radius, or using a poor π approximation like 3 instead of 3.14.
Refer to the figure. A square with side length 10 cm is inscribed in a circle. What is the difference between the area of the circle and the area of the square?
Answer: $50\pi - 100$ square cm The diagonal of the square equals the diameter of the circle. Square diagonal = $10\sqrt{2}$ cm, so circle diameter = $10\sqrt{2}$ cm and radius = $5\sqrt{2}$ cm. Circle area = $\pi(5\sqrt{2})^2 = \pi(25 \cdot 2) = 50\pi$ square cm. Square area = $10^2 = 100$ square cm. Difference = $50\pi - 100$ square cm. Choice A has incorrect calculation and wrong order. Choice C uses wrong radius. Choice D uses side length as radius.
A circular pool has a diameter of $20\text{ ft}$ . A cover needs to match the pool's surface area. What is the area of the pool? Give the exact answer in terms of $\pi$ and an approximate answer using $\pi\approx 3.14$ .
Answer: $100\pi\text{ ft}^2\approx 314\text{ ft}^2$ This question tests applying the area formula A=πr² with diameter given as 20 ft, requiring conversion to radius r=d/2=10 ft, and giving exact and approximate values. So A=π(10)²=100π ft² ≈314 ft² using π≈3.14. For example, a pool with d=20 ft has r=10 ft, A=100π ft² ≈314 ft². Correct application: always convert diameter to radius for area, then square and multiply by π. Common errors include using diameter in place of radius, like π(20)²=400π, or forgetting to halve the diameter. Steps: (1) given d=20 ft, r=10 ft, (2) A=πr², (3) π×100=100π, (4) 100×3.14=314, (5) units ft². Area quadruples when radius doubles, emphasizing the quadratic relationship versus linear for circumference.
A circular garden has an area of $36\pi\text{ m}^2$ . What is the radius of the garden?
Answer: $r=6\text{ m}$ This question tests solving for the radius from the area formula A=πr², rearranging to r=√(A/π), and working with exact π terms. The area A=πr² uses radius squared, so to find r, divide by π and take the square root (for A=36π m², r²=36, r=6 m); remember the quadratic relationship means area quadruples if radius doubles. For example, if A=36π, then πr²=36π → r²=36 → r=6 m, solving for the garden's radius. Correctly apply the formula by dividing both sides by π, then taking the square root of 36 to get r=6. Common errors include forgetting to divide by π before taking the square root, or using circumference formula instead like r=C/(2π). Steps: (1) identify given area, (2) set up πr²=36π, (3) divide by π to get r²=36, (4) take square root r=6, (5) include units m. Mistakes: not canceling π, taking square root of 36π without dividing, or confusing with diameter calculation.
A circular bracelet has a circumference of $31.4\text{ cm}$ (using $\pi\approx 3.14$ ). About what is the radius of the bracelet?
Answer: $r\approx 5\text{ cm}$ This question tests solving for radius from approximate circumference C≈31.4 cm using r=C/(2π) with π≈3.14. Circumference C=2πr, so r=C/(2π) ≈31.4/(6.28)≈5 cm; it's linear, so approximate calculations are straightforward. For example, with C=31.4 cm, r≈31.4/(2×3.14)≈31.4/6.28≈5 cm for the bracelet. Correctly divide 31.4 by 2×3.14=6.28 to get approximately 5. Common errors include using d=C/π≈10 instead of r, or forgetting the 2 in denominator. Steps: (1) identify approximate C, (2) use r=C/(2π), (3) substitute π≈3.14, (4) calculate 31.4/6.28≈5, (5) round to nearest cm. Mistakes: confusing with diameter formula, arithmetic errors like 31.4/3.14=10, or using π=3 giving ≈5.23 but not matching choices.
A bicycle wheel has a diameter of $12\text{ in}$ . About how far does the wheel travel in one full rotation (its circumference)? Give the exact answer in terms of $\pi$ and an approximate answer using $\pi\approx 3.14$ .
Answer: $12\pi\text{ in}\approx 37.7\text{ in}$ This question tests applying the circumference formula C=πd or C=2πr, with diameter 12 in given, and providing exact and approximate values using π≈3.14. The circumference C=πd uses the diameter directly, so for d=12 in, C=12π in ≈37.7 in, or equivalently using radius r=6 in, C=2π×6=12π in. For example, a wheel with d=12 in travels 12π in per rotation, approximately 37.7 in. Correct application involves converting diameter to radius if using the 2πr formula, but here πd is straightforward. Common errors include using C=πr without the 2, giving 6π, or squaring the radius like in area, resulting in wrong values. Steps: (1) identify diameter d=12 in, (2) use C=πd, (3) compute 12π exactly, (4) approximate 12×3.14=37.68 rounded to 37.7, (5) include units in. Circumference is linear with radius, so doubling radius doubles circumference, unlike area which quadruples.
A circular sticker has a radius of $5\text{ cm}$ . What is the area of the sticker? Give the exact answer in terms of $\pi$ and an approximate answer using $\pi\approx 3.14$ .
Answer: $25\pi\text{ cm}^2\approx 78.5\text{ cm}^2$ This question tests applying the circle area formula A=πr², where the radius is given as 5 cm, and providing both exact and approximate values using π≈3.14. The area A=πr² uses the radius squared, so for r=5 cm, A=π×25=25π cm², and approximately 25×3.14=78.5 cm². For example, if a circle has r=5 cm, the area is 25π cm² or about 78.5 cm², which matches the calculation here. To solve correctly, identify the radius, square it, multiply by π for the exact area, and then approximate if needed. Common errors include using the diameter instead of radius, like treating 5 cm as diameter which would give r=2.5 and A=6.25π, or forgetting to square the radius, resulting in A=5π. Steps include: (1) note the given radius r=5 cm, (2) use A=πr², (3) compute π×25=25π exactly, (4) approximate 25×3.14=78.5, (5) add units cm². Remember, area scales quadratically with radius, so if radius doubles, area quadruples, but here it's a direct calculation.
A circle's circumference is $C=18\pi\text{ cm}$ . What is the circle's area? (Hint: First find $r$ from $C=2\pi r$ .)
Answer: $A=81\pi\text{ cm}^2$ This question tests finding area from given circumference C=18π cm by first solving for r using C=2πr, then using A=πr². From C=2πr=18π, r=18π/(2π)=9 cm; then A=π(9)²=81π cm². For example, if C=18π, r=9, A=81π. Correct application: solve for r first, then plug into area formula. Common errors include using d instead of r, or directly setting A=C²/π or similar wrong relations. Steps: (1) given C=18π, (2) r=C/(2π)=9, (3) A=πr²=81π. Since A ∝ r² and C ∝ r, you can also derive A=(C²)/(4π), but step-by-step is safer.
A circular track has a circumference of $20\pi\text{ m}$ . What is the diameter of the track?
Answer: $d=20\text{ m}$ This question tests solving for the diameter from the circumference formula C=πd, rearranging to d=C/π, and handling exact π terms. Circumference C=πd uses diameter, so d=C/π (for C=20π m, d=20 m); since C=2πr, diameter is twice the radius, and circumference doubles if radius doubles. For example, with C=20π m, d=20π/π=20 m, which is the track's diameter. Correctly apply by dividing C by π to get d=20. Common errors include using 2π in the denominator like for radius, or treating it as area and squaring. Steps: (1) identify given circumference, (2) select d=C/π, (3) substitute C=20π, (4) simplify to 20, (5) include units m. Mistakes: confusing with radius formula r=C/(2π) and getting 10 m, not canceling π properly, or using approximate π unnecessarily.
A sprinkler system waters a circular area. When the water pressure is low, it waters a circle with radius 15 feet. When the pressure is high, the radius increases by 40%. If the cost of water is the same per square foot watered, by what percentage does the water cost increase when switching from low to high pressure?
Answer: 96% increase in water cost Low pressure radius = 15 feet, area = $\pi(15)^2 = 225\pi$ square feet. High pressure radius = $15 \times 1.4 = 21$ feet, area = $\pi(21)^2 = 441\pi$ square feet. Percentage increase = $\frac{441\pi - 225\pi}{225\pi} \times 100\% = \frac{216\pi}{225\pi} \times 100\% = \frac{216}{225} \times 100\% = 96\%$ . Choice A incorrectly assumes linear relationship. Choice B uses wrong calculation. Choice C doubles the radius increase percentage.
A circular sign has a diameter of $10\text{ cm}$ . What is the sign's area? Give the exact answer in terms of $\pi$ and an approximate answer using $\pi\approx 3.14$ .
Answer: $25\pi\text{ cm}^2\approx 78.5\text{ cm}^2$ This question tests area A=πr² with diameter 10 cm, so convert to r=5 cm, giving exact and approximate using π≈3.14. A=π(5)²=25π cm² ≈78.5 cm². For example, a sign with d=10 cm has r=5, A=25π ≈78.5. Correct application: halve diameter to get radius, then proceed. Common errors include using d as r, giving π(10)²=100π, or forgetting to square, giving 10π for d. Steps: (1) d=10, r=5, (2) A=πr²=25π, (3) 25×3.14=78.5, (4) units cm². Area quadruples with doubled radius, but here it's direct computation.
A circular garden has a radius of 8 feet. If the owner wants to install a decorative border around the entire perimeter and then cover the garden with mulch, what is the total cost if the border costs $3 per foot and mulch costs $2 per square foot? Use $\pi \approx 3.14$ .
Answer: $552.64 First find the circumference: $C = 2\pi r = 2(3.14)(8) = 50.24$ feet. Border cost: $50.24 \times 3 = \$150.72$ . Then find the area: $A = \pi r^2 = 3.14(8)^2 = 3.14(64) = 200.96$ square feet. Mulch cost: $200.96 \times 2 = \$401.92$ . Total: $150.72 + 401.92 = \$552.64$ . Choice B uses diameter instead of radius for circumference. Choice C forgets to double the radius in circumference formula. Choice D adds an extra calculation error.

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