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.

This question tests solving for diameter from the circumference formula C=πd, given C=20π cm, by rearranging to d=C/π. For C=πd=20π cm, divide both sides by π to get d=20 cm. For example, if C=20π, then d=20, or equivalently r=10 using C=2πr. Correct application: use the direct formula d=C/π without extra steps. Common errors include dividing by 2π instead, giving d=10, or confusing with area and squaring. Steps: (1) given C=20π, (2) use d=C/π, (3) compute 20π/π=20 cm. Since C is proportional to d (or r), doubling diameter doubles circumference, but area would quadruple if radius doubles.

This question tests solving for radius from the area formula $A=\pi r^2$, given $A=36\pi \text{ m}^2$, by rearranging to $r=\sqrt{(A/\pi)}$. For area $A=\pi r^2=36\pi \text{ m}^2$, divide both sides by $\pi$ to get $r^2=36$, so $r=6 \text{ m}$ (positive value). For example, if $A=36\pi$, then $r^2=36$, $r=6$, as in a garden of that size. Correct application: rearrange the formula properly without forgetting to take the square root. Common errors include thinking $r=A/\pi=36$, or using circumference formula instead, or taking square root before dividing by $\pi$. Steps: (1) given $A=36\pi$, (2) $r^2=A/\pi=36$, (3) $r=\sqrt{36}=6 \text{ m}$. Remember the relationship: area is quadratic in r, so for $r=6$, $A=36\pi$, and if r doubles to $12$, A quadruples to $144\pi$.

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.

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.

This question tests applying the circle area formula A=πr², where you use the given radius to calculate both exact and approximate values. For a radius of 5 cm, the area is A=π(5)²=π×25=25π cm², and approximating with π≈3.14 gives 25×3.14=78.5 cm². For example, if the radius were 3 cm, A=π×9=9π≈28.26 cm². To find the area, identify the radius, square it, multiply by π for the exact value, and then use the approximation if needed. A common mistake is using the diameter instead of the radius or forgetting to square the radius, like calculating π×5=5π instead. Steps include: (1) note the radius r=5 cm, (2) compute r²=25, (3) multiply by π for 25π, (4) approximate 25×3.14=78.5, (5) add units cm². Remember, area scales quadratically with radius, so doubling the radius quadruples the area.

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.

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.

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.

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.