2002 AMC 10A Problem 7

Attempt Problem 7 of the 2002 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 10A solutions, or check the answer key.

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7.

If an arc of 4545^\circ on circle AA has the same length as an arc of 3030^\circ on circle B,B, then the ratio of the area of circle AA to the area of circle BB is

49\dfrac{4}{9}

23\dfrac{2}{3}

56\dfrac{5}{6}

32\dfrac{3}{2}

94\dfrac{9}{4}

Answer: A
Concepts:arcarea ratioratio and proportion
Difficulty rating: 1190
Solution:

Equal arc lengths give 453602πrA=303602πrB,\dfrac{45}{360}\cdot 2\pi r_A=\dfrac{30}{360}\cdot 2\pi r_B, so 45rA=30rB45 r_A=30 r_B and rArB=23.\dfrac{r_A}{r_B}=\dfrac{2}{3}.

The ratio of areas is (rArB)2=49.\left(\dfrac{r_A}{r_B}\right)^2=\dfrac{4}{9}.

Thus, the correct answer is A.

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