2002 AMC 12A Problem 7

Below is the professionally curated solution for Problem 7 of the 2002 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12A solutions, or check the answer key.

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Concepts:arccircle arearatio and proportion

Difficulty rating: 1350

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}

Solution:

Equal arc lengths give 453602πRA=303602πRB,\dfrac{45}{360}\cdot 2\pi R_A = \dfrac{30}{360}\cdot 2\pi R_B, so RARB=3045=23.\dfrac{R_A}{R_B} = \dfrac{30}{45} = \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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