1968 AMC 12 Problem 11

Attempt Problem 11 of the 1968 AMC 12 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 1968 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

11.

If an arc of 60∘60^\circ on circle I\mathrm{I} has the same length as an arc of 45∘45^\circ on circle II,\mathrm{II}, the ratio of the area of circle I\mathrm{I} to that of circle II\mathrm{II} is:
16:916:9
9:169:16
4:34:3
3:43:4
none of these
Answer: B
Concepts:arccircle arearatio and proportion
Small hint:
Equate (θ360∘)2πr(\frac{\theta}{360^\circ})2\pi r for the two arcs
Big hint:
Square the resulting radius ratio to get the area ratio
Solution:
Equal arc lengths give 60r1=45r2,60r_1=45r_2, so r1r2=34.\frac{r_1}{r_2}=\frac{3}{4}. Circle areas scale as the squares of the radii, giving A1:A2=9:16.A_1:A_2=9:16.
Therefore, the correct answer is B.
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