2001 AMC 12 Problem 8

Attempt Problem 8 of the 2001 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 2001 AMC 12 solutions, or check the answer key.

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

8.

Which of the cones below can be formed from a 252∘252^\circ sector of a circle of radius 1010 by aligning the two straight sides?
Answer: C
Concepts:conearccircumference
Small hint:
The radius of the sector becomes the slant height of the cone, so the slant height is 1010
Big hint:
The arc length of the sector becomes the base circumference: 252360⋅2π(10)\dfrac{252}{360}\cdot 2\pi(10)
Solution:
When the sector is rolled into a cone, its radius 1010 becomes the slant height, and its arc becomes the base circle.
The arc length is 252360⋅2π(10)=710⋅20π=14π, \dfrac{252}{360}\cdot 2\pi(10) = \dfrac{7}{10}\cdot 20\pi = 14\pi, so the base circumference is 14π14\pi and the base radius is 7.7.
The cone therefore has base radius 77 and slant height 10,10, which is choice C.
Thus, the correct answer is C.
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