Difference between revisions of "2001 AMC 12 Problems/Problem 8"
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Therefore the correct answer is <math>\boxed{\textbf{(C)} \text{ A cone with slant height of } 10 \text{ and radius } 7}</math>. | Therefore the correct answer is <math>\boxed{\textbf{(C)} \text{ A cone with slant height of } 10 \text{ and radius } 7}</math>. | ||
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+ | == Solution 2 == | ||
+ | |||
+ | We already know the slight height is 10. | ||
+ | |||
+ | The formula for the lateral surface area is πrl, l is the slight height. We can find the shape above is (n/360)π r^2. πrl = (n/360) π r^2. | ||
+ | π r 10 = (252/360)10^2. r = (252/360). r = 7 | ||
+ | |||
+ | - mathlover1205 | ||
== See Also == | == See Also == |
Revision as of 22:42, 27 May 2025
- The following problem is from both the 2001 AMC 12 #8 and 2001 AMC 10 #17, so both problems redirect to this page.
Contents
Problem
Which of the cones listed below can be formed from a sector of a circle of radius
by aligning the two straight sides?
Solution
The blue lines will be joined together to form a single blue line on the surface of the cone, hence will be the slant height of the cone.
The red line will form the circumference of the base. We can compute its length and use it to determine the radius.
The length of the red line is . This is the circumference of a circle with radius
.
Therefore the correct answer is .
Solution 2
We already know the slight height is 10.
The formula for the lateral surface area is πrl, l is the slight height. We can find the shape above is (n/360)π r^2. πrl = (n/360) π r^2. π r 10 = (252/360)10^2. r = (252/360). r = 7
- mathlover1205
See Also
2001 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
2001 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.