Difference between revisions of "2001 AMC 12 Problems/Problem 8"
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We now recall that the lateral surface area of a cone with base radius <math>r</math> and slant height <math>l</math> is <math>\pi rl</math>, while the area of the sector is <math>\frac{252^{\circ}}{360^{\circ}} \cdot \pi \cdot 10^2 = \frac{7}{10} \cdot \pi \cdot 10^2 = 70\pi</math>. We must therefore have <cmath>\pi \cdot r \cdot 10 = 70\pi \iff 10r = 70 \iff r = 7,</cmath> so the answer is <math>\boxed{\textbf{(C)} \text{ A cone with slant height of } 10 \text{ and radius } 7}</math>. | We now recall that the lateral surface area of a cone with base radius <math>r</math> and slant height <math>l</math> is <math>\pi rl</math>, while the area of the sector is <math>\frac{252^{\circ}}{360^{\circ}} \cdot \pi \cdot 10^2 = \frac{7}{10} \cdot \pi \cdot 10^2 = 70\pi</math>. We must therefore have <cmath>\pi \cdot r \cdot 10 = 70\pi \iff 10r = 70 \iff r = 7,</cmath> so the answer is <math>\boxed{\textbf{(C)} \text{ A cone with slant height of } 10 \text{ and radius } 7}</math>. | ||
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== See Also == | == See Also == | ||
Latest revision as of 16:26, 10 July 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?
![[asy] import graph; unitsize(1.5cm); defaultpen(fontsize(8pt)); draw(Arc((0,0),1,-72,180),linewidth(.8pt)); draw(dir(288)--(0,0)--(-1,0),linewidth(.8pt)); label("$10$",(-0.5,0),S); draw(Arc((0,0),0.1,-72,180)); label("$252^{\circ}$",(0.05,0.05),NE); [/asy]](http://latex.artofproblemsolving.com/b/b/8/bb8af0c17ff35be1e553d36477a23ce3f026bded.png)
Solution 1
![[asy] import graph; unitsize(1.5cm); defaultpen(fontsize(8pt)); draw(Arc((0,0),1,-72,180),linewidth(.8pt) + red); draw(dir(288)--(0,0)--(-1,0),linewidth(.8pt) + blue); label("$10$",(-0.5,0),S); draw(Arc((0,0),0.1,-72,180)); label("$252^{\circ}$",(0.05,0.05),NE); [/asy]](http://latex.artofproblemsolving.com/2/1/f/21ff3c1e4bdeb7eae01918390e3702a2b8ab2a01.png)
The blue lines will be joined together to form a single blue line on the surface of the cone, so 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
As in Solution 1, we deduce that the slant height of the cone is .
We now recall that the lateral surface area of a cone with base radius 
and slant height 
is 
, while the area of the sector is 
. We must therefore have ![\[\pi \cdot r \cdot 10 = 70\pi \iff 10r = 70 \iff r = 7,\]](http://latex.artofproblemsolving.com/0/4/f/04f7c4f1d7be2f48d2f6e6d41e3dcac99e615f80.png)
so the answer is .
-mathlover1205 -formatted and improved by Sevenoptimus
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. 
