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Difference between revisions of "2001 AMC 12 Problems/Problem 8"

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</asy>
 
</asy>
  
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<math>\text{(A) A cone with slant height of } 10 \text{ and radius } 6</math>
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<math>\textbf{(A)} \text{ A cone with slant height of } 10 \text{ and radius } 6</math>
  
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<math>\text{(B) A cone with height of } 10 \text{ and radius } 6</math>
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<math>\textbf{(B)} \text{ A cone with height of } 10 \text{ and radius } 6</math>
  
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<math>\text{(C) A cone with slant height of } 10 \text{ and radius } 7</math>
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<math>\textbf{(C)} \text{ A cone with slant height of } 10 \text{ and radius } 7</math>
  
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<math>\text{(D) A cone with height of } 10 \text{ and radius } 7</math>
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<math>\textbf{(D)} \text{ A cone with height of } 10 \text{ and radius } 7</math>
  
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<math>\text{(E) A cone with slant height of } 10 \text{ and radius } 8</math>
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<math>\textbf{(E)} \text{ A cone with slant height of } 10 \text{ and radius } 8</math>
  
 
== Solution ==
 
== Solution ==
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</asy>
 
</asy>
  
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The blue lines will be joined together to form a single blue line on the surface of the cone, hence <math>\boxed{10}</math> will be the <math>\boxed{\text{slant height}}</math> of the cone.
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The blue lines will be joined together to form a single blue line on the surface of the cone, hence <math>10</math> 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 red line will form the circumference of the base. We can compute its length and use it to determine the radius.
  
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The length of the red line is <math>\dfrac{252}{360}\cdot 2\pi \cdot 10 = 14\pi</math>. This is the circumference of a circle with radius <math>\boxed{7}</math>.
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The length of the red line is <math>\dfrac{252}{360}\cdot 2\pi \cdot 10 = 14\pi</math>. This is the circumference of a circle with radius <math>7</math>.
  
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Therefore the correct answer is <math>\boxed{\text{C}}</math>.
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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>.
  
 
== See Also ==
 
== See Also ==
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{{AMC12 box|year=2001|num-b=7|num-a=9}}
 
{{AMC12 box|year=2001|num-b=7|num-a=9}}
 
{{AMC10 box|year=2001|num-b=16|num-a=18}}
 
{{AMC10 box|year=2001|num-b=16|num-a=18}}
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{{MAA Notice}}

Latest revision as of 23:32, 29 December 2023

The following problem is from both the 2001 AMC 12 #8 and 2001 AMC 10 #17, so both problems redirect to this page.

Problem

Which of the cones listed below can be formed from a $252^\circ$ sector of a circle of radius $10$ 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]

$\textbf{(A)} \text{ A cone with slant height of } 10 \text{ and radius } 6$

$\textbf{(B)} \text{ A cone with height of } 10 \text{ and radius } 6$

$\textbf{(C)} \text{ A cone with slant height of } 10 \text{ and radius } 7$

$\textbf{(D)} \text{ A cone with height of } 10 \text{ and radius } 7$

$\textbf{(E)} \text{ A cone with slant height of } 10 \text{ and radius } 8$

Solution

[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]

The blue lines will be joined together to form a single blue line on the surface of the cone, hence $10$ 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 $\dfrac{252}{360}\cdot 2\pi \cdot 10 = 14\pi$. This is the circumference of a circle with radius $7$.

Therefore the correct answer is $\boxed{\textbf{(C)} \text{ A cone with slant height of } 10 \text{ and radius } 7}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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