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2001 AMC 12 Problems/Problem 8

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 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]

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

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

Solution 2

As in Solution 1, we deduce that the slant height of the cone is $10$.

We now recall that the lateral surface area of a cone with base radius $r$ and slant height $l$ is $\pi rl$, while the area of the sector is $\frac{252^{\circ}}{360^{\circ}} \cdot \pi \cdot 10^2 = \frac{7}{10} \cdot \pi \cdot 10^2 = 70\pi$. We must therefore have \[\pi \cdot r \cdot 10 = 70\pi \iff 10r = 70 \iff r = 7,\] so the 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC logo.png

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