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2007 AMC 12B Problems/Problem 19: Difference between revisions

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==See Also==
==See Also==
{{AMC12 box|year=2007|ab=B|num-b=11|num-a=13}}
{{AMC12 box|year=2007|ab=B|num-b=18|num-a=20}}

Revision as of 12:55, 24 February 2008

Problem 19

Rhombus $ABCD$, with side length $6$, is rolled to form a cylinder of volume $6$ by taping $\overline{AB}$ to $\overline{DC}$. What is $\sin(\angle ABC)$?

$\mathrm {(A)} \frac{\pi}{9}$ $\mathrm {(B)} \frac{1}{2}$ $\mathrm {(C)} \frac{\pi}{6}$ $\mathrm {(D)} \frac{\pi}{4}$ $\mathrm {(E)} \frac{\sqrt{3}}{2}$

Solution

[asy] pair A=(0,0), B=(6*dir(60)), D=(6,0); pair C=B+D; draw(A--B--C--D--A); draw(B--(3,0)); label("\(A\)",A,SW);label("\(B\)",B,NW);label("\(C\)",C,NE);label("\(D\)",D,SE); label("\(6\)",B/2,NW); label("\(\theta\)",(.8,.5)); label("\(h\)",(3,2.6),E); [/asy]


$V_{Cylinder} = \pi r^2 h$

Where $C = 2\pi r = 6$ and $h=6\sin\theta$

$r = \frac{3}{\pi}$

$V = \pi \left(\frac{3}{\pi}\right)^2\cdot 6\sin\theta$

$6 = \frac{9}{\pi} \cdot 6\sin\theta$

$\sin\theta = \frac{\pi}{9} \Rightarrow \mathrm{(A)}$

See Also

2007 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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
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