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

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==Problem 19==
==Problem==
Rhombus <math>ABCD</math>, with side length <math>6</math>, is rolled to form a cylinder of volume <math>6</math> by taping <math>\overline{AB}</math> to <math>\overline{DC}</math>. What is <math>\sin(\angle ABC)</math>?
Rhombus <math>ABCD</math>, with side length <math>6</math>, is rolled to form a cylinder of volume <math>6</math> by taping <math>\overline{AB}</math> to <math>\overline{DC}</math>. What is <math>\sin(\angle ABC)</math>?


<math>\mathrm {(A)} \frac{\pi}{9}</math>  <math>\mathrm {(B)} \frac{1}{2}</math>  <math>\mathrm {(C)} \frac{\pi}{6}</math>  <math>\mathrm {(D)} \frac{\pi}{4}</math>  <math>\mathrm {(E)} \frac{\sqrt{3}}{2}</math>
<math>\mathrm{(A)}\ \frac{\pi}{9} \qquad \mathrm{(B)}\ \frac{1}{2} \qquad \mathrm{(C)}\ \frac{\pi}{6} \qquad \mathrm{(D)}\ \frac{\pi}{4} \qquad \mathrm{(E)}\ \frac{\sqrt{3}}{2}</math>


==Solution==
==Solution==
<asy>
<asy>
pair A=(0,0), B=(6*dir(60)), D=(6,0);
pair B=(0,0), A=(6*dir(60)), C=(6,0);
pair C=B+D;
pair D=A+C;


draw(A--B--C--D--A);
draw(A--B--C--D--A);
draw(B--(3,0));
draw(A--(3,0));


label("\(A\)",A,SW);label("\(B\)",B,NW);label("\(C\)",C,NE);label("\(D\)",D,SE);
label("\(A\)",A,NW);label("\(B\)",B,SW);label("\(C\)",C,SE);label("\(D\)",D,NE);
label("\(6\)",B/2,NW);
label("\(6\)",A/2,NW);
label("\(\theta\)",(.8,.5));
label("\(\theta\)",(.8,.5));
label("\(h\)",(3,2.6),E);
label("\(h\)",(3,2.6),E);
</asy>
</asy>


 
<math>V_{\mathrm{Cylinder}} = \pi r^2 h</math>
 
<math>V_{Cylinder} = \pi r^2 h</math>


Where <math>C = 2\pi r = 6</math> and <math>h=6\sin\theta</math>
Where <math>C = 2\pi r = 6</math> and <math>h=6\sin\theta</math>
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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}}
{{MAA Notice}}

Latest revision as of 15:35, 15 February 2021

Problem

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

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

Solution

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

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

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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