Difference between revisions of "2007 AMC 12B Problems/Problem 19"

Revision as of 23:51, 24 January 2020

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 $\sin(\angle ABC)$ ?

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

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 $h=6\sin\theta$

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

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

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

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

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

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

@@ Line 18: / Line 18: @@
 </asy>
-<math>V_{Cylinder} = \pi r^2 h</math>
+<math>V_{\mathrm{Cylinder}} = \pi r^2 h</math>
 Where <math>C = 2\pi r = 6</math> and <math>h=6\sin\theta</math>

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Difference between revisions of "2007 AMC 12B Problems/Problem 19"

Revision as of 23:51, 24 January 2020

Problem 19

Solution

See Also