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

m
(Correct point placements in diagram (CCW ordering, not CW))
Line 6: Line 6:
 
==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);

Revision as of 00:19, 7 February 2011

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]

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

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 (ProblemsAnswer KeyResources)
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