Difference between revisions of "2011 AMC 12A Problems/Problem 15"

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== Problem ==
 
== Problem ==
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The circular base of a hemisphere of radius <math>2</math> rests on the base of a square pyramid of height <math>6</math>. The hemisphere is tangent to the other four faces of the pyramid. What is the edge-length of the base of the pyramid?
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<math>
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\textbf{(A)}\ 3\sqrt{2} \qquad
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\textbf{(B)}\ \frac{13}{3} \qquad
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\textbf{(C)}\ 4\sqrt{2} \qquad
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\textbf{(D)}\ 6 \qquad
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\textbf{(E)}\ \frac{13}{2} </math>
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== Solution ==
 
== Solution ==
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=14|num-a=16|ab=A}}
 
{{AMC12 box|year=2011|num-b=14|num-a=16|ab=A}}

Revision as of 01:35, 10 February 2011

Problem

The circular base of a hemisphere of radius $2$ rests on the base of a square pyramid of height $6$. The hemisphere is tangent to the other four faces of the pyramid. What is the edge-length of the base of the pyramid?

$\textbf{(A)}\ 3\sqrt{2} \qquad \textbf{(B)}\ \frac{13}{3} \qquad \textbf{(C)}\ 4\sqrt{2} \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ \frac{13}{2}$

Solution

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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