Difference between revisions of "2011 AMC 12A Problems/Problem 15"
(Created page with '== Problem == == Solution == == See also == {{AMC12 box|year=2011|num-b=14|num-a=16|ab=A}}') |
(→Problem) |
||
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
| + | 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? | ||
| + | |||
| + | <math> | ||
| + | \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} </math> | ||
| + | |||
== 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 
rests on the base of a square pyramid of height 
. The hemisphere is tangent to the other four faces of the pyramid. What is the edge-length of the base of the pyramid?
Solution
See also
| 2011 AMC 12A (Problems • Answer Key • Resources) | |
| 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 | |
