Difference between revisions of "2005 AMC 12B Problems/Problem 6"
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== Solution == | == Solution == | ||
− | Draw height <math>CH</math>. We have that <math>BH=1</math>. From the [[Pythagorean Theorem]], <math>CH=\sqrt{48}</math>. Since <math>CD=8</math>, <math>HD=\sqrt{8^2-48}=\sqrt{16}=4</math>, and <math>BD=HD-1</math>, so <math>BD=\boxed{ | + | Draw height <math>CH</math>. We have that <math>BH=1</math>. From the [[Pythagorean Theorem]], <math>CH=\sqrt{48}</math>. Since <math>CD=8</math>, <math>HD=\sqrt{8^2-48}=\sqrt{16}=4</math>, and <math>BD=HD-1</math>, so <math>BD=3\rightarrow\boxed{A}</math>. |
== See also == | == See also == |
Revision as of 22:42, 8 February 2010
Problem
In , we have
and
. Suppose that
is a point on line
such that
lies between
and
and
. What is
?
Solution
Draw height . We have that
. From the Pythagorean Theorem,
. Since
,
, and
, so
.