Difference between revisions of "2005 AMC 12B Problems/Problem 6"
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| − | {{ | + | \== Problem == |
| − | == | + | In <math>\triangle ABC</math>, we have <math>AC=BC=7</math> and <math>AB=2</math>. Suppose that <math>D</math> is a point on line <math>AB</math> such that <math>B</math> lies between <math>A</math> and <math>D</math> and <math>CD=8</math>. What is <math>BD</math>? |
| + | == 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{3}</math>. | ||
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== See also == | == See also == | ||
* [[2005 AMC 12B Problems/Problem 5 | Previous problem]] | * [[2005 AMC 12B Problems/Problem 5 | Previous problem]] | ||
* [[2005 AMC 12B Problems/Problem 7 | Next problem]] | * [[2005 AMC 12B Problems/Problem 7 | Next problem]] | ||
* [[2005 AMC 12B Problems]] | * [[2005 AMC 12B Problems]] | ||
| + | [[Category:Introductory Geometry Problems]] | ||
Revision as of 14:37, 4 January 2009
\== 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 
.
