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Difference between revisions of "2005 AMC 12B Problems/Problem 6"

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{{duplicate|[[2005 AMC 12B Problems|2005 AMC 12B #6]] and [[2005 AMC 10B Problems|2005 AMC 10B #10]]}}
 
== Problem ==
 
== 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>?
 
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>?
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== See also ==
 
== See also ==
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* [[2005 AMC 12B Problems/Problem 5 | Previous problem]]
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{{AMC10 box|year=2005|ab=B|num-b=9|num-a=11}}
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* [[2005 AMC 12B Problems/Problem 7 | Next problem]]
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{{AMC12 box|year=2005|ab=B|num-b=5|num-a=7}}
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* [[2005 AMC 12B Problems]]
 
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]

Revision as of 17:40, 30 June 2011

The following problem is from both the 2005 AMC 12B #6 and 2005 AMC 10B #10, so both problems redirect to this page.

Problem

In $\triangle ABC$, we have $AC=BC=7$ and $AB=2$. Suppose that $D$ is a point on line $AB$ such that $B$ lies between $A$ and $D$ and $CD=8$. What is $BD$?

$\mathrm{(A)}\ 3 \qquad \mathrm{(B)}\ 2\sqrt{3} \qquad \mathrm{(C)}\ 4 \qquad \mathrm{(D)}\ 5 \qquad \mathrm{(E)}\ 4\sqrt{2}$

Solution

Draw height $CH$. We have that $BH=1$. From the Pythagorean Theorem, $CH=\sqrt{48}$. Since $CD=8$, $HD=\sqrt{8^2-48}=\sqrt{16}=4$, and $BD=HD-1$, so $BD=3\rightarrow\boxed{A}$.

See also

2005 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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 10 Problems and Solutions
2005 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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
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