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{\text{(A)}3}</math>. The thing is, where the heck is point H?
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Draw height <math>CH</math> (Perpendicular line from point C to line AD). 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{\text{(A)}3}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 17:51, 19 January 2017

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 $CH$ (Perpendicular line from point C to line AD). 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 $BD=\boxed{\text{(A)}3}$.

See also

2005 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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