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

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== 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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 +
<math>
 +
\mathrm{(A)}\ 3      \qquad
 +
\mathrm{(B)}\ 2\sqrt{3}      \qquad
 +
\mathrm{(C)}\ 4      \qquad
 +
\mathrm{(D)}\ 5      \qquad
 +
\mathrm{(E)}\ 4\sqrt{2}
 +
</math>
 +
 
== 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{3}</math>.
 
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>.

Revision as of 22:41, 8 February 2010

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 $BD=\boxed{3}$.

See also