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

Revision as of 21:41, 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 $BD=\boxed{3}$ .

Page

Toolbox

Search

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

Revision as of 21:41, 8 February 2010

Problem

Solution

See also

@@ Line 1: / Line 1: @@
 == 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>?
+<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 ==
 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>.