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

Revision as of 09:40, 4 July 2013

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 $BD=3\rightarrow\boxed{A}$ .

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 {{AMC12 box|year=2005|ab=B|num-b=5|num-a=7}}
 [[Category:Introductory Geometry Problems]]
+{{MAA Notice}}

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

Revision as of 09:40, 4 July 2013

Problem

Solution

See also