2005 AMC 12B Problems/Problem 6

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 $BD=\boxed{3}$ .

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2005 AMC 12B Problems/Problem 6

Problem

Solution

See also