2011 CEMC Gauss (Grade 8) Problems/Problem 15

Problem

In the diagram, $AE$ and $BD$ are straight lines that intersect at $C$ . If $BD = 16$ , $AB = 9$ , $CE = 5$ , and $DE = 3$ , then the length of $AC$ is

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$\textbf{(A)}\ 11 \qquad\textbf{(B)}\ 12 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 16$

Solution 1

Using the fact that triangle $CDE$ would have to be a $3-4-5$ right triangle, we can conclude that $CD = 4$ .

Since $BD = BC + CD$ , we have:

$16 = BC + 4$

$BC = 12$

We then see that triangle $ABC$ is a $9-12-15$ right triangle. This means that $AC = \boxed {\textbf {(C) } 15}$ .

~anabel.disher

Solution 2 (similarity)

We see that $m \angle ACB = m \angle DCE$ because they are vertical angles. We can then label side lengths.

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We see that triangle $ABC$ is similar to $EDC$ because two of the angles in the triangles are equal to each other in measure. This allows us to set up equations:

$\frac{AC}{CE} = \frac{AC}{5} = \frac{AB}{DE} = \frac{9}{3} = 3$

We then see that $AC = 5 \times 3 = \boxed {\textbf {(C) } 15}$ .

~anabel.disher

2011 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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2011 CEMC Gauss (Grade 8) Problems/Problem 15

Problem

Solution 1

Solution 2 (similarity)