2005 AMC 8 Problems/Problem 9

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Problem

In quadrilateral $ABCD$, sides $\overline{AB}$ and $\overline{BC}$ both have length 10, sides $\overline{CD}$ and $\overline{DA}$ both have length 17, and the measure of angle $ADC$ is $60^\circ$. What is the length of diagonal $\overline{AC}$? [asy]draw((0,0)--(17,0)); draw(rotate(301, (17,0))*(0,0)--(17,0)); picture p; draw(p, (0,0)--(0,10)); draw(p, rotate(115, (0,10))*(0,0)--(0,10)); add(rotate(3)*p);  draw((0,0)--(8.25,14.5), linetype("8 8"));  label("$A$", (8.25, 14.5), N); label("$B$", (-0.25, 10), W); label("$C$", (0,0), SW); label("$D$", (17, 0), E);[/asy]

$\textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 15.5\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18.5$

Solution

Because $\overline{AD} = \overline{CD}$, $\triangle ADC$ is an equilateral triangle with $\angle DAC = \angle DCA$. Angles in a triangle add up to $180^\circ$, and since $\angle ADC=60^\circ$, the other two angles are also $60^\circ$, and $\triangle ADC$ is an equilateral triangle. Therefore $\overline{AC}=\overline{DA}=\boxed{\textbf{(D)}\ 17}$.

See Also

2005 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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