Difference between revisions of "2005 AMC 8 Problems/Problem 9"

Latest revision as of 21:25, 1 November 2022

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$", (17, 0), E);[/asy]$

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

Solutions

Solution 1

Because $\overline{AD} = \overline{CD}$ , $\triangle ADC$ is an isosceles 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 $\overline{AC}=\overline{DA}=\boxed{\textbf{(D)}\ 17}$ .

Solution 2

We can divide $\overline{CD}$ in half and connect this point to A, dividing $\triangle ADC$ in half. This means the base will be $\frac{17}{2}$ and the hypotenuse will be 17. By using the Pythagorean's Theorem, we see that if the base and height are shared, the hypotenuse should be the same. This tells us that the length of $\overline{AC} = \boxed{(D) 17}$ .

~Champion1234

Solution 3 (Easiest)

Take an equilateral triangle with side length $17$ . $\triangle ADC$ is congruent to this by $SAS$ , hence it is equilateral. The answer is $\boxed{\textbf{(D)}17}$ .

~yofro

Video Solution by OmegaLearn

https://youtu.be/abSgjn4Qs34?t=3034

~pi_is_3.14

@@ Line 31: / Line 31: @@
 ~yofro
+== Video Solution by OmegaLearn ==
+https://youtu.be/abSgjn4Qs34?t=3034
+~pi_is_3.14
 == See Also ==
 {{AMC8 box|year=2005|num-b=8|num-a=10}}
 {{MAA Notice}}

2005 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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All AJHSME/AMC 8 Problems and Solutions

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