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Difference between revisions of "2005 AMC 8 Problems/Problem 9"
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<math> \textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 15.5\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18.5 </math>
 
<math> \textbf{(A)}\ 13.5\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 15.5\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18.5 </math>
  
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==Solution==
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==Solution 1==
 
Because <math>\overline{AD} = \overline{CD}</math>, <math>\triangle ADC</math> is an equilateral triangle with <math>\angle DAC = \angle DCA</math>. Angles in a triangle add up to <math>180^\circ</math>, and since <math>\angle ADC=60^\circ</math>, the other two angles are also <math>60^\circ</math>, and <math>\triangle ADC</math> is an equilateral triangle. Therefore <math>\overline{AC}=\overline{DA}=\boxed{\textbf{(D)}\ 17}</math>.
 
Because <math>\overline{AD} = \overline{CD}</math>, <math>\triangle ADC</math> is an equilateral triangle with <math>\angle DAC = \angle DCA</math>. Angles in a triangle add up to <math>180^\circ</math>, and since <math>\angle ADC=60^\circ</math>, the other two angles are also <math>60^\circ</math>, and <math>\triangle ADC</math> is an equilateral triangle. Therefore <math>\overline{AC}=\overline{DA}=\boxed{\textbf{(D)}\ 17}</math>.
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==Solution 2==
 +
We can divide <math>\overline{CD}</math> in half and connect this point to A, dividing <math>\triangle ADC</math> in half. This means the base will be <math>\frac{17}{2}</math> 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 <math>\overline{AD} = \boxed{(D) 17}</math>.
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~Champion1234
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2005|num-b=8|num-a=10}}
 
{{AMC8 box|year=2005|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:40, 11 November 2019

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 1

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 $\overline{AC}=\overline{DA}=\boxed{\textbf{(D)}\ 17}$.

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 $\overline{AD} = \boxed{(D) 17}$.

~Champion1234

See Also

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

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