Difference between revisions of "2005 AMC 8 Problems/Problem 9"
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− | ==Problem== | + | == Problem == |
In quadrilateral <math> ABCD</math>, sides <math> \overline{AB}</math> and <math> \overline{BC}</math> both have length 10, sides <math> \overline{CD}</math> and <math> \overline{DA}</math> both have length 17, and the measure of angle <math> ADC</math> is <math> 60^\circ</math>. What is the length of diagonal <math> \overline{AC}</math>? | In quadrilateral <math> ABCD</math>, sides <math> \overline{AB}</math> and <math> \overline{BC}</math> both have length 10, sides <math> \overline{CD}</math> and <math> \overline{DA}</math> both have length 17, and the measure of angle <math> ADC</math> is <math> 60^\circ</math>. What is the length of diagonal <math> \overline{AC}</math>? | ||
+ | |||
<asy>draw((0,0)--(17,0)); | <asy>draw((0,0)--(17,0)); | ||
draw(rotate(301, (17,0))*(0,0)--(17,0)); | draw(rotate(301, (17,0))*(0,0)--(17,0)); | ||
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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> | ||
− | ==Solution 1== | + | == Solutions == |
+ | === 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>. | ||
− | ==Solution 2== | + | === 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>. | 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>. | ||
~Champion1234 | ~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 12:33, 19 October 2020
Problem
In quadrilateral , sides and both have length 10, sides and both have length 17, and the measure of angle is . What is the length of diagonal ?
Solutions
Solution 1
Because , is an equilateral triangle with . Angles in a triangle add up to , and since , the other two angles are also , and is an equilateral triangle. Therefore .
Solution 2
We can divide in half and connect this point to A, dividing in half. This means the base will be 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 .
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.