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)); | ||
Line 17: | Line 18: | ||
<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 == |
− | Because <math>\overline{AD} = \overline{CD}</math>, <math>\triangle ADC</math> is an | + | === Solution 1 === |
+ | Because <math>\overline{AD} = \overline{CD}</math>, <math>\triangle ADC</math> is an isosceles 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{ | + | 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{AC} = \boxed{(D) 17}</math>. |
~Champion1234 | ~Champion1234 | ||
− | ==See Also== | + | === Solution 3 (Easiest) === |
+ | Take an equilateral triangle with side length <math>17</math>. <math>\triangle ADC</math> is congruent to this by <math>SAS</math>, hence it is equilateral. The answer is <math>\boxed{\textbf{(D)}17}</math>. | ||
+ | |||
+ | ~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}} | {{AMC8 box|year=2005|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:25, 1 November 2022
Contents
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 isosceles 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
Solution 3 (Easiest)
Take an equilateral triangle with side length .
is congruent to this by
, hence it is equilateral. The answer is
.
~yofro
Video Solution by OmegaLearn
https://youtu.be/abSgjn4Qs34?t=3034
~pi_is_3.14
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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