Difference between revisions of "2012 AMC 12A Problems/Problem 10"
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== Solution == | == Solution == | ||
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<math>AB</math> is the side of length <math>10</math>, and <math>CD</math> is the median of length <math>9</math>. The altitude of <math>C</math> to <math>AB</math> is <math>6</math> because the 0.5(altitude)(base)=Area of the triangle. <math>\theta</math> is <math>\angle CDE</math>. To find <math>\sin{\theta}</math>, just use opposite over hypotenuse with the right triangle <math>\triangle DCE</math>. This is equal to <math>\frac69=\boxed{\textbf{(D)}\ \frac23}</math>. | <math>AB</math> is the side of length <math>10</math>, and <math>CD</math> is the median of length <math>9</math>. The altitude of <math>C</math> to <math>AB</math> is <math>6</math> because the 0.5(altitude)(base)=Area of the triangle. <math>\theta</math> is <math>\angle CDE</math>. To find <math>\sin{\theta}</math>, just use opposite over hypotenuse with the right triangle <math>\triangle DCE</math>. This is equal to <math>\frac69=\boxed{\textbf{(D)}\ \frac23}</math>. | ||
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+ | ===Solution 2=== | ||
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+ | It is a well known fact that a median divides the area of a triangle into two smaller triangles of equal area. Therefore, the area of <math>\triangle BCD = 15</math> in the above figure. Therefore, <math>\frac{1}{2} \cdot 5 \cdot 9 \cdot \sin{\theta} = 15</math>. Solving for <math>\theta</math> gives <math>\theta = \frac{2}{3}</math>. \boxed{D}. | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2012|ab=A|num-b=9|num-a=11}} | {{AMC12 box|year=2012|ab=A|num-b=9|num-a=11}} |
Revision as of 00:16, 14 February 2012
Problem
A triangle has area , one side of length , and the median to that side of length . Let be the acute angle formed by that side and the median. What is ?
Solution
Solution 1
is the side of length , and is the median of length . The altitude of to is because the 0.5(altitude)(base)=Area of the triangle. is . To find , just use opposite over hypotenuse with the right triangle . This is equal to .
Solution 2
It is a well known fact that a median divides the area of a triangle into two smaller triangles of equal area. Therefore, the area of in the above figure. Therefore, . Solving for gives . \boxed{D}.
See Also
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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 AMC 12 Problems and Solutions |