Difference between revisions of "2013 AMC 10A Problems/Problem 15"
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<math> \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18 </math> | <math> \textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18 </math> | ||
− | ==Solution 1 ( | + | ==Solution 1 (Process of Elimination)== |
The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes, | The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes, |
Revision as of 16:19, 2 February 2017
Two sides of a triangle have lengths and . The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?
Solution 1 (Process of Elimination)
The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes, therefore the length of the side perpendicular to that altitude will be between and . The only answer choice that meets this requirement is .
Solution 2
Let the height to the side of length be , the height to the side of length 10 be , the area be , and the height to the unknown side be .
Because the area of a triangle is , we get that and , so, setting them equal, . From the problem, we know that . Substituting, we get that Thus, the side length is going to be .
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 10 Problems and Solutions |
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