2013 AMC 10A Problems/Problem 15
Contents
Problem
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 .
Solution 3 (Answer Choices)
Let be the height of triangle when the base is and is the height of the triangles when the base is . This means the height for when the triangles has the rd side length, the height would be giving us the following equation:
For , we can plug in the answer choices and check when the ratio of for and is the same because we can observe that only for one value, the ratio remains constant. From brute force we get that meaning that is the rd base. ~ Batmanstark
Video Solution
~savannahsolver
See Also
2013 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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All AMC 10 Problems and Solutions |
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