Difference between revisions of "2013 AMC 10A Problems/Problem 15"
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<cmath>h_{3} = 1.25h_{1}.</cmath> | <cmath>h_{3} = 1.25h_{1}.</cmath> | ||
Thus, the side length is going to be <math>\frac{2A}{1.25h_{1}} = \frac{15}{\frac{5}{4}} = \boxed{\textbf{(D) }12}</math>. | Thus, the side length is going to be <math>\frac{2A}{1.25h_{1}} = \frac{15}{\frac{5}{4}} = \boxed{\textbf{(D) }12}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/27TpizTnMeM | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Revision as of 15:03, 25 December 2020
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
.
Video Solution
~savannahsolver
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.