Difference between revisions of "2014 AMC 10A Problems/Problem 9"
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We find that the area of the triangle is <math>6\times \sqrt{3}=6\sqrt{3}</math>. By the [[Pythagorean Theorem]], we have that the length of the hypotenuse is <math>\sqrt{(2\sqrt{3})^2+6^2}=4\sqrt{3}</math>. Dropping an altitude from the right angle to the hypotenuse, we can calculate the area in another way. | We find that the area of the triangle is <math>6\times \sqrt{3}=6\sqrt{3}</math>. By the [[Pythagorean Theorem]], we have that the length of the hypotenuse is <math>\sqrt{(2\sqrt{3})^2+6^2}=4\sqrt{3}</math>. Dropping an altitude from the right angle to the hypotenuse, we can calculate the area in another way. | ||
| − | Let <math>h</math> be the third height of the triangle. We have <math>4\sqrt{3}h=2\times 6\sqrt{3}=12\sqrt{3}\implies h=\boxed{\textbf{(C)} 3}</math> | + | Let <math>h</math> be the third height of the triangle. We have <math>4\sqrt{3}h=2\times 6\sqrt{3}=12\sqrt{3}\implies h=\boxed{\textbf{(C)}\ 3}</math> |
==See Also== | ==See Also== | ||
Revision as of 23:16, 6 February 2014
Problem
The two legs of a right triangle, which are altitudes, have lengths 
and 
. How long is the third altitude of the triangle?
Solution
We find that the area of the triangle is 
. By the Pythagorean Theorem, we have that the length of the hypotenuse is 
. Dropping an altitude from the right angle to the hypotenuse, we can calculate the area in another way.
Let 
be the third height of the triangle. We have 
See Also
| 2014 AMC 10A (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 AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
