Difference between revisions of "2014 AMC 10A Problems/Problem 9"
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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> | ||
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==See Also== | ==See Also== |
Revision as of 21:47, 20 January 2015
Problem
The two legs of a right triangle, which are altitudes, have lengths and . How long is the third altitude of the triangle?
Solution 1
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 |
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