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 | + | Let <math>h</math> be the third height of the triangle. We have <math>4\sqrt{3}h=2\sqrt{3}\times 6=12\sqrt{3}\implies h=\boxed{\textbf{(C)}\ 3}</math> |
==Solution 2== | ==Solution 2== |
Revision as of 03:01, 6 February 2018
Contents
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
Solution 2
By the Pythagorean Theorem, we have that the length of the hypotenuse is . Notice that we now have a 30-60-90 triangle, with the angle between sides and equal to . Dropping an altitude from the right angle to the hypotenuse, we see that our desired height is (We can also check from the other side).
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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