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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+ | ==Solution 2== | ||
+ | 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>. Notice that we now have a 30-60-90 triangle, with the angle between sides <math>2\sqrt{3}</math> and <math>4\sqrt{3}</math> equal to <math>60^{\circ}</math>. Dropping an altitude from the right angle to the hypotenuse, we see that our desired height is \boxed{\textbf{(C)}\ 3} (We can also check from the other side). | ||
==See Also== | ==See Also== |
Revision as of 21:43, 24 January 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 \boxed{\textbf{(C)}\ 3} (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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