Difference between revisions of "2014 AMC 10A Problems/Problem 9"
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<math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math> | <math> \textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5 </math> | ||
| − | ==Solution | + | ==Solution== |
| − | 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>\frac{6\times 2\sqrt{3}}{2} =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>\frac{4\sqrt{3}h}{2}=6\sqrt{3}\implies h = \frac{6 \cdot 2}{4} \implies h=\boxed{\textbf{(C)}\ 3}</math> |
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| + | Note: The third altitude of a right triangle is always the product of the lengths of the two legs divided by the hypotenuse. | ||
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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 <math>\boxed{\textbf{(C)}\ 3}</math> (We can also check from the other side). | ||
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| + | ==Video Solution (CREATIVE THINKING)== | ||
| + | https://youtu.be/ZvoMN5tJHBk | ||
| + | |||
| + | ~Education, the Study of Everything | ||
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| + | ==Video Solution== | ||
| + | https://youtu.be/cd0yW4k4Fo8 | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
Latest revision as of 18:41, 31 August 2024
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 
Note: The third altitude of a right triangle is always the product of the lengths of the two legs divided by the hypotenuse.
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).
Video Solution (CREATIVE THINKING)
https://youtu.be/ZvoMN5tJHBk
~Education, the Study of Everything
Video Solution
~savannahsolver
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. 
