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= | + | 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> |
==Solution 2== | ==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). | |
− | + | ==Video Solution== | |
+ | https://youtu.be/cd0yW4k4Fo8 | ||
− | + | ~savannahsolver | |
− | |||
− | |||
==See Also== | ==See Also== | ||
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{{AMC10 box|year=2014|ab=A|num-b=8|num-a=10}} | {{AMC10 box|year=2014|ab=A|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category: Introductory Geometry Problems]] |
Latest revision as of 10:38, 7 September 2021
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).
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.