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2014 AMC 10A Problems/Problem 9: Difference between revisions

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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 1==
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\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>\frac{4\sqrt{3}h}{2}=6\sqrt{3}\implies h = \frac{6 \cdot 2}{4} \implies h=\boxed{\textbf{(C)}\ 3}</math>
 
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==
==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).
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).
 
==Solution 3 (Simple)==
From solution 1, we know the 3rd altitude of a right triangle is the product of the 2 legs that are not the hypotenuse, divided by the length of the hypotenuse. The product of the 2 legs of this triangle is <math>12\sqrt{3}</math>. By the [[Pythagorean Theorem]], the length of the hypotenuse is <math>\sqrt{(2\sqrt{3})^2+6^2}=4\sqrt{3}</math>. The length of the third altitude is <math>\frac{12\sqrt{3}}{4\sqrt{3}}=\boxed{\textbf{(C)}\ 3}</math>
~[[shunyipanda]]
 
==Video Solution (CREATIVE THINKING)==
https://youtu.be/ZvoMN5tJHBk
 
~Education, the Study of Everything
 
 
 
 
==Video Solution==
https://youtu.be/cd0yW4k4Fo8
 
~savannahsolver


==See Also==
==See Also==

Latest revision as of 18:15, 23 October 2025

Problem

The two legs of a right triangle, which are altitudes, have lengths $2\sqrt3$ and $6$. How long is the third altitude of the triangle?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution 1

We find that the area of the triangle is $\frac{6\times 2\sqrt{3}}{2} =6\sqrt{3}$. By the Pythagorean Theorem, we have that the length of the hypotenuse is $\sqrt{(2\sqrt{3})^2+6^2}=4\sqrt{3}$. Dropping an altitude from the right angle to the hypotenuse, we can calculate the area in another way.

Let $h$ be the third height of the triangle. We have $\frac{4\sqrt{3}h}{2}=6\sqrt{3}\implies h = \frac{6 \cdot 2}{4} \implies h=\boxed{\textbf{(C)}\ 3}$

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 $\sqrt{(2\sqrt{3})^2+6^2}=4\sqrt{3}$. Notice that we now have a 30-60-90 triangle, with the angle between sides $2\sqrt{3}$ and $4\sqrt{3}$ equal to $60^{\circ}$. 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).

Solution 3 (Simple)

From solution 1, we know the 3rd altitude of a right triangle is the product of the 2 legs that are not the hypotenuse, divided by the length of the hypotenuse. The product of the 2 legs of this triangle is $12\sqrt{3}$. By the Pythagorean Theorem, the length of the hypotenuse is $\sqrt{(2\sqrt{3})^2+6^2}=4\sqrt{3}$. The length of the third altitude is $\frac{12\sqrt{3}}{4\sqrt{3}}=\boxed{\textbf{(C)}\ 3}$ ~shunyipanda

Video Solution (CREATIVE THINKING)

https://youtu.be/ZvoMN5tJHBk

~Education, the Study of Everything



Video Solution

https://youtu.be/cd0yW4k4Fo8

~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

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