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Difference between revisions of "2014 AMC 10A Problems/Problem 9"

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==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.
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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>
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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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==Solution 2==
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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==
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https://youtu.be/cd0yW4k4Fo8
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~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}}
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[[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 $2\sqrt3$ 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

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 $\frac{4\sqrt{3}h}{2}=6\sqrt{3}\implies h = \frac{6 \cdot 2}{4} \implies h=\boxed{\textbf{(C)}\ 3}$

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).

Video Solution

https://youtu.be/cd0yW4k4Fo8

~savannahsolver

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
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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