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 1==
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==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>
  
 
==Solution 2==
 
==Solution 2==
Directly following from the formula for the area of a triangle, <math>\frac{bh}{2}</math>, we have that, for any right triangle, the product of the two legs of the triangle is equal to the product of the hypotenuse and the altitude to the hypotenuse. (This is always true because the two legs can be the base and the height of the triangle, and so can the hypotenuse and the altitude of the hypotenuse.)
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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).
  
To find the hypotenuse, we can apply Pythagoras to obtain <cmath>\sqrt{(2\sqrt{3})^2+6^3}</cmath> <cmath>\implies \sqrt{12+36}</cmath> <cmath>\implies \sqrt{48}=4\sqrt{3}</cmath>
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==Video Solution==
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https://youtu.be/cd0yW4k4Fo8
  
Let <math>a</math> be the length of the third altutude - the altitude to the hypotenuse. We have <cmath>2\sqrt3 \times 6 = a\times 4\sqrt3</cmath> <cmath>\implies a=\frac{2\sqrt3\times6}{4\sqrt{3}}</cmath> <cmath>\implies \frac{6}{2}=\boxed{\textbf{(C)}\ 3}</cmath>
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~savannahsolver
 
 
(Solution by bestwillcui1)
 
  
 
==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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