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Difference between revisions of "2015 AMC 12B Problems/Problem 11"

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==Solution==
 
==Solution==
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Clearly the line and the coordinate axes form a right triangle. Since the x-intercept and y-intercept are 5 and 12 respectively, 5 and 12 are two sides of the triangle that are not the hypotenuse, and are thus two of the three heights. In order to find the third height, we can use different equations of the area of the triangle. The using the lengths we know, the area of the triangle is <math>\tfrac{1}{2} \times 5 \times 12= 30</math>. We can use the hypotenuse as another base to find the third height. Using the distance formula, the length of the hypotenuse is <math>\sqrt{5^2+12^2}=13</math>. Then <math>\tfrac{1}{2} \times 13 \times h=30</math>, and <math>h = \tfrac{60}{13}</math>. Then the sum of all the heights is <math>5+12+\frac{60}{13}=\boxed{\textbf{(E)}\; \frac{281}{13}}</math>.
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Clearly the line and the coordinate axes form a right triangle. Since the x-intercept and y-intercept are 5 and 12 respectively, 5 and 12 are two sides of the triangle that are not the hypotenuse, and are thus two of the three heights. In order to find the third height, we can use different equations of the area of the triangle. Using the lengths we know, the area of the triangle is <math>\tfrac{1}{2} \times 5 \times 12= 30</math>. We can use the hypotenuse as another base to find the third height. Using the distance formula, the length of the hypotenuse is <math>\sqrt{5^2+12^2}=13</math>. Then <math>\frac{1}{2} \times 13 \times h=30</math>, and so <math>h = \frac{60}{13}</math>. Therefore the sum of all the heights is <math>5+12+\frac{60}{13}=\boxed{\textbf{(E)}\; \frac{281}{13}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2015|ab=B|num-a=12|num-b=10}}
 
{{AMC12 box|year=2015|ab=B|num-a=12|num-b=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:52, 5 March 2015

Problem

The line $12x+5y=60$ forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?

$\textbf{(A)}\; 20 \qquad\textbf{(B)}\; \dfrac{360}{17} \qquad\textbf{(C)}\; \dfrac{107}{5} \qquad\textbf{(D)}\; \dfrac{43}{2} \qquad\textbf{(E)}\; \dfrac{281}{13}$

Solution

Clearly the line and the coordinate axes form a right triangle. Since the x-intercept and y-intercept are 5 and 12 respectively, 5 and 12 are two sides of the triangle that are not the hypotenuse, and are thus two of the three heights. In order to find the third height, we can use different equations of the area of the triangle. Using the lengths we know, the area of the triangle is $\tfrac{1}{2} \times 5 \times 12= 30$. We can use the hypotenuse as another base to find the third height. Using the distance formula, the length of the hypotenuse is $\sqrt{5^2+12^2}=13$. Then $\frac{1}{2} \times 13 \times h=30$, and so $h = \frac{60}{13}$. Therefore the sum of all the heights is $5+12+\frac{60}{13}=\boxed{\textbf{(E)}\; \frac{281}{13}}$.

See Also

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 12 Problems and Solutions

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