Difference between revisions of "2015 AMC 10B Problems/Problem 13"
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==Solution== | ==Solution== | ||
− | We find the x-intercepts and the y-intercepts to find the intersections of the axes and the line. If <math>x=0</math>, then <math>y=12</math>. If <math>y</math> is <math>0</math>, then <math>x=5</math>. Our three vertices are <math>(0,0)</math>, <math>(5,0)</math>, and <math>(0,12)</math>. Two of our altitudes are <math>5</math> and <math>12</math>, and since it is a 5-12-13 right triangle, the hypotenuse is <math>13</math>. Since the area of the triangle is <math>30</math>, so our final altitude is <math>\frac{30(2)}{13}=\frac{60}{13}</math>. The sum of our altitudes is <math>\frac{60+156+65}{13}=\boxed{\textbf{(E)} \dfrac{281}{13}}</math>. | + | We find the x-intercepts and the y-intercepts to find the intersections of the axes and the line. If <math>x=0</math>, then <math>y=12</math>. If <math>y</math> is <math>0</math>, then <math>x=5</math>. Our three vertices are <math>(0,0)</math>, <math>(5,0)</math>, and <math>(0,12)</math>. Two of our altitudes are <math>5</math> and <math>12</math>, and since it is a 5-12-13 right triangle, the hypotenuse is <math>13</math>. Since the area of the triangle is <math>30</math>, so our final altitude is <math>\frac{30(2)}{13}=\frac{60}{13}</math>. The sum of our altitudes is <math>\frac{60+156+65}{13}=\boxed{\textbf{(E)} \dfrac{281}{13}}</math>. Note that there is no need to calculate the final answer after we know that the third altitude has length <math>\frac{60}{13}</math> since <math>E</math> is the only choice with a denominator of <math>13</math>. |
==See Also== | ==See Also== |
Revision as of 18:01, 1 January 2018
Problem
The line forms a triangle with the coordinate axes. What is the sum of the lengths of the altitudes of this triangle?
Solution
We find the x-intercepts and the y-intercepts to find the intersections of the axes and the line. If , then . If is , then . Our three vertices are , , and . Two of our altitudes are and , and since it is a 5-12-13 right triangle, the hypotenuse is . Since the area of the triangle is , so our final altitude is . The sum of our altitudes is . Note that there is no need to calculate the final answer after we know that the third altitude has length since is the only choice with a denominator of .
See Also
2015 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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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