Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2013 AMC 10A Problems/Problem 15"
(→‎Solution 1 (cheap way))
Line 9: Line 9:
 
The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes,
 
The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes,
 
therefore the length of the side perpendicular to that altitude will be between <math>10</math> and <math>15</math>. The only answer choice that meets this requirement is <math>\boxed{\textbf{(D) }12}</math>.
 
therefore the length of the side perpendicular to that altitude will be between <math>10</math> and <math>15</math>. The only answer choice that meets this requirement is <math>\boxed{\textbf{(D) }12}</math>.
 +
 +
 +
-credit goes to infuzion17
  
 
==Solution 2 (actually solving)==
 
==Solution 2 (actually solving)==

Revision as of 22:39, 8 February 2013

Problem

Two sides of a triangle have lengths $10$ and $15$. The length of the altitude to the third side is the average of the lengths of the altitudes to the two given sides. How long is the third side?


$\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 18$

Solution 1 (cheap way)

The shortest side length has the longest altitude perpendicular to it. The average of the two altitudes given will be between the lengths of the two altitudes, therefore the length of the side perpendicular to that altitude will be between $10$ and $15$. The only answer choice that meets this requirement is $\boxed{\textbf{(D) }12}$.


-credit goes to infuzion17

Solution 2 (actually solving)

Let the height to the side of length 15 be $h_{1}$, the height to the side of length 10 be $h_{2}$, the area be $A$, and the height to the unknown side be $h_{3}$.

Because the area of a triangle is $\frac{bh}{2}$, we get that $15(h_{1}) = 2A$ and $10(h_{2}) = 2A$, so, setting them equal, $h_{2} = \frac{3h_{1}}{2}$. Now, we know that $2h_{3} = h_{1} + h_{2}$. Substituting, we get that

$h_{3} = 1.25h_{1}$.

Thus, the side length is going to be $\frac{2A}{1.25h_{1}} = \frac{15}{\frac{5}{4}} = \boxed{\textbf{(D) }12}$.

See Also

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_10A_Problems/Problem_15&oldid=51006"