Difference between revisions of "2002 AMC 10A Problems/Problem 13"
Slackroadia (talk | contribs) (→Problem) |
m (→Problem) |
||
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
| − | + | Given a triangle with side lengths 15, 20, and 25, find the triangle's shortest altitude | |
<math>\text{(A)}\ 6 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 12.5 \qquad \text{(D)}\ 13 \qquad \text{(E)}\ 15</math> | <math>\text{(A)}\ 6 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 12.5 \qquad \text{(D)}\ 13 \qquad \text{(E)}\ 15</math> | ||
Revision as of 21:43, 8 January 2018
Problem
Given a triangle with side lengths 15, 20, and 25, find the triangle's shortest altitude
Solution
Solution 1
This is a Pythagorean triple (a 3-4-5 actually) with legs 15 and 20. The area is then 
. Now, consider an altitude drawn to any side. Since the area remains constant, the altitude and side to which it is drawn are inversely proportional. To get the smallest altitude, it must be drawn to the hypotenuse. Let the length be x; we have 
, so 
and x is 12. Our answer is then 
.
Solution 2
By Heron's formula, the area is 
, hence the shortest altitude's length is 
.
See Also
| 2002 AMC 10A (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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
