Difference between revisions of "2002 AMC 10A Problems/Problem 13"
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===Solution 2=== | ===Solution 2=== | ||
| − | By [[Heron's formula]], the area is <math>150</math>, hence the shortest altitude's length is <math>2\cdot\frac{150}{25}=\boxed{12\Rightarrow \ | + | By [[Heron's formula]], the area is <math>150</math>, hence the shortest altitude's length is <math>2\cdot\frac{150}{25}=\boxed{12\Rightarrow \textbf{(B)}}</math>. |
===Solution 3 (similarity rule)=== | ===Solution 3 (similarity rule)=== | ||
Revision as of 13:20, 8 November 2021
Contents
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 
and 
. 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 
; we have 
, so 
and is 12. Our answer is then 
.
Solution 2
By Heron's formula, the area is 
, hence the shortest altitude's length is 
.
Solution 3 (similarity rule)
The sides are in the ratio of 3-4-5, and are therefore a Pythagorean Triple and a part of a right triangle.
We know two altitudes, the legs 15 and 20. All that is left is to check the altitude from the right angle vertex to the hypotenuse. There is a rule that states that the three triangles formed when the altitude is dropped (the big triangle and two smaller ones it gets split into) are similar! We take the triangle with hypotenuse 20. The ratio of sides between that triangle and the bigger one is 4:5. Since the side of length 15 in the big triangle corresponds with our altitude, we take 
to get our answer 
or 
.
Solution 4 (fast calculation)
The triangle forms a 
right triangle, so we can first draw a triangle with legs 
and 
and hypotenuse 
. The two altitudes and are known, and the area is 
; therefore 
, and 
, which is clearly less than or . Now scale the triangle back up by multiplying every dimension by . The smallest altitude is then 
= or
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. 
