Difference between revisions of "2002 AMC 10A Problems/Problem 13"

Revision as of 17:23, 21 January 2014

Problem

Give a triangle with side lengths 15, 20, and 25, find the triangle's smallest height.

$\text{(A)}\ 6 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 12.5 \qquad \text{(D)}\ 13 \qquad \text{(E)}\ 15$

Solution

Solution 1

This is a Pythagorean triple (a 3-4-5 actually) with legs 15 and 20. The area is then $\frac{(15)(20)}{2}=150$ . 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 $\frac{(x)(25)}{2}=\frac{(15)(20)}{2}$ , so $25x=300$ and x is 12. Our answer is then $\boxed{\text{(B)}\ 12}$ .

Solution 2

By Heron's formula, the area is $150$ , hence the shortest altitude's length is $2\cdot\frac{150}{25}=\boxed{12\Rightarrow \text{(B)}}$ .

Solution 3

Going off of Solution 1, we see a  triangle and we know that the shortest altitude is . Therefore, the answer is .

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.

@@ Line 11: / Line 11: @@
 ===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 \text{(B)}}</math>.
+===Solution 3===
+ Going off of Solution 1, we see a <math>15, 20, 25</math> triangle and we know that the shortest altitude is <math>\frac{(15)(20)}{25}</math>. Therefore, the answer is <math>\boxed{\text{(B)}\ 12}</math>.
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