2002 AMC 10A Problems/Problem 13

Problem

Given a triangle with side lengths 15, 20, and 25, find the triangle's shortest altitude.

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

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 $2\cdot\frac{150}{25}=\boxed{12\Rightarrow \text{(B)}}$ .

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
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2002 AMC 10A Problems/Problem 13

Contents

Problem

Solution

Solution 1

Solution 2

See Also