2002 AMC 10A Problems/Problem 13

Revision as of 18:24, 21 January 2014 by Smart99 (talk | contribs) (Solution 3)

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

See Also

2002 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AMC 10 Problems and Solutions

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