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

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 $15\cdot\frac{4}{5}$ to get our answer ${12}$ or $\boxed{\text{(B)}\ 12}$.

See Also

2002 AMC 10A (ProblemsAnswer KeyResources)
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

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