2008 iTest Problems/Problem 37

Problem

A triangle has sides of length $48, 55$ , and $73$ . Let a and b be relatively prime positive integers such that $a/b$ is the length of the shortest altitude of the triangle. Find the value of $a+b$ .

Solutions

Solution 1

Note that $48^2 + 55^2 = 73^2$ , so the side lengths are part of a right triangle. Also note that the shortest altitude is the one that is perpendicular to the longest side. Let $x$ be the shortest altitude. Using the area formula, $\frac{73 \cdot x}{2} = \frac{48 \cdot 55}{2}$ $x = \frac{2640}{73}$ Thus, $a+b=\boxed{2713}$ .

Solution 2

Using Heron's Formula, the area of the triangle is $\sqrt{88 \cdot 15 \cdot 40 \cdot 33}$ Note that the shortest altitude is the one that is perpendicular to the longest side. Let $x$ be the shortest altitude. Using the area formula, $\frac{73x}{2} = 1320$ $x = \frac{2640}{73}$ Thus, $a+b=\boxed{2713}$ .

2008 iTest (Problems)
Preceded by: Problem 36	Followed by: Problem 38
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2008 iTest Problems/Problem 37

Contents

Problem

Solutions

Solution 1

Solution 2

See Also