Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Problems - Page 76: 1


The solution is $$h=\frac{4\sqrt{c^2-16}}{c}.$$

Work Step by Step

Let $a=4$ be the one leg of this triangle and $b$ the other leg and $c$ be the hypotenuse. Using the expression for the area of such a triangle written in terms of $a$ and $b$ and in terms of $c$ and $h$ where $h$ is the altitude we have $$\frac{1}{2}ab=\frac{1}{2}ch\Rightarrow h=\frac{ab}{c}.$$ Now from Pythagorean theorem we have $$c^2=a^2+b^2\Rightarrow b=\sqrt{c^2-a^2}.$$ Putting this into the expression for altitude $$h=\frac{a\sqrt{c^2-a^2}}{c}.$$ Putting $a=4$ $$h=\frac{4\sqrt{c^2-16}}{c.}$$
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