Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Review Exercises - Page 657: 13

Answer

$60\sqrt 5\approx 134.2$ meters

Work Step by Step

Let's note by $h$ the height o the building. We draw a right triangle with legs $h$ and $2h$ and hypotenuse of length $300$. We determine $h$ using the Pythagorean theorem: $$\begin{align*} h^2+(2h)^2&=300^2\\ h^2+4h^2&=90,000\\ 5h^2&=90,000\\ h^2&=\dfrac{90,000}{5}\\ h^2&=18,000\\ h&=\pm\sqrt{18,000}=\pm 60\sqrt 5\approx \pm 134.2. \end{align*}$$ Because $h>0$, the only solution is $60\sqrt 5\approx 134.2$ meters.
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