Intermediate Algebra for College Students (7th Edition)

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

Chapter 5 - Section 5.7 - Polynomial Equations and Their Applications - Exercise Set - Page 393: 83

Answer

$30\frac{1}{8}$ ft.

Work Step by Step

The tree and the segment connecting the base of the tree to the point on the ground where the wire is anchored are the legs of a right triangle, while the wire is its hypotenuse. Let's note the height of the tree by $x$. The distance on the ground between the base of the tree and the point where the wire is anchored is $15$ feet. The hypotenuse is $x+4$ feet. Apply the Pythagorean Theorem. $hypotenuse^2=leg_1^2+leg_2^2$ $\Rightarrow (x+4)^2=x^2+15^2$ Square $(x+4)$ and $15$. $\Rightarrow x^2+8x+16=x^2+225$ Subtract $x^2+16$ from both sides. $\Rightarrow x^2+8x+16-x^2-16=x^2+225-x^2-16$ Simplify. $\Rightarrow 8x=209 $ Divide both sides by $8$. $\Rightarrow \frac{8x}{8}=\frac{209}{8} $ Simplify. $\Rightarrow x=\frac{209}{8} $ Length of the wire is $=x+4$ $=\frac{209}{8}+4$ $=\frac{209+32}{8}$ $=\frac{241}{8}$ $=30\frac{1}{8}$ ft.
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