Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 8 - Right Triangles and Trigonometry - Chapter Review - Page 535: 4


$x = 2\sqrt {113}$

Work Step by Step

We can find the third side, $x$, by using the Pythagorean theorem, which states that $a^2 + b^2 = c^2$, where $a$ and $b$ are the legs of the right triangle and $c$ is the hypotenuse. Let's plug in what we know into the Pythagorean theorem: $14^2 + 16^2 = x^2$ Evaluate the exponents: $196 + 256= x^2$ Add to simplify: $452 = x^2$ Rewrite $452$ as the product of a perfect square and another factor: $x^2 = 4 • 113$ Take the positive square root to solve for $x$: $x = 2\sqrt {113}$
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