Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 5 - Review Exercises - Page 263: 40d

Answer

x=2$\sqrt {14}$ y=13

Work Step by Step

x$^2$=(6$\sqrt 2$)$^2$-4$^2$ x$^2$=72-16 x$^2$=56 x=2$\sqrt {14}$ The right triangle in the center has legs of length 3 and 4 which makes it a pythagorean triple with a hypotenuse of 5. y$^2$=12$^2$+5$^2$ y$^2$=169 y=13
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