Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 5 - Review Exercises - Page 263: 30a

Answer

DC=8$\frac{1}{3}$

Work Step by Step

BD$^2$+AD$^2$=AB$^2$ AB=$\sqrt {34}$ $\frac{DC+BD}{AB}$=$\frac{AB}{BD}$ $\frac{DC+3}{\sqrt {34}}$=$\frac{\sqrt {34}}{3}$ 3DC+9=34 3DC=25 DC=8$\frac{1}{3}$
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