Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 5 - Section 5.5 - Special Right Triangles - Exercises - Page 248: 32

Answer

BD=4

Work Step by Step

BC=AC$\sqrt 3$ BC=2$\sqrt 3$($\sqrt 3$) BC=6 DC$\sqrt 3$=AC DC$\sqrt 3$=2$\sqrt 3$ DC=2 BD=BC-CD BD=6-2 BD=4
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