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: 27

Answer

DC=2$\sqrt 3$ DB=4$\sqrt 3$

Work Step by Step

2AC=AB AC=6 CB=AC$\sqrt 3$ CB=6$\sqrt 3$ CD$\sqrt 3$=AC CD=$\frac{6\sqrt 3}{3}$ CD=2$\sqrt 3$ DB=CB-CD DB=6$\sqrt 3$-2$\sqrt 3$ DB=4$\sqrt 3$
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