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: 31

Answer

BD=4$\sqrt 3$$\approx$6.93

Work Step by Step

2AC=AB AC=6 BC=AC$\sqrt 3$ BC=6$\sqrt 3$ DC$\sqrt 3$=AC DC=2$\sqrt 3$ BD=BC-DC BD=6$\sqrt 3$-2$\sqrt 3$ BD=4$\sqrt 3$$\approx$6.93
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.