Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 7 - Review Exercises - Page 338: 24


a) BG = 12 b) GE = 2 c) DG = $8\sqrt3$

Work Step by Step

a) A theorem states that BG equals 2/3 of BF. Thus: $BG = 18 \times (2/3) = 12 $ b) GE is half of AG. Thus: $GE = .5(4) = 2$ c) Using the same method as in b), we know that DG is twice the value of CG. Thus: $ DG = 2 (4\sqrt{3}) = 8\sqrt3$
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.