Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Chapter 5 - Section 5.6 - Segments Divided Proportionally - Exercises - Page 266: 6

Answer

FG=4$\frac{2}{7}$ GH=3$\frac{3}{7}$ EH=13$\frac{5}{7}$

Work Step by Step

$\frac{FG}{BC}$=$\frac{EF}{AB}$ $\frac{FG}{5}$=$\frac{6}{7}$ 7FG=30 FG=4$\frac{2}{7}$ $\frac{GH}{CD}$=$\frac{EF}{AB}$ $\frac{GH}{4}$=$\frac{6}{7}$ 7GH=24 GH=3$\frac{3}{7}$ $\frac{EH}{AD}$=$\frac{EF}{AB}$ $\frac{EH}{16}$=$\frac{6}{7}$ 7EH=96 EH=13$\frac{5}{7}$
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