Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 5 - Section 5.6 - Segments Divided Proportionally - Exercises - Page 267: 5

Answer

EF=4$\frac{1}{6}$ FG=3$\frac{1}{3}$ GH=2$\frac{1}{2}$

Work Step by Step

AD=AB+BC+CD AD=5+4+3 AD=12 $\frac{EF}{EH}$=$\frac{AB}{AD}$ $\frac{EF}{10}$=$\frac{5}{12}$ 12EF=50 EF=4$\frac{1}{6}$ $\frac{FG}{EH}$=$\frac{BC}{AD}$ $\frac{FG}{10}$=$\frac{4}{12}$ 12FG=40 FG=3$\frac{1}{3}$ (AD)GH=(EH)CD 12GH=30 GH=2$\frac{1}{2}$
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