Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 5 - Section 5.2 - Similar Polygons - Exercises - Page 224: 23

Answer

BC$\approx$14.5

Work Step by Step

$\frac{AC}{EA}$=$\frac{BC}{DE}$ $\frac{20}{20-BC}$=$\frac{BC}{4}$ 20BC-BC$^2$=80 BC$^2$-20BC+80=0 BC=$\frac{20+\sqrt {400-320}}{2}$ BC=10+.5$\sqrt {80}$ BC$\approx$14.5
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