Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 9 - Section 9.3 - Cylinders and Cones - Exercises - Page 417: 23a

Answer

L=6$\pi$$\sqrt {85}$$\approx$173.78 in$^2$

Work Step by Step

l$^2$=7$^2$+6$^2$ l$^2$=49+36=85 l=$\sqrt {85}$ L=.5($\sqrt {85}$)(12$\pi$) L=6$\pi$$\sqrt {85}$$\approx$173.78 in$^2$
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