Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10: 9781285195698
ISBN 13: 978-1-28519-569-8

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$
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.